# IV. Algebra

*The first rule of intelligent tinkering is to save all the parts.*

Paul R. Ehrlich (1932– )

Many important aspects of serious mathematics have their roots in the world of arithmetic. However, when we implement an arithmetical procedure by
combining numbers with very different meanings to produce *a single numerical output,* it becomes almost impossible to see how the separate
ingredients contribute to the final answer. In other words, calculating exclusively with numbers contravenes Paul Ehrlich’s “first rule
of intelligent tinkering”. This is why in Chapters 1 and 2 we stressed the
need to move beyond blind calculation, and to begin to think *structurally* - even when calculating purely with numbers. *Algebra* can be
seen as a remarkable way of “tinkering with numbers”, so that we not only “keep all the parts”, but manage to *keep them
separate* (by giving them different names), and hence can see clearly what contribution each ingredient variable makes to the final output. To
benefit from this feature of algebra, we need to learn to “read” algebraic expressions, and to interpret what they are telling us - in
much the same way that we learn to read numbers (so that, where appropriate, 100 is seen as 10^{2}, and 10 is seen as 1 + 2 + 3 + 4).

Before algebra proper was invented (around 1600), the ability to extract the general picture lying hidden inside each calculation was restricted to
specialists. The ancient Babylonians (1700–1500 BC) described their general procedures as *recipes*, presented in the context of problems
involving particular numbers. But they did this in such a way as to demonstrate convincingly that whoever formulated the procedure had managed to see
“the general in the particular”. The ancient Greeks used a geometrical setting to reveal generality, and encoded what we would see as
“algebraic” methods in geometrical language. In the 9^{th} century AD, Arabs such as Al-Khwarizmi (c.780–c.850), managed
to encapsulate generality using a very limited kind of algebra, without the full symbolical language that would emerge later. The *abacists,*
such as Paolo dell’Abbaco (1282–1374) who featured in Chapter **3**, clearly saw that the power and
spirit of mathematics was rooted in this generality. But modern algebraic symbolism - in particular, the idea that to express generality we need to
use letters to represent not only variables, but also important *parameters* (such as the coefficients *a, b, c* in a general quadratic
*ax*^{2} + *bx + c)* - had to wait for the inscrutable writings of Viète (1540–1603), and especially for Fermat
(1601–1665) and Descartes (1596–1650) who simplified and extended Viète’s ideas in the 1630s.

Within a generation, the huge potential of this systematic use of symbols was revealed by the triumphs of Newton (1642–1727), Leibniz
(1646–1716), and others in the years before 1700. Later, the refinements proposed by Euler (1707–1783) in his many writings throughout
the 18^{th} century, made this new language and its discoveries accessible to us all - much as Stevin’s (1548–1620) version of
place value for numbers made calculation accessible to Everyman.

Our coverage of algebra is necessarily selective. We focus on a few ideas that are needed in what follows, and which should ideally be familiar - but with an emphasis that may be less familiar. When working algebraically, the key mathematical messages are mostly implicit in the manipulations themselves. Hence many of the additional comments in this chapter are to be found as part of the solutions, rather than within the main text.

## 4.1. Simultaneous linear equations and symmetry

**Problem 92** Dad took our new baby to the clinic to be weighed. But the baby would not stay still and caused the needle on the
scales to wobble. So Dad held the baby still and stood on the scales, while nurse read off their combined weight: 78kg. Then nurse held the baby,
while Dad read off their combined weight: 69kg. Finally Dad held the nurse, while the baby read off their combined weight: 137kg. How heavy was the
baby?

The situation described in Problem **92** is representative of a whole class of problems, where the given information
incorporates a certain *symmetry, *which the solver would be wise to respect. Hence one should hesitate before applying systematic brute force
(as when using the information from one weighing to substitute for one of the three unknown weights – a move which effectively reduces the
number of unknowns, but which fails to respect the symmetry in the data).

A similar situation arises in certain puzzles like the following.

**Problem 93** Numbers are assigned (secretly) to the vertices of a polygon. Each edge of the polygon is then labelled with the sum
of the numbers at its two end vertices.

(a) If the polygon is a triangle *ABC*, and the labels on the three sides are *c *(on *AB*), b (on *AC*), and *a* (on
*BC*), what were the numbers written at each of the three vertices?

(b) If the polygon is a quadrilateral *ABCD,* and the labels on the four sides are *w* (on AB), *x* (on *BC), y* (on
*CD*), and *z* (on *DA*), what numbers were written at each of the four vertices?

(c) the polygon is a pentagon *ABCDE,* and the labels on the five sides are *d* (on *AB*), e (on *BC*), *a* (on
*CD*), *b* (on *DE*), and *c* (on *EA*), what numbers were written at each of the five vertices?

In case any reader is inclined to dismiss such problems as “artificial puzzles”, it may help to recall two familiar instances
(Problems **94** and **96**) which give rise to precisely the above situation.

**Problem 94** In the triangle *ABC* with sides of lengths *a* (opposite *A*), *b *(opposite *B*), and
*c* (opposite *C*), we want to locate the three points where the *incircle* touches the three sides - at point *P* (on *BC*),
*Q* (on *CA*), and *R* (on *AB*). To this end, let the two tangents to the incircle from A (namely *AQ* and *AR)* have
length *x,* the two tangents from B (namely *BP* and *BR) *have length *y,* and the two tangents from *C* (namely CP and
*CQ)* have length *z*. Find the values of *x, y, z* in terms of *a, b, c.*

The second instance requires us first to review the basic properties of midpoints in terms of vectors.

**Problem 95**

(a) Write down the coordinates of the midpoint M of the line segment joining *Y =* (a, b) and *Z =* (c, d). Justify your answer.

(b) Position a general triangle *XYZ* so that the vertex *X* lies at the origin (0,0). Suppose that Y then has coordinates *(a, b)*
and *Z* has coordinates (c, d). Let *M* be the midpoint of *XY*, and *N* be the midpoint of *XZ*. Prove the *Midpoint
Theorem,* namely that

“

MNis parallel toYZand half its length”.

(c) Given any quadrilateral *ABCD,* let *P* be the midpoint of *AB*, let *Q* be the midpoint of *BC*, let *R* be the
midpoint of *CD*, and let *S* be the midpoint of *DA*. Prove that *PQRS* is always a parallelogram.

**Problem 96**

(a) Suppose you know the position vectors **p, q, r** corresponding to the midpoints of the three sides of a triangle. Can you
reconstruct the vectors **x, y, z** corresponding to the three vertices?

(b) Suppose you know the vectors **p, q, r, s** corresponding to the midpoints of the four sides of a quadrilateral.
Can you reconstruct the vectors **w, x, y, z** corresponding to the four vertices?

(c) Suppose you know the vectors **p, q, r, s, t** corresponding to the midpoints of the five sides of a
pentagon. Can you reconstruct the vectors **v, w, x, y, z** corresponding to the five vertices?

The previous five problems explore a common structural theme - namely the link between certain sums (or averages) and the original, possibly unknown, data. However this algebraic link was in every case embedded in some practical, or geometrical, context. The next few problems have been stripped of any context, leaving us free to focus on the underlying structure in a purely algebraic, or arithmetical, spirit.

**Problem 97** Solve the following systems of simultaneous equations.

(a) (i) *x + y* = 1, *y + z* = 2, *x + z* = 3

(ii) *uv =* 2, *vw =* 4, uw *=* 8

(b) (i) *x + y* = 2, *y + z* = 3, *x + z* = 4

(ii) *uv* = 6, *vw* = 10, *uw* = 15

(iii) *uv* = 6, *vw* = 10, *uw* = 30

(iv) *uv* = 4, *vw* = 8, *uw* = 16

**Problem 98** Use what you know about solving two simultaneous linear equations in two unknowns to construct the general positive
solution to the system of equations:

$${u}^{a}{v}^{b}=m,\text{\hspace{1em}}{u}^{c}{v}^{d}=n.$$

Interpret your result in the language of *Cramer’s Rule.* (Gabriel Cramer (1704–1752)).

**Problem 99**

(a) For which values *b*, *c* does the following system of equations have a unique solution?

$$x+y+z=3,\text{\hspace{1em}}xy+yz+zx=b,\text{\hspace{1em}}{x}^{2}+{y}^{2}+{z}^{2}=c$$

(b) For which values *a, b, c* does the following system of equations have a unique solution?

$$x+y+z=a,\text{\hspace{1em}}xy+yz+zx=b,\text{\hspace{1em}}{x}^{2}+{y}^{2}+{z}^{2}=c$$

## 4.2. Inequalities and modulus

The transition from school to university mathematics is in many ways marked by a shift from simple variables, equations and functions, to conditions and analysis involving inequalities and modulus.

**Problem 100** What is | − *x*| equal to: *x* or −*x*? (What if *x* is
negative?)

### 4.2.1 Geometrical interpretation of modulus, of inequalities, and of modulus inequalities

**Problem 101**

(a) Mark on the coordinate line all those points *x* in the interval [0,1) which have the digit “1” immediately after the
decimal point in their decimal expansion. What fraction of the interval [0,1) have you marked?

**Note:** “[0,1)” denotes the set of all points *between* 0 and 1, together with 0, but not including 1. [0,1] denotes the
interval including *both* endpoints; and (0,1) denotes the interval *excluding* both endpoints.

(b) Mark on the interval [0,1) all those points *x* which have the digit “1” in *at least one* decimal place. What fraction
of the interval [0,1) have you marked?

(c) Mark on the interval [0,1) all those points *x* which have a digit “1” in at least one position of their *base* 2
expansion. What fraction of the interval [0,1) have you marked?

(d) Mark on the interval [0,1) all those points *x* which have a digit “1” in at least one position of their *base* 3
expansion. What fraction of the interval [0,1) have you marked?

**Problem 102** Mark on the coordinate line all those points *x* for which *two *of the following inequalities are true,
and *five* are false:

$$x>1,x>2,x>3,x>4,x>5,x>6,x>7.$$

**Problem 103** Mark on the coordinate line all those points *x* for which

$$|x-5|=3.$$

**Problem 104**

(a) Mark on the coordinate line all those points *x* for which *two* of the following inequalities are true, and *five* are
false:

$$\left|x\right|>1,\left|x\right|>2,\left|x\right|>3,\left|x\right|>4,\left|x\right|>5,\left|x\right|>6,\left|x\right|>7.$$

(b) Mark on the coordinate line all those points *x* for which *two* of the following inequalities are true, and *five* are
false:

$$|x-1|>1,|x-2|>2,|x-3|>3,|x-4|>4,\text{\hspace{0.17em}}|x-5|>5,|x-6|>6,|x-7|>7.$$

**Problem 105** Mark on the coordinate line all those points *x* for which

$$|x+1|+|x+2|=2.$$

**Problem 106** Find numbers *a* and *b* with the property that the set of solutions of the inequality

$$|x-a|<b$$

consists of the interval (−1, 2).

**Problem 107**

(a) Mark on the coordinate plane all points (*x*, *y*) satisfying the inequality

$$|x-y|<3.$$

(b) Mark on the coordinate plane all points (*x*, *y*) satisfying the inequality

$$|x-y+5|<3.$$

(c) Mark on the coordinate plane all points (*x*, *y*) satisfying the inequality

$$|x-y|<|x+y|.$$

### 4.2.2 Inequalities

**Problem 108** Suppose real numbers *a*, *b, c, d* satisfy $\frac{a}{b}<\frac{c}{d}$ .

(i) Prove that

$$\frac{a}{b}<\frac{\left(\frac{a}{b}+\frac{c}{d}\right)}{2}<\frac{c}{d}.$$

(ii) If b, d > 0, prove that

$$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}.$$

**Problem 109 (Farey series)** When the fully cancelled fractions in [0,1] with denominator *≤ n* are arranged in increasing order,
the result is called the *Farey series* (or *Farey sequence*) *of order n.*

Order 1: $\frac{0}{1}<\frac{1}{1}$

Order 2: $\frac{0}{1}<\frac{1}{2}<\frac{1}{1}$

Order 3: $\frac{0}{1}<\frac{1}{3}<\frac{1}{2}<\frac{2}{3}<\frac{1}{1}$

Order 4: $\frac{0}{1}<\frac{1}{4}<\frac{1}{3}<\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{1}{1}$

(a) Write down the full Farey series (or sequence) of order **7**.

(b) (i) Imagine the points 0.1, 0.2,0.3,..., 0.9 dividing the interval [0,1] into ten subintervals of length $\frac{1}{10}$ . Now insert the eight points corresponding to

$$\frac{1}{9},\frac{2}{9},\frac{3}{9},\dots ,\frac{8}{9}.$$

Into which of the ten subintervals do they fall?

(ii) Imagine the *n* points

$$\frac{1}{n+1},\frac{2}{n+1},\frac{3}{n+1},\dots ,\frac{n}{n+1}$$

dividing the interval [0,1] into *n* + 1 subintervals of length $\frac{1}{n+1}$ . Now insert the *n* − 1 points

$$\frac{1}{n},\frac{2}{n},\frac{3}{n},\dots ,\frac{n-1}{n}.$$

Into which of the *n* + 1 subintervals do they fall?

(iii) In passing from the Farey series of order *n* to the Farey series of order *n +* 1, we insert fractions of the form $\frac{k}{n+1}$ between certain pairs of adjacent fractions in the Farey series of order *n*. If $\frac{a}{b}<\frac{c}{d}$ are adjacent
fractions in the Farey series of order *n*, prove that, when adding fractions for the Farey series of order *n* + 1, **at most one**
fraction is inserted between $\frac{a}{b}\text{\hspace{0.17em}}\mathrm{a}\mathrm{n}\mathrm{d}\text{\hspace{0.17em}}\frac{c}{d}$ .

(c) **Note:** It is worth struggling to prove the two results in part (c). But do not be surprised if they prove to be elusive – in which
case, be prepared to simply use the result in part (c)(ii) to solve part (d).

(i) In the Farey series of order *n* the first two fractions are $\frac{0}{1}<\frac{1}{n}$ , and the last two fractions are $\frac{n-1}{n}<\frac{1}{1}$ . Prove that every other adjacent pair of fractions $\frac{a}{b}<\frac{c}{d}$ in the Farey series of order *n* satisfies *bd > n.*

(ii) Let $\frac{a}{b}<\frac{c}{d}$ be adjacent fractions in the Farey series of order *n*. Prove (by induction on *n*) that
*bc* − *ad =* 1.

(d) Prove that if

$$\frac{a}{b}<\frac{c}{d}<\frac{e}{f}$$

are three successive terms in any Farey series, then

$$\frac{c}{d}=\frac{a+e}{b+f}.$$

**Problem 110** Solve the following inequalities.

(a) $x+\frac{1}{x}<2$

(b) $x\u2a7d1+\frac{2}{x}$

(c) $\sqrt{x}<x+\frac{1}{4}$

**Problem 111**

(a) The sum of two positive numbers equals 5. Can their product be equal to 7?

(b) (**Arithmetic mean, Geometric mean, Harmonic mean, Quadratic mean**) Prove that, if *a,b >* 0, then

$$\begin{array}{l}\frac{2}{\left[\frac{1}{a}+\frac{1}{b}\right]}=\frac{2ab}{a+b}\u2a7d\sqrt{ab}\u2a7d\frac{a+b}{2}\u2a7d\sqrt{\frac{{a}^{2}+{b}^{2}}{2}}\\ (\text{HM}\u2a7d\text{GM}\u2a7d\text{AM}\u2a7d\text{QM})\end{array}$$

**Problem 112** The two hundred numbers

$$1,2,3,4,5,\dots ,200$$

are written on the board. Students take turns to replace two numbers *a*, b from the current list by their sum divided by $\sqrt{2}$ . Eventually one number is left on the
board. Prove that the final number must be less than 2000.

## 4.3. Factors, roots, polynomials and surds

**Problem 113**

(a) (i) Find a prime number which is one less than a square.

(ii) Find another such prime.

(b) (i) Find a prime number which is one more than a square.

(ii) Find another such prime.

(c) (i) Find a prime number which is one less than a cube.

(ii) Find another such prime.

(d) (i) Find a prime number which is one more than a cube.

(ii) Find another such prime.

**Problem 114** Factorise *x*^{4} + 1 as a product of two quadratic polynomials with real coefficients.

### 4.3.1 Standard factorisations

The challenge to factorise unfamiliar expressions, may at first leave us floundering. But if we assume that each such problem is solvable with the tools at our disposal, we then have no choice but to fall back on the standard tools we have available (in particular, the standard factorisation of a difference of two squares, in which “cross terms” cancel out). The next problem extends this basic repertoire of standard factorisations.

**Problem 115**

(a)(i) Factorise *a*^{3} − *b ^{3}.*

(ii) Factorise *a*^{4} − *b*^{4} as a product of one linear factor and one factor of degree 3, and as a product of
two linear factors and one quadratic factor.

(iii) Factorise *a ^{n}* −

*b*as a product of one linear factor and one factor of degree

^{n}*n −*1.

(b)(i) Factorise *a*^{3} + *b*^{3}.

(ii) Factorise *a*^{5} + *b*^{5} as a product of one linear factor and one factor of degree 4.

(iii) Factorise *a ^{2n+1} +*

*b*

^{2n+1}as a product of one linear factor and one factor of degree 2

*n*.

Problem **115** develops the ideas that were implicit in Problem **113**. The clue lies in
Problem **113**(a), and in the comment made in the main text in Chapter **1** (after
Problem **4** in Chapter 1), which we repeat here:

“The last part [of Problem

113(a)] is included to emphasise a frequently neglected message:Words and images are part of the way we communicate.

But most of us cannotcalculatewith words and images.To make use of mathematics, we must routinely translate

wordsintosymbols.So “numbers” need to be represented by symbols, and points in a geometric diagram need to be properly labelled before we can begin to calculate, and to reason, effectively.”

As soon as one reads the words “one less than a square”, one should instinctively translate this into the form
“*x*^{2} *−* 1”. Bells will then begin to ring; for it is impossible to forget the factorisation

$${x}^{2}-1=(x-1)(x+1).$$

And it follows that:

for a number that factorises in this way to be prime, the smaller factorx− 1 must be equal to 1;

$\therefore x=2$ , so there is only one such prime.

The integer factorisations in Problem **113**(c) − namely

$${3}^{3}-1=2\times 13,{4}^{3}-1=3\times 21,{5}^{3}-1=4\times 31,{6}^{3}-1=5\times 43,\dots $$

may help one to remember (or to discover) the related factorisation

$${x}^{3}-1=(x-1)\left({x}^{2}+x+1\right).$$

∴ For a number that factorises in this way to be prime, the smaller factor “x − 1” must be equal to 1;

∴x= 2, so there is only one such prime.

Problem **113** parts (a) and (c) highlight the completely general factorisation (Problem **115**(a)(iii)):

$${x}^{n}-1=(x-1)\left({x}^{n-1}+{x}^{n-2}+\cdots +{x}^{2}+x+1\right).$$

This family of factorisations also shows that we should think about the factorisation of *x*^{2} − 1 as (** x**
−

**1**)(

*x*+ 1), with the uniform factor (

**−**

*x***1**) first (rather than as (

*x*+ 1)(

*x*− 1)). Similarly, the results of Problem

**115**show that we should think of the familiar factorisation of

*a*

^{2}−

*b*

^{2}as

^{(a − b)(a}+ b), (not as (a ' b)(a − b), but always with the factor

*(a*− b) first).

The integer factorisations in Problem **113**(d) − namely

$${3}^{3}+1=4\times 7,{4}^{3}+1=5\times 13,{5}^{3}+1=6\times 21,{6}^{3}+1=7\times 31,{7}^{3}+1=8\times 43,\dots $$

may help one to remember (or to discover) the related factorisation

$${x}^{3}+1=(x+1)\left({x}^{2}-x+1\right)$$

∴ For such a number to be prime, one of the factors must be equal to 1.

This time one has to be more careful, because the first bracket may not be the “smaller factor” – so there are two cases to consider:

(i) if

x+ 1 = 1 , thenx= 0, andx^{3}+ 1 = 1 is not prime;(ii) if

x^{2}−x= 1 thenx= 0 orx= 1, sox= 1 and we obtain the prime 2 as the only solution.

The factorisation for *x*^{3} + 1 works because “3 is **odd**”, which allows the alternating +/− signs to end
in a “+” as required. Hence Problem **113**(d)(iii) highlights the completely general factorisation **for odd
powers:**

$${x}^{2n+1}+1=(x+1)\left({x}^{2n}-{x}^{2n-1}+{x}^{2n-2}-\cdots +{x}^{2}-x+1\right)$$

You probably know that there is no standard factorisation of *x*^{2} + 1, or of *x*^{4} + 1 (but see Problem **114** above).

**Problem 116**

(a) Derive a *closed formula* for the sum of the geometric series

$$1+r+{r}^{2}+{r}^{3}+\cdots +{r}^{n}.$$

(The meaning of *closed formula* was discussed in the **Note** to the solution to Problem **54**(b) in Chapter 2.)

(b) Derive a closed formula for the sum of the geometric series

$$a+ar+a{r}^{2}+a{r}^{3}+\cdots +a{r}^{n}$$

We started this subsection by looking for prime numbers of the form *x*^{2} *−* 1. A simple-minded approach to the
distribution of prime numbers might look for formulae that generate primes - all the time, or infinitely often, or at least much of the time. In Chapter **1** (Problem **25**) you showed that no prime of the form 4*k* + 3 can be
“represented” as a sum of two squares (i.e. in the form “*x*^{2} + *y*^{2}”), and we remarked
that every other prime can be so represented in exactly one way. It is true (but not obvious) that roughly half the primes fall into the second
category; so it follows that substituting integers for the two variables in the polynomial *x*^{2} + y^{2} produces a prime
number infinitely often.

**Problem 117** Experiment suggests, and Goldbach (1690–1764) showed in 1752 that no polynomial in one variable, and with
integer coefficients, can give prime values for all integer values of the variable. But Euler (1707–1783) was delighted when he discovered the
quadratic

$$f(x)={x}^{2}+x+41.$$

Clearly *f* (0) = 41 is prime. And *f* (1) = 43 is also prime. What is the first positive integer *n* for which *f* (n) is
**not** prime?

Problem **117** should be seen as a particular instance of the question as to whether prime numbers can be captured by a
*polynomial* with integer coefficients, and in particular by a *quadratic.* The next two problems consider the simplest instances of
representing prime numbers by expressions involving *exponentials* (that is, where the variable is in the *exponent*).

**Problem 118**

(a)(i) Suppose *a ^{n} −* 1 =

*p*is a prime. Prove that

*a =*2 and that

*n*must itself be prime.

(ii) How many primes are there among the first five such numbers

$${2}^{2}-1,{2}^{3}-1,{2}^{5}-1,{2}^{7}-1,{2}^{11}-1?$$

(b) (i) Suppose *a ^{n} +* 1 =

*p*is a prime. Prove that either

*a =*1, or

*a*must be even and that

*n*must then be a power of 2.

(ii) In the simplest case, where *a* = 2, how many primes are there among the first five such numbers

$${2}^{1}+1,{2}^{2}+1,{2}^{4}+1,{2}^{8}+1,{2}^{16}+1?$$

Primes of the form 2* ^{p}* − 1 are called

*Mersenne primes*(after Marin Mersenne (1588–1648)). We now know at least fifty such primes (with the exponent

*p*ranging up to around 80 million). Finding new primes is not in itself important, but the search for Mersenne primes has been used as a focus for many new developments in programming, and in number theory.

Primes of the form 2* ^{n}* + 1 are called

*Fermat primes*(after Pierre de Fermat (1601–1665)). The story here is very different. We now refer to the number 2

^{n}+ 1 with

*n = 2*as the

^{k}*k*. You showed in Problem

^{th}Fermat number f_{k}**118**(as Fermat did himself) that

*f*f

_{0},_{1}, f

_{2},

*f*are all prime. Fermat then rather rashly claimed that

_{3}, f_{4}*f*is always prime. However, Euler showed (100 years later) that the very next Fermat number

_{n}*f*fails to be prime. And despite all the power of modern computers, we have still not found another Fermat number that is prime!

_{5}### 4.3.2 Quadratic equations

The general solution of quadratic equations dates back to the ancient Babylonians ($\approx 1700\text{BC}$ ). Our modern understanding depends on two facts:

- an equation of the form
*x*where^{2}= a*a*> 0, has exactly two solutions: $x=\pm \sqrt{a}$ ; - any product
*X . Y*is equal to 0 precisely when one of the two factors*X*,*Y*is equal to 0.

**Problem 119** Solve the following quadratic equations:

(a) ${x}^{2}-3x+2=0$

(b) ${x}^{2}-1=0$

(c) ${x}^{2}-2x+1=0$

(d) ${x}^{2}+\sqrt{2}x-1=0$

(e) ${x}^{2}+x-\sqrt{2}=0$

(f) ${x}^{2}+1=0$

(g) ${x}^{2}+\sqrt{2}x+1=0$

**Problem 120** Let

$$p(x)={x}^{2}+\sqrt{2}x+1.$$

Find a polynomial *q*(*x*) such that the product *p*(*x*)*q*(*x*) coefficients.

**Problem 121**

(a) I am thinking of two numbers, and am willing to tell you their sum *s* and their product p. Express the following procedure algebraically
and explain why it will always determine my two unknown numbers.

Halve the sum

s, and square the answer.Then subtract the product

pand take the square root of the result, to get the answer.Add “the answer” to half the sum and you have one unknown number; subtract “the answer” from half the sum and you have the other unknown number.

(b) I am thinking of the length of one side of a square. All I am willing to tell you are two numbers *b* and *c*, where when I add
*b* times the side length to the area I get the answer c. Express the following procedure algebraically and explain why it will always determine
the side length of my square.

Take one half of

b, square it and add the result toc.Then take the square root.

Finally subtract half of b from the result.

(c) A regular pentagon *ABCDE* has sides of length 1.

(i) Prove that the diagonal AC is parallel to the side *ED.*

(ii) If *AC* and BD meet at *X*, explain why *AXDE* is a rhombus.

(iii) Prove that triangles *ADX* and *CBX* are similar.

(iv) If *AC* has length *x*, set up an equation and find the exact value of *x.*

Problem **121**(a), (b) link to Problem **111**(a) (and to Problem **129** below), in relating the roots and the coefficients of a quadratic. If we forget for the moment that the coefficients are usually known,
while the roots are unknown, then we see that if *α* and *β* are the roots of the quadratic

$${x}^{2}+bx+c,$$

then

$$(x-\alpha )(x-\beta )={x}^{2}+bx+c,$$

so

$$\alpha +\beta =-b\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\alpha \beta =c.$$

In other words, the two coefficients *b, c* are equal to the two simplest *symmetric* expressions in the two roots *α* and
*β.* Part (a) of the next problem is meant to suggest that all other symmetric expressions in *α* and *β* (that
is, any expression that is unchanged if we swap *α* and *β*) can then be written in terms of b and *c.* The full result
proving this fact is generally attributed to Isaac Newton (1642–1727). Part (b) suggests that, provided one is willing to allow case
distinctions, something similar may be true of *anti-symmetric* expressions (where the effect of swapping *a* and ft is to multiply the
expression by “−1”).

**Problem 122** Let *α* and *β* be the roots of the quadratic equation

$${x}^{2}+bx+c=0.$$

(a) (i) Write *α*^{2} + *β*^{2} in terms of *b* and *c* only.

(ii) Write *α*^{2}*β* + *β*^{2}*α* in terms of *b* and *c* only.

(iii) Write *α*^{3} + *β*^{3} − 3*αβ* in terms of *b* and *c* only.

(b) (i) Write *α* − *β* in terms of b and c only.

(ii) Write *α*^{2}*β* − *β*^{2}*α* in terms of *b* and *c*
only.

(iii) Write *α*^{3} − *β*^{3} in terms of *b* and *c* only.

**Problem 123 (Nested surds, simplification of surds)**

(a)(i) For any positive real numbers *a, b,* prove that $$\sqrt{a}+\sqrt{b}=\sqrt{a+b+\sqrt{4ab}}$$

(ii) Simplify $\sqrt{5+\sqrt{24}}$ .

(b) (i) Find a similar formula for $\sqrt{a}-\sqrt{b}$ .

(ii) Simplify $\sqrt{5-\sqrt{16}}$ and $\sqrt{6-\sqrt{20}}$.

**Problem 124 (Integer polynomials with a given root)** We know that *α* = 1 is a root of the polynomial equation
*x*^{2} − 1 = 0; that $\alpha =\sqrt{2}$ is a root of *x*^{2} − 2 = 0; and that $\alpha =\sqrt{3}$ is a root of *x*^{2} − 3 = 0.

(a) Find a quadratic polynomial with integer coefficients which has

$$\alpha =1+\sqrt{2}$$

as a root.

(b) Find a quadratic polynomial with integer coefficients which has

$$\alpha =1+\sqrt{3}$$

as a root.

(c) Find a polynomial with integer coefficients which has

$$\alpha =\sqrt{2}+\sqrt{3}$$

as a root. What are the other roots of this polynomial?

(d) Find a polynomial with integer coefficients which has

$$\alpha =\sqrt{2}+\frac{1}{\sqrt{3}}$$

as a root. What are the other roots of this polynomial?

**Problem 125**

(a) Prove that the number $\sqrt{2}+\sqrt{3}$ is irrational.

(b) Prove that the number $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational.

**Problem 126 (Polynomial long division)** Find

(i) the quotient and the remainder when we divide *x*^{10} + 1 by *x*^{3} − 1

(ii) the remainder when we divide *x*^{2013} + 1 by *x*^{2} − 1

(iii) the quotient and the remainder when we divide *x ^{m}* + 1 by

*x*− 1, for $m>n\u2a7e1$ .

^{n}**Problem 127** Find the remainder when we divide *x*^{2013} + 1 by *x*^{2} + *x* + 1.

## 4.4. Complex numbers

Up to this point, the chapter and solutions have largely avoided mentioning *complex numbers*. However, the present chapter would be
incomplete were we not to interpret some of the earlier material in terms of complex numbers. Readers who have already met complex numbers will
probably still find much in the next two sections that is new. Those for whom complex numbers are as yet unfamiliar should muddle through as best they
can, and may then be motivated to learn more in due course.

We already know that the square *x*^{2} of any real number *x* is ⩾ 0.

- If
*a =*0, then the equation*x*^{2}=*a*has exactly one root, namely*x*= 0; - if
*a >*0, then the equation*x*^{2}=*a*has exactly two roots – namely ±$\sqrt{a}$, where $\sqrt{a}$ denotes the root that is*positive*; - if
*a <*0, then the equation*x*^{2}=*a*has no real roots.

And that is where the matter would have rested.

From a modern perspective, we can see that *complex numbers* are implicit in the formula for the roots of a quadratic equation: complex
numbers become explicit as soon as the coefficients of a quadratic *ax ^{2} + bx + c* give rise to a negative

*discriminant*

*b*

^{2}− 4ac < 0.

But this may not have been quite how complex numbers were discovered. Contrary to oft-repeated myths, complex numbers may not have forced
themselves on our attention by someone asking about “solutions to the *quadratic* equation *x*^{2} = −1”. As
long as we inhabit the domain of *real *numbers, we can be sure that no known number *x* could possibly have such a square, so we are
unlikely to go in search of it.

New ideas in the history of mathematics tend to emerge when a fresh analysis of *familiar* entities forces us to consider the possible
existence of some previously unsuspected universe. In the time from the ancient world up to the fifteenth century, the idea of “number”,
and of calculation, was restricted to the world of *real* (usually positive) numbers. In such a world, quadratic equations with non-real
solutions simply could not arise.

However, in the Brave New World of the Renaissance, where novelty, exploration, and discovery were part of the *Zeitgeist*, a general method
for solving *cubic* equations was part of the as-yet-undiscovered “wild west” of mathematics, part of the mathematical New World
which invited exploration. Notice that this was not a wildly speculative venture (like trying to solve the meaningless equation
“*x*^{2} = − 1”), since a cubic polynomial **always** has at least one real root. After three thousand years in
which little progress had been made, the first half of the sixteenth century witnessed an astonishing burst of progress, resulting in the solution not
only of cubic equations, but also of quartic equations. We postpone the details until Section 4.5. All we note here is that,

the general method for solving cubic equations published in 1545, was given as a procedure, illustrated by examples, that showed how to find

genuinelyrealsolutions to equations of the third degree having genuinely real (and positive) coefficients.

The procedure clearly worked. And it proceeded as follows:

Construct the real solution

xas the sumx=u + vof two intermediate answers u and v - where the two summandsuandvsometimes turned out to be what we would call “conjugate complex numbers”,whose imaginary parts cancelled out, leaving a real result for the required rootx.

Those who devised the procedure had no desire to leave the *real* domain: they were focused on a problem in the *real* domain (a cubic
equation with *real *coefficients, having a *real* root), and devised a general procedure to find that genuinely *real* root. But the
procedure they discovered led the solver on a journey that sometimes “passed into the complex domain”, before returning to the real
domain! (See Problem **129**.)

Working with complex numbers depends on two skills - one very familiar, and one less so.

- The familiar skill is a willingness to work with a “number”
*in terms of its properties only,*without wishing to evaluate it.

We are thoroughly familiar with this when we work with $\frac{2}{3}$ and other fractions: we know that $\frac{2}{3}=2\times \frac{1}{3}$ ; and all we know about $\frac{1}{3}$ is that “whenever we have 3 copies of $\frac{1}{3}$, we can simplify this to 1”. Much the same happens when we first learn to work with $\sqrt{2}$ , where we carry out such calculations as ${(1+\sqrt{2})}^{2}=3+2\sqrt{2}$ , based only on collecting up like terms and the fact that $\sqrt{2}\times \sqrt{2}$ can always be replaced by 2.

- The less familiar skill is easily overlooked. When, for whatever reason, we decide to allow solutions to the equation
*x*^{2}= − 1, three things need to be understood.− First, these new solutions come

**in pairs**: if*i is one solution of*−1 then −*x*^{2}=*i*is another (because (−1)*x*(−1) = 1 means that (−*x*)^{2}=*x*^{2 }for all “numbers”*x*.− Second, the equation

*x*^{2}= − 1 has**exactly two solutions**- one the negative of the other (if*x*and*y*are both solutions, then*x*^{2}=*y*^{2}, so*x*^{2}−*y*^{2}= (*x*−*y*)(*x + y*) = 0, so either*x*=*y*, or*x*= −*y*).− Third,

*we have no way of telling these two solutions apart*: we know that each is the negative of the other, but there is no way of singling out one of them as “the main one” (as we could when defining the square root of a positive real such as 2). We can call them ±*i*, but they are*each as good as the other*. This important fact is often undermined by referring to one of these roots as $\sqrt{-1}$ (as if it were the dominant partner), and to the other as −$\sqrt{-1}$ (as if it were somehow just the “negative” of the main root).

The truth is that “$\sqrt{-1}$ ” is a serious abuse of notation, because there is no way to extend the definition of the *function*
“$\sqrt{}$ ” in the way
that this implies: when we try to “take square roots” of negative (or complex) numbers, the output is inescapably
“two-valued”, so “$\sqrt{}$” is no longer a function. The two roots of *x*^{2} = −1 are like Tweedledum and Tweedledee: we know there
are two of them, and we know how they are related; but we have no way of distinguishing them, or of singling one of them out.

Once we accept this, we can write complex numbers in the form *a + bi, *where *a* and b are real numbers (just as we used to write
numbers in the form $a+b\sqrt{2}$ where *a* and b are rational numbers). And we can proceed to add, subtract, multiply, and divide such expressions, and
then collect up the “real” and “imaginary” parts to tidy up the answer.

**Problem 128**

(a) Write the inverse (*a* + *bi*)^{−1} in the form *c + di.*

(b) Write down a quadratic equation with real coefficients, which has *a + bi* as one root (where *a* and b are real numbers).

**Problem 129** Divide 10 into two parts, whose product is 40.

Problem **129** appears in Chapter XXXVII of Girolamo Cardano’s (1501–1576) book *Ars Magna* (1545).
Having previously presented the general methods for solving quadratic, cubic, and quartic equations, he honestly confronts the phenomenon that his
method for solving cubic equations (see Problem **135**) produces the required real (and positive) solution *x* as a sum
of *complex* conjugates *u* and *v* - involving not only *negative* numbers, but * square roots of negative numbers.*
After presenting the formal solution of Problem

**129**, and having shown that the calculation works exactly as it should, he adds the bemused remark:

“So progresses arithmetic subtlety,

the end of which … is as refined as it is useless.”

Arithmetic with complex numbers in the form *a + bi* is done by carrying out the required operations, and then collecting up the
“real” and “imaginary” parts as separate components - just as with adding vectors (*a, b*). We treat the two parts as
Cartesian coordinates, and so identify the complex number *a + bi* with the point (*a*, *b*) in the complex plane.

The “Cartesian” representation *a + bi* is very convenient for *addition.* But the essential definition (and significance)
of complex numbers is rooted in *multiplication.* And for multiplication it is often much better to work with complex numbers written in *polar
form.* Suppose we mark the complex number *w = a + bi* in the complex plane.

The *modulus* |w| of *w* (often denoted by *r*) is the distance $r=\sqrt{{a}^{2}+{b}^{2}}$ of the complex number *a* + 6*i* from the origin in the complex plane.

The angle θ, measured anticlockwise from the positive real axis to the line joining the complex number *w* to the origin, is called the
*argument,* $Arg(w)=\theta $ , of *w*.

It is then easy to check that *a = r* cos θ, *b = r* sin θ, and that

$$w=r(\mathrm{cos}\theta +i\mathrm{sin}\theta )$$

This is the *polar form* for *w*. Instead of focusing on the Cartesian coordinates *a*, *b*, the polar form pinpoints *w*
in terms of

- its
*length,*or*modulus,**r*(which specifies the circle, with centre at the origin, on which the complex number*w*lies), and - the
*argument*0 (which tells us where on this circle*w*is to be found).

**Problem 130**

(a) Given two complex numbers in polar form:

$$w=r(\mathrm{cos}\theta +i\mathrm{sin}\theta ),z=s(\mathrm{cos}\varphi +i\mathrm{sin}\varphi )$$

show that their product is precisely

$$wz=rs(\mathrm{cos}(\theta +\varphi )+i\mathrm{sin}(\theta +\varphi )).$$

(b) (**de Moivre’s Theorem**: Abraham de Moivre (1667–1754)) Prove that

$${(\mathrm{cos}\theta +i\mathrm{sin}\theta )}^{n}=\mathrm{cos}(n\theta )+i\mathrm{sin}(n\theta ).$$

(c) Prove that, if

$$z=r(\mathrm{cos}\theta +i\mathrm{sin}\theta )$$

satisfies *z ^{n}* = 1 for some integer

*n*, then

*r*= 1.

The last three problems in this subsection look more closely at “roots of unity” − that is, roots of the polynomial equation
*x ^{n}* = 1. In the

*real*domain, we know that:

(i) when *n* is odd, the equation *x ^{n}* = 1 has exactly one root, namely

*x*= 1; and

(ii) when *n* is even, the equation *x ^{n}* = 1 has just two solutions, namely $x=\pm 1$.

In contrast, in the *complex* domain, there are *n* “n^{th} roots of unity”. Problem **130**(c) shows that these “roots of unity” all lie on the unit circle, centered at the origin. And if we put $n\theta =2k\pi $ in Problem **130**(b) we see that the *n* *n*^{th} roots of unity include the point “1 =
cos0 + *i* sin0”, and are then equally spaced around that circle with $\theta =\frac{2k\pi}{n}(1\u2a7dk\u2a7dn-1)$ , and
form the vertices of a regular *n*-gon.

**Problem 131**

(a) Find all the complex roots of unity of degree 3 (that is, the roots of *x*^{3} = 1) in surds form.

(b) Find all the complex roots of unity of degree 4 in surd form.

(c) Find all the complex roots of unity of degree 6 in surd form.

(d) Find all the complex roots of unity of degree 8 in surd form.

**Problem 132** Use Problem **131**(d) to factorise *x*^{4} + 1 as a product of four linear
factors, and hence as a product of two quadratic polynomials with real coefficients.

**Problem 133**

(a) Find all the complex roots of unity of degree 5 in surd form.

(b) Factorise *x ^{5} −* 1 as a product of one linear and two quadratic polynomials with real coefficients.

## 4.5. Cubic equations

The first recorded procedure for finding the positive roots of any given quadratic equation dates from around 1700 BC (ancient Babylonian). A corresponding procedure for cubic equations had to wait until the early sixteenth century AD. The story is a slightly complicated one - involving public contests, betrayal, and much else besides.

In Section 4.4 we saw that the cubic equation *x*^{3} = 1 has three solutions - two of which are complex
numbers. But in the sixteenth century, even negative numbers were viewed with suspicion, and complex numbers were still unknown. Moreover, symbolical
algebra had not yet been invented, so everything was carried out in words: constants were “numbers”; a given multiple of the unknown was
referred to as so many “things”; a given multiple of the square of the unknown was simply referred to as “squares”; and so
on.

In short, we know that an improved method for sometimes finding a (positive) unknown which satisfied a cubic equation was devised by Scipione del
Ferro (1465–1526) around 1515. He kept his method secret until just before his death, when he told his student Antonio del Fiore (1506-??).
Niccolo Tartaglia (1499–1557) then made some independent progress in solving cubic equations. At some stage (around 1535) Fiore challenged
Tartaglia to a public “cubic solving contest”. In preparing for this event, Tartaglia managed to improve on his method, and he seems to
have triumphed in the contest. Tartaglia naturally hesitated to divulge his method in order to preserve his superiority, but was later persuaded to
communicate what he knew to Girolamo Cardano (1501–1576) after Cardano promised not to publish it (either never, or not before Tartaglia
himself had done so). Cardano improved the method, and his student Ferrari (1522–1565) extended the idea to give a method for solving quartic
equations - all of which Cardano then published, contrary to his promise, but with full attribution to the rightful discoverers, in his groundbreaking
book *Ars Magna* (1545 -just two years after Copernicus (1473–1543) published his *De revolutionibus* ...). Problem **134** illustrates the necessary first move in solving any cubic equation. Problem **135** then
illustrates the general method in a relatively simple case.

**Problem 134**

(a) Given the equation *x*^{3} + 3*x*^{2} − 4 = 0, choose a constant *a,* and then change variable by
substituting *y = x + a* to produce an equation of the form *y*^{3} + *ky =* constant.

(b) In general, given any cubic equation *ax*^{3} + *bx*^{2} + *c*^{x} + *d* = 0 with $a\ne 0$ , show how to change
variable so as to reduce this to a cubic equation with no quadratic term.

**Problem 135** The equation *x*^{3} + 3*x*^{2} − 4 = 0 clearly has “*x* = 1” as
a positive solution. (The other two solutions are *x* = −2, and *x* = −2 - a repeated root; however negatives were viewed with
suspicion in the sixteenth century, so this root might well have been ignored.) Try to understand how the following sequence of moves “finds
the root *x* = 1”:

(i) substitute *y = x* + 1 to get a cubic equation in *y* with no term in *y*^{2};

(ii) imagine *y = u + v* and interpret the identity for

$${(u+v)}^{3}={u}^{3}+3uv(u+v)+{v}^{3}$$

as your cubic equation in *y*;

(iii) solve the simultaneous equations “3*uv* = 3”, “*u*^{3} + *v*^{3} = 2” (not by
guessing, but by substituting $v=\frac{1}{u}$ from the first equation into the second to get a quadratic equation in “*u*^{3}”, which
you can then solve for *u*^{3} before taking cube roots);

(iv) then find the corresponding value of *v*, hence the value of *y* = *u* + *v*, and hence the value of *x*.

The simple method underlying Problem **135** is in fact completely general. Given any cubic equation

$$a{x}^{3}+b{x}^{2}+cx+d=0\text{\hspace{1em}}(\mathrm{with}\text{\hspace{0.17em}}a\ne 0)$$

we can divide through by a to reduce this to

$${x}^{3}+p{x}^{2}+qx+r=0$$

with leading coefficient = 1. Then we can substitute $y=x+\frac{p}{3}$ and reduce this to a cubic equation in *y*

$${y}^{3}-3{\left(\frac{p}{3}\right)}^{2}y+qy+\left[r+2{\left(\frac{p}{3}\right)}^{3}-q\left(\frac{p}{3}\right)\right]=0$$

which we can treat as having the form

$${y}^{3}-my-n=0.$$

So we can set *y = u + v* (for some unknown *u* and *v* yet to be chosen), and treat the last equation as an instance of the
identity

$${(u+v)}^{3}-3uv(u+v)-\left({u}^{3}+{v}^{3}\right)=0$$

which it will become if we simply choose *u* and *v* to solve the simultaneous equations

$$3uv=m,\text{\hspace{1em}}{u}^{3}+{v}^{3}=n.$$

We can then solve these equations to find *u*, then *v* - and hence find *y = u + v* and $x=y-\frac{p}{3}$ .

## 4.6. An extra

Back in Chapter **1**, Problem **6** we introduced the Euclidean algorithm for
integers. The same idea was extended to polynomials with integer coefficients in Problem **126**. In both these settings one
starts with a domain (whether the set of integers, or the set of all polynomials with integer coefficients) where there is a notion of divisibility:
given two elements *m*, *n* in the relevant domain, we say

“

ndividesm” if there exists an elementqin the domain such thatm=qn.

The next problem invites you to think how one might extend the Euclidean algorithm to a new domain, namely the *Gaussian integers*
ℤ[*i*] − the set of all complex numbers *a + bi* in which the real and imaginary “coordinates” *a *and
*b* are integers.

**Problem 136** Complex numbers *a + bi,* where both *a* and b are integers, are called Gaussian integers. Try to
formulate a version of the “division algorithm” for “division with remainder” (where the remainder is always “less
than” the divisor in some sense) for pairs of Gaussian integers. Extend this to construct a version of the Euclidean algorithm to find the HCF
of two given Gaussian integers.

*It is a profoundly erroneous truism … that we should cultivate the habit of thinking what we are doing.*

*The precise opposite is the case. Civilisation advances by extending the number of important operations which we can perform without thinking
about them.*

Alfred North Whitehead (1861–1947)

## 4.7. Chapter **4**: Comments and solutions

**92.** Answer: Humour aside, this is a common situation.

We know $d+b$, $n+b$, $d+n$, rather than the values of *d*, *b*, *n*.

The key is to exploit the symmetry in the given data, rather than solving blindly. Adding all three two-way totals gives $2(d+b+n)=284$ whence $d+b+n=142$. We can then subtract the given value of $d+n=137$ to get the value of $b=5$.

**93.**

(a) Let the numbers at the three vertices be *A*, *B*, *C*. Adding shows that

so

$$\begin{array}{c}A=\genfrac{}{}{0.1ex}{}{a+b+c}{2}-(B+C)=\genfrac{}{}{0.1ex}{}{b+c-a}{2}\hfill \end{array}$$etc.

(c) **Note:** We postpone the “solution” of part (b), and address part (c) first. Let the numbers at the five vertices be
*A*, *B*, *C*, *D*, *E*. Adding shows that

so

$$\begin{array}{c}\mathit{A}=\frac{\mathit{d}+\mathit{e}+\mathit{a}+\mathit{b}+\mathit{c}}{2}-(\mathit{B}+\mathit{C})-(\mathit{D}+\mathit{E})\hfill \\ \phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}=\frac{\mathit{d}-\mathit{e}-\mathit{a}-\mathit{b}-\mathit{c}}{2}\hfill \end{array}$$etc.

(b) The second part is different. The four given edge-values do not determine the four unknown vertex-values. It may look as though four pieces of
information should suffice to find four unknowns; but there is a catch: the sum of the numbers on the two opposite edges *AB* and *CD* is
just the sum of the numbers at the four vertices, and so is equal to the sum of the numbers on the edges *BC* and *DA*. Hence one of the
given edge-values is determined by the other three.

**Note:** This distinction between polygons with an odd and an even number of vertices would arise in exactly the same way if each edge was
labelled with the average (“half the sum”) of the numbers at its two end vertices.

**94.** $a=BC=BP+PC=y+z$; $b=x+z$; $c=x+y$. Hence

so

$$\mathit{x}+\mathit{y}+\mathit{z}=\frac{\mathit{a}+\mathit{b}+\mathit{c}}{2}.$$So

$$\begin{array}{c}\mathit{x}=\frac{\mathit{a}+\mathit{b}+\mathit{c}}{2}-(\mathit{y}+\mathit{z})\hfill \\ \phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}=\frac{\mathit{b}+\mathit{c}-\mathit{a}}{2}\hfill \end{array}$$etc.

**95.**

(a) Let

$$\mathit{M}=\left(\frac{\mathit{a}+\mathit{c}}{2},\frac{\mathit{b}+\mathit{d}}{2}\right).$$The shift, or vector, from (*a*, *b*) to (*c*, *d*) goes

“along

c–ain thex-direction” and “upd–bin they-direction”.

Draw the ordinate through *Y* and the abscissa through *Z*, to meet at *P*, so creating a right angled triangle with legs *Y P*
of length $|c-a|$ and
*PZ* of length $|d-b|$.
The midpoint of *Y P* clearly lies halfway along *Y P* at

and the midpoint of *PZ* clearly lies halfway up *PZ* at

Then $\mathrm{\u25b3}\mathit{YSM}$ and $\mathit{\u25b3MTZ}$ are both right-angled triangles and are congruent (by
RHS congruence). Hence $YM=MZ$, so *M* is the midpoint of *Y Z*.

(b)

$$\mathit{M}=\left(\frac{\mathit{a}}{2},\frac{\mathit{b}}{2}\right),\phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}\mathit{N}=\left(\frac{\mathit{c}}{2},\frac{\mathit{d}}{2}\right)$$so vector

$$\mathit{MN}=\left(\frac{\mathit{c}-\mathit{a}}{2},\frac{\mathit{d}-\mathit{b}}{2}\right)=\frac{1}{2}\mathit{BC}.$$(c) Note: We use the result from part (b), but not the method from part (b). By part (b) applied to $\mathit{\u25b3BAC}$, *PQ* is half the length of *AC* and parallel
to *AC*. By part (b) applied to $\mathit{\u25b3DAC}$, *SR* is
half the length of *AC* and parallel to *AC*. Hence *PQ* is parallel to *SR*.

Similarly one can prove (applying part (b) twice – first to $\mathit{\u25b3ABD}$, and then to $\mathit{\u25b3CBD}$) that
*PS* is parallel to *QR*.

Hence *PQRS* is a parallelogram.

**96.**

(a) $\mathrm{p}=\frac{1}{2}(\mathrm{x}+\mathrm{y}\mathrm{),}\phantom{\rule{0.33em}{0ex}}\mathrm{q}=\frac{1}{2}(\mathrm{y}+\mathrm{z}\mathrm{),}\phantom{\rule{0.33em}{0ex}}\mathrm{r}=\frac{1}{2}(\mathrm{z}+\mathrm{x})$, so

$$\begin{array}{c}\mathrm{p\; +\; q\; +\; r\; =\; x\; +\; y\; +\; z.}\hfill \end{array}$$Hence

$$\begin{array}{c}\mathrm{x}=(\mathrm{p}+\mathrm{q}+\mathrm{r})-(\mathrm{y}+\mathrm{z})=\mathrm{p}-\mathrm{q}+\mathrm{r}\hfill \\ \mathrm{y}=(\mathrm{p}+\mathrm{q}+\mathrm{r})-(\mathrm{x}+\mathrm{z})=\mathrm{p}+\mathrm{q}-\mathrm{r}\hfill \\ \mathrm{z}=(\mathrm{p}+\mathrm{q}+\mathrm{r})-(\mathrm{x}+\mathrm{y})=\mathrm{q}+\mathrm{r}-\mathrm{p}.\hfill \end{array}$$(b)

$$\mathrm{p}=\frac{1}{2}(\mathrm{w}+\mathrm{x}\mathrm{),}\phantom{\rule{0.33em}{0ex}}\mathrm{q}=\frac{1}{2}(\mathrm{x}+\mathrm{y}\mathrm{),}\phantom{\rule{0.33em}{0ex}}\mathrm{r}=\frac{1}{2}(\mathrm{y}+\mathrm{z}\mathrm{),}\phantom{\rule{0.33em}{0ex}}\mathrm{s}=\frac{1}{2}(\mathrm{z}+\mathrm{w})$$so

$$\begin{array}{c}\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s}=\mathrm{w}+\mathrm{x}+\mathrm{y}+\mathrm{z}.\hfill \end{array}$$Hence

$$\mathrm{w}=(\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s})-(\mathit{x}+\mathrm{y}+\mathrm{z}\mathrm{);}$$but there is no obvious way to pin down (x + y + z).

In fact different quadrilaterals may give rise to the same four “midpoints”. (It is an interesting exercise to identify the family of quadrilaterals corresponding to a given set of four midpoints.)

(c) As in parts (a) and (b),

$$p=\frac{1}{2}(v+w),q=\frac{1}{2}(w+x),r=\frac{1}{2}(x+y),s=\frac{1}{2}(y+z),t=\frac{1}{2}(z+v)$$

Hence

$$p+q+r+s+t=v+w+x+y+z$$

so

$$\begin{array}{l}v=(p+q+r+s+t)-(w+x)-(y+z)=p-q+r-s+t\\ w=(p+q+r+s+t)-(x+y)-(z+v)=p+q-r+s-t\\ x=(p+q+r+s+t)-(v+w)-(y+z)=-p+q+r-s+t\\ y=(p+q+r+s+t)-(w+x)-(z+v)=p-q+r+s-t\\ z=(p+q+r+s+t)-(v+w)-(x+y)=-p+q-r+s+t\end{array}$$.

**97.**

(a)(i) As in Problems **93**-**95** we instinctively add to get

$$2(x+y+z)=6$$

so

$$x+y+z=3.$$

Hence

$$\begin{array}{l}x=3-(y+z)=1\\ y=3-(x+z)=0\\ z=3-(x+y)=2.\end{array}$$

(ii) The same idea (replacing addition by multiplication) leads to

$$2\times 4\times 8=64=uv\cdot vw\cdot wu={(uvw)}^{2}$$

so $uvw=\pm 8$ . Hence

$$\begin{array}{l}u=\frac{uvw}{vw}=\frac{\pm 8}{4}=\pm 2\\ v=\frac{uvw}{uw}=\frac{\pm 8}{8}=\pm 1\\ w=\frac{uvw}{uv}=\frac{\pm 8}{2}=\pm 4.\end{array}$$

$\therefore (u,v,w)=(2,1,4)\text{\hspace{0.17em}}or\text{\hspace{0.17em}}(-2,-1,-4).$

**Note:** Alternatively, we may notice that *u, v, w* are either all positive, or all negative. If we restrict in the first instance to
purely positive solutions, then we may set *u* = 2^{x}, *v* = 2^{y}, *w* = 2^{z}, translate (ii) into (i), and
conclude that (*x, y, z*) = (1, 0, 2), so that (*u, v, w*) = (2, 1, 4). We must then remember the negative solution (*u, v, w*) =
(−2, −1, −4).

(b) (i) As in (a)(i) we add to get 2(*x* + *y* + *z*) = 9, so $x+y+z=\frac{9}{2}$ . Hence

$$\begin{array}{l}x=\frac{9}{2}-(y+z)=\frac{3}{2}\\ y=\frac{9}{2}-(x+z)=\frac{1}{2}\\ z\text{}=\frac{9}{2}-(x+y)=\frac{5}{2}.\end{array}$$

(ii) The same idea leads to

$$6\times 10\times 15=900=uv\cdot vw\cdot wu={(uvw)}^{2}$$,

so $uvw=\pm 30$ .

Hence

$$\begin{array}{l}u=\frac{uvw}{vw}=\frac{\pm 30}{10}=\pm 3\\ v=\frac{uvw}{uw}=\frac{\pm 30}{15}=\pm 2\\ w=\frac{uvw}{uv}=\frac{\pm 30}{6}=\pm 5.\end{array}$$

Either *u*, *v*, *w* are all positive, or all negative.

$\therefore (u,v,w)=(3,2,5)\text{\hspace{0.17em}}or\text{\hspace{0.17em}}(-3,-2,-5).$

(iii) The same idea leads to

$$6\times 10\times 30=2\times 900=vw\cdot wu={(uvw)}^{2},$$

so $uvw=\pm 30\sqrt{2}.$ Hence

$$\begin{array}{l}u=\frac{uvw}{vw}=\frac{\pm 30\sqrt{2}}{10}=\pm 3\sqrt{2}\\ v=\frac{uvw}{uw}=\frac{\pm 30\sqrt{2}}{15}=\pm 2\sqrt{2}\\ w=\frac{uvw}{uv}=\frac{\pm 30\sqrt{2}}{6}=\pm 5\sqrt{2}.\end{array}$$

Either *u, v*, *w* are all positive, or all negative.

$\therefore (u,v,w)=(3\sqrt{2},2\sqrt{2},5\sqrt{2})\text{\hspace{0.17em}}or\text{\hspace{0.17em}}(-3\sqrt{2},-2\sqrt{2},-5\sqrt{2}).$

(iv) We could of course repeat the same method.

Or we could again look in the first instance for positive solutions, notice that 4 = 2^{2}, 8 = 2^{3}, 16 = 2^{4}, and take
logs (to base 2). Then

$$\begin{array}{l}{\mathrm{log}}_{2}u+{\mathrm{log}}_{2}v=2\\ {\mathrm{log}}_{2}v+{\mathrm{log}}_{2}w=3\\ {\mathrm{log}}_{2}u+{\mathrm{log}}_{2}w\text{}=\mathrm{4,}\end{array}$$

so (from part (i)) any positive solution satisfies

$${\mathrm{log}}_{2}u=\frac{3}{2},{\mathrm{log}}_{2}v=\frac{1}{2},{\mathrm{log}}_{2}w=\frac{5}{2},$$

so

$$(u,v,w)=(2\sqrt{2},\sqrt{2},4\sqrt{2}).$$

We must then remember to include

$$(u,v,w)=(-2\sqrt{2},-\sqrt{2},-4\sqrt{2}).$$

**98.** The simplest idea is to take logs, and reduce the system to a familiar linear system:

$$\begin{array}{ll}a\cdot \mathrm{log}u+b\cdot \mathrm{log}v\hfill & =\mathrm{log}m\hfill \\ c\cdot \mathrm{log}u+d\cdot \mathrm{log}v\hfill & =\mathrm{log}n.\hfill \end{array}$$

Multiplying the first equation by c and subtracting it from the second equation multiplied by *a* gives:

$$\mathrm{log}v=\frac{a\cdot \mathrm{log}n-c\cdot \mathrm{log}m}{ad-bc}.$$

Multiplying the first equation by *d* and subtracting *b* times the second equation gives:

$$\mathrm{log}u=\frac{d\cdot \mathrm{log}m-b\cdot \mathrm{log}n}{ad-bc}.$$

If the numerators and denominators are expressed in determinant form, we get the 2 × 2 version of *Cramer’s Rule.* The original
unknowns *u, v* can then be obtained by taking suitable powers.

What emerges looks interesting:

$$\begin{array}{l}u={m}^{\frac{d}{ad-bc}}\cdot {n}^{-\frac{b}{ad-bc}}\\ v\text{}={m}^{-\frac{c}{ad-bc}}\cdot {n}^{\frac{a}{ad-bc}}\end{array}$$

but it is not clear how it generalises.

**99.**

(a) *x* + *y* + *z* = 3 is the equation of a plane through the three points (3, 0, 0), (0, 3, 0), (0,0, 3).

*x*^{2} + *y*^{2} + *z*^{2} = *c* is the equation of a sphere, centered at the origin, with radius
$\sqrt{c}.$ . The sphere misses
the plane completely when *c* < 3, meets the plane in a single point when *c* = 3, and cuts the plane in a circle *C* when
*c* > 3 (the circle lying in the positive octant provided *c* < 9).

If *xy* + *yz* + *zx* = *b* meets this intersection at all, then any permutation of the three coordinates *x, y, z*
produces another point which also satisfies the other two equations (since they are both symmetrical). Hence for the system to have a unique solution,
the circle *C* must contain a point with *x = y = z*. Hence *c* = 3, and *b* = 3, and the unique solution is

$$(x,y,z)=(1,1,1).$$

(b) We must have $c\u2a7e0$ for
any solution. If *c* = 0, then for a unique solution, we must have *x = y = z* = 0, so *a = b* = 0. If we exclude this case, then we
may assume that *c* > 0.

$$x+y+z=a$$

is the equation of a plane through the three points (*a*, 0, 0), (0, *a*, 0), (0, 0, *a*).

$${x}^{2}+{y}^{2}+{z}^{2}=c$$

is the equation of a sphere, centre the origin, with radius $\sqrt{c}$, which misses the 2 2 plane completely when $c<\frac{{a}^{2}}{3}$ , meets the plane in
a single point when $c=\frac{{a}^{2}}{3}$ , and cuts the plane in a circle C when $c>\frac{{a}^{2}}{3}$ (the circle lying in the positive octant provided c < *a*^{2}).

If

$$xy+yz+zx=b$$

meets this intersection at all, then any permutation of the three coordinates *x, y, z* produces another point which also satisfies the other
two equations (since they are both symmetrical). Hence for the system to have a unique solution, the circle *C* must contain a point with *x =
y = z*. Hence that point is $x=y=z=\frac{a}{3}$ , so $c=\frac{{a}^{2}}{3}=b$ , and the unique solution is

$$(x,y,z)=\left(\frac{a}{3},\frac{a}{3},\frac{a}{3}\right).$$

**100.** $|-x|$ is never negative. If $x\u2a7e0$ , then $|-x|=x$ ; if *x* is negative,
then − *x* is positive, so $|-x|=-x$ .

**Note:** We need to learn to see both *x* and − *x* as *algebraic* entities, with *x* as a placeholder (which may
well be negative, in which case − *x* would be positive).

**101.**

(a) The interval [0.1, 0.2). We have marked exactly $\frac{1}{10}$ of the interval [0,1).

(b) This needs a little thought. First we mark the interval [0.1, 0.2), of length $\frac{1}{10}$. Then we mark 9 smaller intervals

$$[0.01,0.02),[0.21,0.22),\dots ,[0.91,0.92)$$

of total length $9\cdot {\left(\frac{1}{10}\right)}^{2}$ . Then 9^{2}
smaller intervals

$$[0.001,0.002),[0.021,0.022),\dots ,[0.991,0.992)$$

of total length ${9}^{2}\cdot {\left(\frac{1}{10}\right)}^{3}$ . And so on.

$$\begin{array}{ccc}\left[0.1\mathrm{,0}.2\right)\hfill & \hfill & \hfill \\ \hfill & \cup \hfill & \left[0.01\mathrm{,0}.02\right)\cup \left[0.21\mathrm{,0}.22\right)\cup \left[0.31\mathrm{,0}.32\right)\cup \left[0.41\mathrm{,0}.42\right)\hfill \\ \hfill & \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\cup \phantom{\rule{0.167em}{0ex}}\left[0.51\mathrm{,0}.52\right)\cup \left[0.61\mathrm{,0}.62\right)\cup \left[0.71\mathrm{,0}.72\right)\hfill \\ \hfill & \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\cup \phantom{\rule{0.167em}{0ex}}\left[0.81\mathrm{,0}.82\right)\cup \left[0.91\mathrm{,0}.92\right)\hfill \\ \hfill & \cup \hfill & \left[0.001\mathrm{,0}.002\right)\cup \left[0.021\mathrm{,0}.022\right)\cup \left[0.031\mathrm{,0}.032\right)\cup \cdots \hfill \\ \hfill & \cup \hfill & \cdots \hfill \end{array}$$It would seem that a vast number of points are left *unm,a,rked* - namely, every point whose decimal representation uses only 0s, 2s, 3s, 4s,
5s, 6s, 7s, 8s, and 9s. However, **the total length** of the *marked* intervals is given by adding:

$$\frac{1}{10}+9\cdot {\left(\frac{1}{10}\right)}^{2}+{9}^{2}\cdot {\left(\frac{1}{10}\right)}^{3}+{9}^{3}\cdot {\left(\frac{1}{10}\right)}^{4}+{9}^{4}\cdot {\left(\frac{1}{10}\right)}^{5}+{9}^{5}\cdot {\left(\frac{1}{10}\right)}^{6}+\cdots $$

That is an infinite geometric series with first term $a=\frac{1}{10}$ and common ratio $r=\frac{9}{10}$, and hence with sum = 1. In other words, the **total length of what remains unmarked is zero.**

(c) (0, 1): every real number except 0 has an expansion in base 2 with a “1” in some position. So this time *nothing is left
unmarked,* (except 0). Hence the complement of the set of marked points consists simply of one point, namely {0}. So it is not surprising that the
total of all the marked intervals has length 1.

(d) First we mark the interval [0.1,0.2), of length $\frac{1}{3}$. Then we mark 2 smaller intervals

$$[0.01,0.02),[0.21,0.22)$$

of total length $2\cdot {\left(\frac{1}{3}\right)}^{2}$ . Then 2^{2} smaller
intervals

$$[0.001,0.002),[0.021,0.022),[0.201,0.202),[0.2211,0.222)$$

of total length ${2}^{2}\cdot {\left(\frac{1}{3}\right)}^{3}$ . And so on.

$$\begin{array}{l}[0.1,0.2)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cup \text{\hspace{1em}}[0.01,0.02)\cup [0.21,0.22)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cup [0.001,0.002)\cup [0.021,0.022))\cup [0.201,0.202)\cup [0.2201,0.02202)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cup \cdots \end{array}$$

The set of marked points is the complement of the famous *Cantor set* (Georg Cantor (1845–1918)) and has total length

$$\frac{1}{3}+2\cdot {\left(\frac{1}{3}\right)}^{2}+{2}^{2}\cdot {\left(\frac{1}{3}\right)}^{3}+{2}^{3}\cdot {\left(\frac{1}{3}\right)}^{4}+{2}^{4}\cdot {\left(\frac{1}{3}\right)}^{5}+{2}^{5}\cdot {\left(\frac{1}{3}\right)}^{6}+\cdots $$

This is an infinite geometric series with first term $a=\frac{1}{3}$ and common ratio $r=\frac{2}{3}$, and so has sum = 1.

Hence, **the total length of what remains unmarked is zero**.

**Note:** The set described in (d) leaves as its complement a collection of points – the *Cantor set* – which consists of the
“endpoints” of the intervals that have been removed; these are points whose base 3 expansion involves only 0s and 2s. This
complement:

(i) is “the same size” as the whole interval [0,1] (since if we interpret the 2s in the base 3 expansion as 1s, we get a
correspondence between the set of “endpoints” and the set of all possible *base* 2 expansions for real numbers in [0,1]);

(ii) is “nowhere dense” (since every pair of points in the complement is separated by some interval)

(iii) has total length = 0.

**102.** (2, 3]. Each inequality implies all the ones before it. Hence the two which are true must be the first two. Hence
$x\u2a7d3$ , and *x* >
2.

**103.** If $x-5\u2a7e0$ , then we must solve $x-5=3$ ; so $x=8$ ; if $x-5<0$ ,
we must solve $x-5=-3$ , so $x=2$ .

**Note:** The fact that $\left|x\right|$ denotes the positive value of the pair {x, − *x*} can be rephrased as: $\left|x\right|$ is equal to the distance from *x* to 0.

In the same way, $|x-5|$ denotes the positive member of the pair

$$\{x-5,-(x-5)\}$$

so $|x-5|$ is equal to the distance from $x-5$ to 0 (i.e. the distance from
*x* to 5). This is a very important way to think about expressions like $|x-5|$.

**104.**

(a) $[-3,-2)\cup (2,3]$ . (Each inequality implies all those that go before it. So we need
solutions to $\left|x\right|>1$ **and** $\left|x\right|>2$ , which satisfy $\left|x\right|\u2a7d3$

(b) (4, 6]. (Each inequality implies the one before it. To see this, think in terms of distances: we want points *x* whose distance from 1 is
> 1, whose distance from 2 is > 2, etc.. So we need to find points *x* which solve the first two inequalities, but not the third.
Points in the half line $(-\infty ,0)$ satisfy all seven inequalities, so we are left
with (4, 6].)

**105.** \left\{-\frac{5}{2},-\frac{1}{2}\right\} . (We need all points *x* for
which

“the distance from

xto −1” plus “the distance fromxto −2”

equals 2. This excludes all points between −2 and −1, for which the sum is equal to 1; for points between $-\frac{5}{2}$ and $-\frac{1}{2}$ the sum is < 2; for points in $\left(-\infty ,-\frac{5}{2}\right)$ or $\left(-\frac{1}{2},\infty \right)$ the sum is > 2.)

**106.** $a=\frac{1}{2}$ , $b=\frac{3}{2}$. (For solutions to exist, we must have *b* > 0. The solutions of the given inequality then consist of
all *x* such that

“the distance from

xto a is less thanb”

that is, all *x* in the interval (*a* − *b*, *a* + *b*). Hence a − *b* = −1, *a + b* =
2.)

**107.**

(a) “The difference between the *x*- and *y*-coordinates is < 3”, means that the point (*x, y*) lies in the
infinite strip between the lines *x* − *y* = −3 and *x* − *y* = 3.

(b) Shifting the origin of coordinates to (−5, 0) changes the *x*-coordinate to “*X = x* + 5” and leaves the
y-coordinate unaffected (so *Y = y*). In this new frame we want “|*X* − *Y*| < 3”, so the required points
lie in the infinite strip between the lines *X* − *Y* = −3 and *X* − *Y* = 3; that is, between the lines
$x-y+5=-3$ and $x-y+5=3$ .

(b) *x* > 0 and *y* > 0, or *x* < 0 and *y* < 0. (For any solution at all, we must have $|x+y|>0$ , which excludes points on the line *x +
y* = 0. Divide both sides by $|x+y|$ and simplify to get

$$\left|1-\frac{2y}{x+y}\right|<1$$

In other words:

$$0<\frac{2y}{x+y}<2$$.

If *y* > 0, then *x + y* > 0, so 2*x* + 2*y* > 2*y*, whence *x* > 0 (so “x
> 0 and *y* > 0”).

If *y* < 0, then *x + y* < 0, so 2*x* + 2*y* < 2*y*, whence *x* < 0 (so “*x*
< 0 and *y* < 0”).

If *x* > 0 and *y* > 0, or *x* < 0 and *y* < 0, then clearly $|x-y|$ < $|x+y|$ .)

**108.** Let

$$x=\frac{a}{b}<\frac{c}{d}=y.$$

(i) Since *x* < *y*, it follows that

$$x-\frac{x}{2}=\frac{x}{2}<\frac{y}{2},$$

so $x<\frac{x+y}{2}$ ; moreover $\frac{x}{2}<y-\frac{y}{2}$ , so $\frac{x+y}{2}<y$ .

(ii) Since $\frac{a}{b}<\frac{c}{d}$ and $b,d>0$, we can multiply both sides by *bd* to get *ad* < *bc.
*Therefore

$$a(b+d)=ab+ad<ba+bc=b(a+c),$$

and

$$(a+c)d=ad+cd<bc+dc=(b+d)c.$$

$\therefore \frac{a}{b}<\frac{a+c}{b+d}$ , and
$\frac{a+c}{b+d}<\frac{c}{d}$ (since *b*, *d*, and
*b* + *d* are all > 0, so we can divide the first inequality by *b(b* + *d)* and the second by *d(b* + *d)*).

**109.**

(a)

$$\frac{0}{1}<\frac{1}{7}<\frac{1}{6}<\frac{1}{5}<\frac{1}{4}<\frac{2}{7}<\frac{1}{3}<\frac{2}{5}<\frac{3}{7}<\frac{1}{2}<\frac{4}{7}<\frac{3}{5}<\frac{2}{3}<\frac{5}{7}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6}<\frac{6}{7}<\frac{1}{1}$$

(b) (i) It is tempting simply to consider the decimals

$$\frac{1}{9}=0.111\dots ,\frac{2}{9}=0.222\dots ,\frac{3}{9}=0.333\dots ,\dots ,\frac{8}{9}=0.888\dots $$

in order to conclude that these fractions miss the first and last subinterval, and then fall one in each of the remaining subintervals. In preparation for part (ii), it is better to observe that

* $\frac{1}{10}<\frac{1}{9}$ and $\frac{8}{9}<\frac{9}{10}$ , so none of the
9^{th}s land up in the first or last subintervals;

* then rewrite

$$\frac{1}{9}=\frac{1}{10}+\frac{1}{90},\frac{2}{9}=\frac{2}{10}+\frac{2}{90},\frac{3}{9}=\frac{3}{10}+\frac{3}{90},\dots ,\frac{8}{9}=\frac{8}{10}+\frac{8}{90}$$

and notice that

$$\frac{0}{10}<\frac{1}{90}<\cdots <\frac{8}{90}<\frac{1}{10},$$

so that, for $1\u2a7dm\u2a7d9$ ,

$$\frac{m}{10}<\frac{m}{9}<\frac{m+1}{10};$$

hence **exactly one** 9^{th} goes in each of the other eight subintervals.

(ii) Notice that

* $\frac{1}{n+1}<\frac{1}{n}$ and $\frac{n-1}{n}<\frac{n}{n+1}$ , so none of the n^{th}s land up in the first or last subintervals;

* then rewrite

$$\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)},\frac{2}{n}=\frac{2}{n+1}+\frac{2}{n(n+1)},\dots ,\frac{n-1}{n}=\frac{n-1}{n+1}+\frac{n-1}{n(n+1)}$$

and notice that

$$\frac{0}{n+1}<\frac{1}{n(n+1)}<\cdots <\frac{n-1}{n(n+1)}<\frac{1}{n+1},$$

so, for $1\u2a7dm\u2a7dn$,

$$\frac{m}{n+1}<\frac{m}{n}<\frac{m+1}{n+1};$$

hence **exactly one** *n*^{th} goes in each of the other *n* − 1 subintervals.

(iii) Suppose two (or more) fractions are inserted between$\frac{a}{b}$ and $\frac{c}{d}$. Then these two fractions would have to be successive multiples of $\frac{1}{n+1}$; but then they would have a multiple of $\frac{1}{n}$ between them (by part (ii)), and this would be a term of the Farey series of order *n* sitting between $\frac{a}{b}$ and $\frac{c}{d}$. Since there is no such
term, at most one fraction can be inserted between $\frac{a}{b}$ and $\frac{c}{d}$.

(c) **Note 1:** This problem was included because the idea of Farey series seems so simple, and their curious properties are so intriguing.
While this remains true, it turns out that Farey series also have something different, and slightly unexpected, to teach us about “the essence
of mathematics”. Part of us expects that simple-looking results should have short and accessible proofs - even though we know that
*Fermat’s Last Theorem* shows otherwise. In the case of Farey series, the relevant properties can be proved in ways that should be
accessible (in principle); but the proofs are not easy. So do not be upset if, after all your efforts, you land up trying to absorb the solution given
here - and the underlying idea that

simple objects and “elementary” proofs can sometimes be more intricate than one anticipates.

**Note 2:** If

$$\frac{a}{b}<\frac{c}{d}$$

are consecutive terms in a Farey series, then *“bc − ad”* must be an integer > 0. The fact that this difference is
always equal to 1 is easily checked in any *particular* case, but it is unclear exactly why this is *necessarily* true (rather than an
accident) - or even how one would go about proving it. Every treatment of Farey series has to find its own way round this difficulty. We give the
simplest proof we can (in the sense that it assumes no more than we have already used: a little about numbers and some algebra). But it is not at all
“easy”. We indicate a different approach in the “**Notes**” at the end of part (d).

(i) [The fact that, except for the two end intervals, we have bd > *n* will be needed in the proof of part (c)(ii).]

We proceed by mathematical induction on *n* (the “order” of the Farey series) - a technique which we have already met in Chapter **2** (Problems **54**-**59**, **76**) and which will be addressed more fully in Chapter **6**.

* When *n* = 1, the Farey series of order *n* is just:

$$\frac{0}{1}<\frac{1}{1}$$

and this subinterval is both the first and the last, so the claim is “vacuously true” (because there is “nothing to
check”). When *n* = 2, the Farey series of order *n* is:

$$\frac{0}{1}<\frac{1}{2}<\frac{1}{1},$$

and again the only subintervals are the first and the last, so again there is nothing to check.

* We now *suppose that we know the claim is true* for the Farey series of order *k*, for some *k* > 1, and show that it must
then also be true for the Farey series of order *k* + 1. (Since we know it is true for *n* = 2, it will then be true for *n* = 3; and
once we know it is true for *n* = 3, it must then be true for *n* = 4; and so on.)

To show that the claim is true for the Farey series of order *k* + 1, we consider any adjacent pair of fractions

$$\frac{a}{b}<\frac{c}{d}$$

(*other than the first pair and the last pair*) in the Farey series of order *k* + 1.

**Claim** *bd* > *k* + 1.

**Proof** Note first that, since we are avoiding the two end subintervals, both *b* and d are > 1.

Suppose first that the pair $\frac{a}{b}<\frac{c}{d}$ are not adjacent in the previous Farey series of order *k*. Then at least
one of the two fractions has been inserted in creating the Farey series of order *k* + 1, and so has denominator = *k* + 1. (The fractions
inserted are precisely those with denominator “*k* + 1” which cannot be reduced by cancelling.) Hence the product

$$bd\u2a7e2(k+1)>k+1.$$

Thus we may assume that the pair a $\frac{a}{b}<\frac{c}{d}$ were already adjacent in the Farey series of order *k*. But then by our
“induction hypothesis” (namely that the desired result is already known to be true for the Farey series of order *k*), we know that
*bd* > *k*. If bd > *k* + 1, then the pair $\frac{a}{b}<\frac{c}{d}$ satisfies the required condition. Hence
we only have to worry about the possibility that *bd* = *k* + 1. Suppose that *bd* = *k* + 1. Then the interval $\frac{a}{b}<\frac{c}{d}$ has length

$$\frac{bc-ad}{bd}=\frac{r}{k+1}$$

for some positive integer *r* = *bc* − *ad*.

If *r* > 1, then the interval would have length > $\frac{1}{k+1}$ , so $\frac{a}{b}<\frac{c}{d}$ would
**not** be successive terms in the series (for we would have inserted some additional term when moving from the Farey series of order k to the
Farey series of order *k* + 1).

Hence we can be sure that *r* = 1, that the subinterval $\frac{a}{b}<\frac{c}{d}$ has length exactly $\frac{1}{k+1}$ . Now successive fractions with denominator *k* + 1 differ by exactly $\frac{1}{k+1}$, so some fraction with denominator *k* + 1 must
lie in this subinterval. Since no additional fraction is inserted between them in passing from the series of order *k* to the series of order
*k* + 1, and $\frac{a}{b}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\frac{c}{d}$ must both be
“cancelled versions” of two successive fractions with denominator *k* + 1. But then, by part (b)(ii), there would have to be a
fraction with denominator *k* in the interval $\frac{a}{b}<\frac{c}{d}$ - which is not the case.

Therefore the possibility *bd = k* + 1 does not in fact occur.

So we can be sure that in every case, *bd > k* + 1.

Hence whenever the result is true for the Farey series of order *k,* it must then also be true for the Farey series of order *k* + 1.

It follows that the result is true for the Farey series of order *n*, for all $n\u2a7e1$ .

(ii) We proceed by induction on *n.*

* If $\frac{a}{b}<\frac{c}{d}$ are adjacent fractions in the Farey series of order 1, then $\frac{a}{b}=\frac{0}{1}$ and $\frac{c}{d}=\frac{1}{1}$ , so *bc − ad =* 1.

* Now suppose that, for some $k\u2a7e1$ , we already know that the result holds for any adjacent pair in the Farey series of order *k.*

Let $\frac{a}{b}<\frac{c}{d}$ be adjacent fractions in the Farey series of order *k* + 1.

If $\frac{a}{b}<\frac{c}{d}$ were already adjacent fractions in the Farey series of order *k* (i.e. if no fraction has been
inserted between $\frac{a}{b}$ and
$\frac{c}{d}$ in passing from the
series of order *k* to the series of order *k* + 1), then we already know (by the induction hypothesis) that *bc − ad =* 1.

Thus we may concentrate on the case where $\frac{a}{b}<\frac{c}{d}$ are not adjacent fractions in the Farey series of order *k.* By
(b)(iii), at most one fraction with denominator *k* + 1 is inserted between any two adjacent fractions in the Farey series of order *k*, so
we have either

$$\frac{a}{b}<\frac{c}{d}<\frac{e}{f},$$

with $\frac{a}{b}<\frac{e}{f},$ being adjacent fractions in the Farey series of order *k* (so *be − af* =
1), or

$$\frac{e}{f}<\frac{a}{b}<\frac{c}{d},$$

with $\frac{e}{f}<\frac{c}{d}$ being adjacent fractions in the Farey series of order *k* (so *fc − ed =* 1). We
consider the first of these possibilities (the second is entirely similar).

Suppose

$$\frac{a}{b}<\frac{c}{d}<\frac{e}{f},$$

with $\frac{a}{b}<\frac{e}{f}$ being adjacent fractions in the Farey series of order *k.* By part (i) we know that $bf\u2a7ek+1$
; and by induction we know that *be − af =* 1. Hence the interval $\frac{a}{b}<\frac{e}{f}$ has length at most $\frac{1}{k+1}.$ We have to prove that *bc* − *ad* = 1.

Let *bc − ad = r >* 0, and *ed − fc = s >* 0.

Then *sa* + *re = c,* and *sb* + *rf* = *d*.

In particular, *HCF(r,s) =* 1 (since *HCF(c,d) =* 1).

Hence $\frac{c}{d}$ belongs to the family

$\begin{array}{cc}\phantom{\rule{6.0em}{0ex}}& S=\{\genfrac{}{}{0.1ex}{}{xa+ye}{xb+yf}:\text{where}x,y\text{are any positive integers with}HCF(x,y)=1\}.\hfill \end{array}$Since everything is positive, easy algebra shows that

$$\frac{a}{b}<\frac{xa+ye}{xb+yf}<\frac{e}{f},$$

so every element of *S* lies between $\frac{a}{b}$ and $\frac{e}{f}$.

As long as we choose *x, y* such that *HCF*_{(x,y)} = 1, any common factor of *xa + ye* and *xb + yf* would also
divide both

$$b(xa+ye)-a(xb+yf)=(be-af)y=y,$$

and

$$e(xb+yf)-f(xa+ye)=(be-af)x=x.$$

Hence

$$HCF(xa+ye,xb+yf)=1$$

so each element of *S* is in lowest terms (i.e. no further cancelling is possible).

We have shown that “$\frac{c}{d}$ belongs to the family S”, and that *all* elements of *S *fit *between* $\frac{a}{b}$ and $\frac{e}{f}$; which are *adjacent* fractions in the Farey series of order *k*. So none of
the elements of *S* can have arisen before the series of order *k* + 1. But each fraction in *S* arises at some stage in a Farey
series.

And the first to *arise* (because it has the smallest denominator) is "$\frac{a+e}{b+f}$"

Hence

$$\frac{c}{d}=\frac{a+e}{b+f},$$

so *r* = *s* = 1, and *bc* − *ad* = 1 as
required. QED

(d) Let

$$\frac{a}{b}<\frac{c}{d}<\frac{e}{f}$$

be three successive terms in any Farey sequence. By (c) we know that *bc−ad* = 1, and that *de* − *cf* = 1. In
particular, *bc* - *ad* = *de* - *cf*, so

$$\frac{c}{d}=\frac{a+e}{b+f}.$$

**Note 1:** It may not be clear why we are proving this result “again” - since it appeared in the final line of the solution to
part (c). However, in part (c) the statement that

$$\frac{c}{d}=\frac{a+e}{b+f}$$

was arrived at *within the induction step,* and so was *subject to other assumptions. *In contrast, now that the result in part (c) has
been clearly established, we can use it to prove part (d) without any hidden assumptions.

**Note 2:** If we represent each fraction $\frac{a}{b}$ in the Farey series of order *n* by the point *(b,* a), then each point lies in the right angled
triangle joining (0, 0), (*n,* 0), and (*n,n*), and each fraction in the Farey series is equal to the gradient of the line, or vector,
joining the origin to the integer lattice point (*b,a*). The ordering of the fractions in the Farey series corresponds to the sequence of
increasing gradients, from $\frac{0}{1}$ up to $\frac{1}{1}$. If
$\frac{a}{b}$ and $\frac{e}{f}$ are adjacent fractions in
some Farey series, then the result in (d) says that the next fraction to be inserted between them is $\frac{a+e}{b+f}$
corresponding to the vector sum of (*b, a*) and (*f, e*). And the result in (c) says that the area of the parallelogram with vertices (0,
0), (*b, a*), (*f, e*), (*b + f, a + e*) is equal to 1 (see Problem **57**(b)). Hence the result in (c) reduces
to the fact that

TheoremAny parallelogram, whose vertices are integer lattice points (i.e. points (b,a) where both coordinates are integers), and with no additional lattice points inside the parallelogram or on the four sides, has area 1.

**110.**

(a) Suppose that *x* satisfies $x+\frac{1}{x}<2$ . Then $x\ne 0$ (or $\frac{1}{x}$ is not defined).

$\therefore \frac{{x}^{2}+1}{x}<2.$

If *x* > 0, then *x*^{2} − 2*x* + 1 = (*x* − 1)^{2} which has no solutions.

$\therefore x<0$ in
which case *x* satisfies $x+\frac{1}{x}<0<2$ , so every *x* < 0 is a solution of the original
inequality

(b) Suppose *x* satisfies $x\u2a7d1+\frac{2}{x}$ . Again $x\ne 0$ (or $\frac{1}{x}$ is not defined).

(i) If *x* > 0, then *x*^{2} − *x* − 2 = (*x* − 2)(*x* + 1) < 0.

$\therefore -1\u2a7dx\u2a7d2$ (and *x* > 0); hence
$0<x\u2a7d2$ , and all such *x* satisfy the original inequality.

(ii) If *x* < 0, then ${x}^{2}-x-2\u2a7e0$ , so $(x-2)(x+1)\u2a7e0$ .

$\therefore \text{\hspace{0.17em}}either\text{\hspace{0.17em}}x\u2a7d-1$ , or $x\u2a7e2$ (and
*x* < 0); hence $x\u2a7d-1$ , and all such *x* satisfy the original inequality.

(c) Suppose *x* satisfies $\sqrt{x}<x+\frac{1}{4}$ .

$\therefore 4{(\sqrt{x})}^{2}-4\sqrt{x}+1>0$

$\therefore {(2\sqrt{x}-1)}^{2}>0$ ,so $x\ne \frac{1}{4}$ and all such *x* satisfy the original inequality.

**111.**

(a) If *a + b* = 5 and *ab* = 7, then *a, b* are solutions of

$$(x-a)(x-b)={x}^{2}-5x+7=0.$$

But the roots of this quadratic equation are

$$\frac{5\pm \sqrt{25-28}}{2}=\frac{5\pm \sqrt{-3}}{2},$$

so *a* and *b* cannot be “positive reals”.

(b) We abbreviate the “arithmetic mean” by AM, the “geometric mean” by GM, the “harmonic mean” by HM, and the “quadratic mean” by QM.

$${(\sqrt{a}-\sqrt{b})}^{2}\u2a7e0$$

so

$$a+b\u2a7e2\sqrt{ab}$$

therefore

$$\sqrt{ab}\u2a7e\frac{2ab}{a+b}\text{\hspace{1em}}(\text{GM}\u2a7e\text{HM})$$

and

$$\sqrt{ab}\u2a7d\frac{a+b}{2}\text{\hspace{1em}}(\text{GM}\u2a7d\text{AM}).$$

Also

$${\left(\frac{a-b}{2}\right)}^{2}\u2a7e0,$$

so

$$\frac{{a}^{2}+{b}^{2}}{4}\u2a7e\frac{2ab}{4}$$

whence

$$\frac{{a}^{2}+{b}^{2}}{2}\u2a7e\frac{{a}^{2}+{b}^{2}+2ab}{4}={\left(\frac{a+b}{2}\right)}^{2}.$$

Therefore

$$\sqrt{\frac{{a}^{2}+{b}^{2}}{2}}\u2a7e\frac{a+b}{2}\text{\hspace{1em}}(\text{QM}\u2a7e\text{AM}).$$ QED

**112.** [This delightful problem was devised by Oleksiy Yevdokimov.] We need to find something which remains constant, or
which does not increase, when we replace two terms *a*, *b* by $\frac{a+b}{\sqrt{2}}$ .

**Idea:** If the two terms a, b were replaced each time by their sum a + b, then the sum of all the numbers in the list would be unchanged, so
we could be sure that the final number after 199 such moves would have to be

$$1+2+3+\cdots +200=\frac{200\times 201}{2}.$$

**This doesn’t work here**. However, in the spirit of this section on inequalities, one may ask:

What happens to the sum of the squares of the terms in the list aftereach move?

When we move from one list to the next, only two terms are affected, and for these two terms, the previous sum of squares is replaced by ${\left(\frac{a+b}{\sqrt{2}}\right)}^{2}$. How does this affect the sum of all squares on the list?

We know that ${a}^{2}+{b}^{2}\u2a7e2ab$ for all a, b. And it is easy to see that this is equivalent to:

$${a}^{2}+{b}^{2}\u2a7e{\left(\frac{a+b}{\sqrt{2}}\right)}^{2}$$

So when we replace two terms *a, b* by $\frac{a+b}{\sqrt{2}}$ , the sum of the squares of all the terms in
the list *never increases.* Hence the final term is less than or equal to the square

root of the **initial** sum of squares

$$\begin{array}{ccc}{1}^{2}+{2}^{2}+{3}^{2}+\cdots +20{0}^{2}\hfill & =\hfill & \genfrac{}{}{0.1ex}{}{200\times 201\times 401}{6}\phantom{\rule{2.00em}{0ex}}\left(\text{by Problem 62}\right)\hfill \\ \hfill & <\hfill & \genfrac{}{}{0.1ex}{}{200\times 300\times 400}{6}\hfill \\ \hfill & =\hfill & 4\times 1{0}^{6}.\hfill \end{array}$$

$\therefore \text{\hspace{0.17em}}thefinaltermis\text{\hspace{0.17em}}<\sqrt{4\times {10}^{6}}=2000.$

**113.**

(a) (i) 3 = 2^{2} - 1.

(ii) It seems hard to find another.

(b) (i) 2 = 1^{2} + 1.

(ii) 5 = 2^{2} + 1 (or 17 = 4^{2} + 1; or 37 = 6^{2} + 1; or 101 = 10^{2} + 1; or …). In other words, there
seem to be lots.

**Note:** At first sight primes of this form “keep on coming”. Given that we now know (see Problem **76**)
that the list of all prime numbers “goes on for ever”, it is natural to ask: Are there infinitely many prime numbers “one more
than a square”? Or does the list run out?

This is one of the simplest questions one can ask to which the answer is **not yet known!**

(c) (i) 7 = 2^{3} - 1.

(ii) It seems hard to find another.

(d) (i) 2 = 1^{3} + 1.

(ii) It seems hard to find another.

**Note:** Parts (a), (c) and (d) should make one suspicious - provided one notices that:

(a) 63 = 7 × 9, 143 = 11 × 13, 323 = 17 × 19;

(c) 511 = 7 × 73, 1727 = 11 × 157;

(d) 217 = 7 × 31, 513 = 9 × 57, 1001 = 7 × 143.

This problem is so instructive that its solution is discussed in the main text following Problem **115**.

**114.**

$${x}^{4}+1=\left({x}^{2}+\sqrt{2}\cdot x+1\right)\left({x}^{2}-\sqrt{2}\cdot x+1\right).$$

(Suppose

$${x}^{4}+1=\left({x}^{2}+ax+b\right)\left({x}^{2}+cx+d\right).$$

It is natural to try *b* = *d* = 1 in order to make the constant term *bd* = 1, and then to try c = −a (so that the
coefficients of *x*^{3} and of *x* are both 0). It then remains to choose the value of *a* so that the total coefficient
“2 − *a*^{2}” of all terms in *x*^{2} is equal to 0: that is, $a=\sqrt{2}$ ).

**115.**

(a)(i)

$${a}^{3}-{b}^{3}=(a-b)\left({a}^{2}+ab+{b}^{2}\right).$$

(ii)

$$\begin{array}{ccc}{a}^{4}-{b}^{4}\hfill & =\hfill & (a-b)({a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3})\hfill \\ \hfill & =\hfill & ({a}^{2}-{b}^{2})({a}^{2}+{b}^{2})\hfill \\ \hfill & =\hfill & (a-b)(a+b)({a}^{2}+{b}^{2}).\hfill \end{array}$$

(iii)

$${a}^{n}-{b}^{n}=(a-b)\left({a}^{n-1}+{a}^{n-2}b+{a}^{n-3}{b}^{2}+\cdots +a{b}^{n-2}+{b}^{n-1}\right).$$

**Note:** The general factorisation

$${x}^{n}-1=(x-1)\left({x}^{n-1}+{x}^{n-2}+\cdots +{x}^{2}+x+1\right)$$

provides a fresh slant on the test for divisibility by 9 in base 10, or in general for divisibility by b − 1 in base b (see Problem **51**):

“an integer written in base b is divisible by b − 1 precisely when its digit sum is divisible by b − 1”.

(b) (i)

$${a}^{3}+{b}^{3}=(a+b)\left({a}^{2}-ab+{b}^{2}\right).$$

(ii)

$${a}^{5}+{b}^{5}=(a+b)\left({a}^{4}-{a}^{3}b+{a}^{2}{b}^{2}-a{b}^{3}+{b}^{4}\right).$$

(iii)

$${a}^{2n+1}+{b}^{2n+1}=(a+b)\left({a}^{2n}-{a}^{2n-1}b+{a}^{2n-2}{b}^{2}-{a}^{2n-3}{b}^{3}+\cdots -a{b}^{2n-1}+{b}^{2n}\right).$$

**116.**

(a) Replace a by 1, b by r, and *n* by *n* + 1 in the answer to **115**(a)(iii), to see that:

$$1+r+{r}^{2}+{r}^{3}+\cdots +{r}^{n}=\frac{1-{r}^{n+1}}{1-r}.$$

(b) Multiply the closed formula in (a) by “a” to see that:

$$a+ar+a{r}^{2}+a{r}^{3}+\cdots +a{r}^{n}=a\cdot \frac{1-{r}^{n+1}}{1-r}.$$

**117.** When *x* = 40,

$$f(x)={x}^{2}+(x+40)+1={40}^{2}+2\times 40+1={41}^{2}$$

is not prime. So the sequence of prime outputs must stop some time before *f* (40). But it in fact keeps going as long as it possibly could,
so that

$$f(0),f(1),f(2),\dots ,f(39)$$

are all prime. (This may explain Euler’s delight.)

**Note:** The links between polynomials with integer coefficients (even lowly quadratics) and prime numbers are still not fully understood. For
example, you might like to look up *Ulam’s spiral.* (Ulam (1909–1984) plotted the positive integers in a square spiral and found
the primes arranging themselves in curious patterns that we still do not fully understand.)

Interest in the connections between polynomials and primes was revived in the second half of the 20^{th} century. It was eventually proved
that there exists a polynomial in 10 variables, with *integer* coefficients, which takes both positive and negative values when the variables run
through all possible non-negative *integer *values, but which does so in such a way that it generates all the primes as the set of positive
outputs.

**118.**

(a) (i) For

$${a}^{n}-1=(a-1)\left({a}^{n-1}+{a}^{n-2}+\cdots +a+1\right)$$

to be prime, the smaller factor must be = 1, so *a* = 2.

If *n* is not prime, we can factorise *n* = *rs,* with *r, s* > 1. Then

$${2}^{n}-1={2}^{rs}-1={\left({2}^{r}\right)}^{s}-1=\left({2}^{r}-1\right)\left({2}^{r(s1)}+{2}^{r(s2)}+\cdots +2+1\right);$$

Hence 2* ^{n}* 1 also factorises, so could not be prime. Hence

*n*must be prime.

(ii) 2^{2} − 1 = 3, 2^{3} − 1 = 7, 2^{5} − 1 = 31, 2^{7} − 1 = 127 are all prime;
2^{11} − 1 = 2047 = 23 × 89 is not.

**Note:** This is a simple example of the need to distinguish carefully between the statement

“if 2

− 1 is prime, then^{n}nis prime” (which is true),

and its converse

“if

nis prime, then 2− 1 is prime” (which is false).^{n}

(b)(i) Suppose that a > 1. Then *a*^{n} + 1 > 2; so for *a*^{n} + 1 to be prime, it must be odd, so
*a* must be even.

If *n* has an odd factor *m* > 1, we can write *n* = *km.* Then

$$\begin{array}{cccc}{a}^{n}+1\hfill & =\hfill & \phantom{\rule{6em}{0ex}}& {a}^{km}+1\hfill \\ \hfill & =\hfill & {\left({a}^{k}\right)}^{m}+1\hfill \\ \hfill & =\hfill & ({a}^{k}+1)({a}^{k(m-1)}-{a}^{k(m-2)}+\cdots -{a}^{k}+1).\hfill \end{array}$$

Since *m* is odd and > 1, we have $m\u2a7e3$ . It is then easy to show that

$${a}^{k}+1\u2a7d{a}^{k(m-1)}-{a}^{k(m-2)}+\cdots -{a}^{k}+1.$$

And since *a* > 1, neither factor = 1, so *a ^{n}* + 1 can never be prime. Hence

*n*can have no odd factor > 1, which is the same as saying that

*n*= 2

^{r}must be a power of 2.

(ii) 2^{1} + 1 = **3**, 2^{2} + 1 = **5**, 2^{4} + 1 = **17**, 2^{8} + 1 = **257**, 2^{16} + 1
= **65 537** are all prime. (The very next such expression

$${2}^{32}+1=4294967297=641\times 6700417$$

is not prime.)

**Note:** The sad tale of Fermat’s claim that “all Fermat numbers are prime” shows that mathematicians are not exempt from
the obligation to distinguish carefully between a statement and its converse!

**119.**

(a) *x*^{2} − 3*x* + 2 = (*x* − 2)(*x* − 1) = 0 precisely when one of the brackets = 0; that is,
*x* = 2, or *x* = 1.

(b) *x*^{2} − 1 = (*x* − 1)(*x* + 1) = 0 precisely when *x* = 1 or *x* = −1.

(c) *x*^{2} − 2*x* + 1 = (*x* − 1)^{2} = 0 precisely when *x* = 1 (a repeated root).

(d) ${x}^{2}+\sqrt{2}x-1=0$ requires us

− to complete the square

$${x}^{2}+\sqrt{2}x-1={\left(x+\frac{\sqrt{2}}{2}\right)}^{2}-1-\frac{1}{2}.$$

so

$$x+\frac{\sqrt{2}}{2}=\pm \sqrt{\frac{3}{2}},$$

− or to use the quadratic formula:

$$x=\frac{-\sqrt{2}\pm \sqrt{2+4}}{2}.$$

(e) ${x}^{2}+x-\sqrt{2}=0$ requires us

− to complete the square

$${x}^{2}+x-\sqrt{2}={\left(x+\frac{1}{2}\right)}^{2}-\sqrt{2}-\frac{1}{4},$$

so

$$x+\frac{1}{2}=\pm \sqrt{\sqrt{2}+\frac{1}{4}}$$

− or to use the quadratic formula:

$$x=\frac{-1\pm \sqrt{1+4\sqrt{2}}}{2}.$$

(f) *x*^{2} + 1 = 0 yields $x=\pm \sqrt{-1}.$

(g) ${x}^{2}+\sqrt{2}x+1=0$ yields

$$x=\frac{-\sqrt{2}\pm \sqrt{2-4}}{2}=\frac{-\sqrt{2}\pm \sqrt{-2}}{2}.$$

**120.** $q(x)={x}^{2}-\sqrt{2}\cdot x+1$ . (There is no obvious magic method here. However, it should be natural to try to
insert a term $\sqrt{2}$ in
*q*(*x*) to “resolve” the term $\sqrt{2}$ in *p*(*x*); and the familiar cancelling of cross terms in (*a* + *b*)(*a* − *b*) should
then suggest the possible benefit of trying $q(x)={x}^{2}-\sqrt{2}\cdot x+1.$ )

**Note:** *p*(*x*)*q*(*x*) = *x*^{4} + 1 (see Problem **114**).

**121.**

(a) Let the two unknown numbers be *α* and *β* Then $s=\alpha +\beta $ , and $p=\alpha \beta $. “The square of half the sum" ${\left(\frac{s}{2}\right)}^{2}={\left(\frac{\alpha +\beta}{2}\right)}^{2}$ .

Subtracting $p=\alpha \beta $ produces ${\left(\frac{\alpha -\beta}{2}\right)}^{2}$ whose “square root” will be either $\frac{\alpha -\beta}{2}$ , or $-\left(\frac{\alpha -\beta}{2}\right)$ − whichever is positive

Adding this to “half the sum” gives one root; subtracting gives the other root.

(b) Let the length of one side be *x*. We are told that *x*^{2} + *bx* = *c*.

“Take half of

b, square it, and add the result toc”

translates as:

“Rewrite the equation as: ${\left(x+\frac{b}{2}\right)}^{2}=c+{\left(\frac{b}{2}\right)}^{2}$ .”

That is, we have “completed the square” ${\left(x+\frac{b}{2}\right)}^{2}$ . If we now take the (positive) square root and subtract $\frac{b}{2}$ , we get a single value for x, which determines the side length of my square as required.

If the same method is applied to the general quadratic equation

$$a{x}^{2}+bx+c=0.$$

with the extra initial step of “multiply through by $\frac{1}{a}$ ”, we produce first

$${x}^{2}+\frac{b}{a}x+\frac{c}{a}=0,$$

then

$${\left(x+\frac{b}{2a}\right)}^{2}+\left(\frac{c}{a}-{\left(\frac{b}{2a}\right)}^{2}\right)=0,$$

then

$$x+\frac{b}{2a}=\pm \sqrt{{\left(\frac{b}{2a}\right)}^{2}-\frac{c}{a}}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a},$$

which leads to the familiar quadratic formula.

(c) See Problem **3**(c)(iv). *AD* : *CB = DX* : *BX*, so *x* : 1 = 1 : (*x* − 1). Hence
*x*^{2} − *x* − 1 = 0. If we use the quadratic formula derived in the answer to part (b) above, and realise that
*x* > 1, then we obtain $x=\frac{1+\sqrt{5}}{2}$.

**Note:** The procedure given in (a) dates back to the ancient Babylonians (~ 1700 BC) and later to the ancient Greeks (~ 300 BC). Both cultures
worked *without *algebra. The Babylonians gave their verbal procedures as recipes in words, in the context of particular examples. The Greeks
expressed everything geometrically. In modern language, if we denote the unknown numbers by α and β, then

$$(x-\alpha )(x-\beta )={x}^{2}-(\alpha +\beta )x+\alpha \beta .$$

Being told the sum and product is therefore the same as being given the coefficients of a quadratic equation, and being asked to find the two roots.

Our method for factorizing a quadratic involves a mental process of ‘inverse arithmetic’, where we juggle possibilities in search of
*α* and *β*, when all we know are the coefficients (that is, the sum *α* + *β*, and the product
*αβ*).

The procedure in (b) also dates back to the ancient Babylonians, and is essentially our process of completing the square. It was given as a
procedure, without our algebraic notation. The Babylonians seem not to have been hampered (as the Greeks were) by the fact that it makes no sense to
add a *length* and an *area*! They worded things geometrically, but seem to have understood that they were really playing *numerical*
games (an idea which European mathematicians found elusive right up to the time of Descartes (1590–1656)).

Similarly, the modern use of symbols - allowing one to represent either positive or negative quantities - was widely resisted right into the
nineteenth century. What we would write as a single family of quadratic equations, *ax ^{2}* +

*bx*+ c = 0, had to be split into separate cases where two

*positive*quantities were equated. For example, the groundbreaking book

*Ars Magna*in which Cardano (1501–1576) explained how to solve cubic and quartic equations begins with quadratics - where his procedure distinguishes four different cases: “squares equal to numbers”, “squares equal to things”, “squares and things equal to numbers”, “squares and numbers equal to things”.

**122.**

(a)(i) ${\alpha}^{2}+{\beta}^{2}={(\alpha +\beta )}^{2}-2\alpha \beta ={b}^{2}-2c$ .

(ii) ${\alpha}^{2}\beta +{\beta}^{2}\alpha =\alpha \beta (\alpha +\beta )=c\xb7(-b)=-bc$ .

(iii) We rearrange

$$\begin{array}{ccc}{\alpha}^{3}+{\beta}^{3}-3\alpha \beta \hfill & =\hfill & (\alpha +\beta )({\alpha}^{2}-\alpha \beta +{\beta}^{2})-3\alpha \beta \hfill \\ \hfill & =\hfill & (-b)\cdot ({b}^{2}-3c)-3c\hfill \\ \hfill & =\hfill & -{b}^{3}+3bc-3c.\hfill \end{array}$$

[Alternatively: ${\alpha}^{3}+{\beta}^{3}={(\alpha +\beta )}^{3}-3\alpha \beta (\alpha +\beta )$ , etc.]

(b) (i) [Cf **121**(a).] ${(\alpha -\beta )}^{2}={(\alpha +\beta )}^{2}-4\alpha \beta $.

Therefore

$$\alpha -\beta =\sqrt{{b}^{2}-4c}\text{\hspace{0.17em}}if\text{\hspace{0.17em}}\alpha \u2a7e\beta ,$$

and

$$\alpha -\beta =-\sqrt{{b}^{2}-4c}\text{\hspace{0.17em}}if\text{\hspace{0.17em}}\alpha <\beta .$$

(ii)

$${\alpha}^{2}\beta -{\beta}^{2}\alpha =-\alpha \beta (\alpha -\beta )=-c\sqrt{{b}^{2}-4c}\text{\hspace{0.17em}}if\text{\hspace{0.17em}}\alpha \u2a7e\beta ,$$

and

$$\alpha -\beta =-\sqrt{{b}^{2}-4c}\text{\hspace{0.17em}}if\text{\hspace{0.17em}}\alpha <\beta .$$

(iii) ${\alpha}^{3}-{\beta}^{3}=(\alpha -\beta )\left({\alpha}^{2}+\alpha \beta +{\beta}^{2}\right)$.

Therefore

$${\alpha}^{3}-{\beta}^{3}=\left[\sqrt{{b}^{2}-4c}\right]\left({b}^{2}-c\right)\text{\hspace{1em}}\text{\hspace{0.17em}}if\text{\hspace{0.17em}}\text{\hspace{1em}}\alpha \u2a7e\beta ,$$

and

$${\alpha}^{3}-{\beta}^{3}=\left[-\sqrt{{b}^{2}-4c}\right]\left({b}^{2}-c\right)\text{\hspace{1em}}\text{\hspace{0.17em}}if\text{\hspace{0.17em}}\alpha <\beta .$$

**123.**

(a)(i) $\sqrt{a}+\sqrt{b}$ and $\sqrt{a+b+\sqrt{4ab}}$ are both positive. And it is easy to check that they have the same square:

$${(\sqrt{a}+\sqrt{b})}^{2}=a+b+2\sqrt{ab},$$

and

$${(\sqrt{a+b+\sqrt{4ab}})}^{2}=a+b+\sqrt{4ab}.$$

Hence

$$\sqrt{a}+\sqrt{b}=\sqrt{a+b+\sqrt{4ab}}.$$

(ii) 5 = 2 + 3, and 24 = 4 × 2 × 3;

Therefore

$$\sqrt{2+3+\sqrt{4\times 2\times 3}}=\sqrt{2}+\sqrt{3}$$

(which is easy to check).

(b) (i) **Claim** If $a\u2a7eb(\ne 0)$ , then

$$\sqrt{a}-\sqrt{b}=\sqrt{a+b-\sqrt{4ab}}.$$

**Proof** $\sqrt{a}-\sqrt{b}$ and $\sqrt{a+b-\sqrt{4ab}}$ are
$both\text{\hspace{0.17em}}\u2a7e\text{\hspace{0.17em}}0$ (Why?). And it is
easy to check that

$${(\sqrt{a}-\sqrt{b})}^{2}=a+b-2\sqrt{ab},$$

and

$${(\sqrt{a+b-\sqrt{4ab}})}^{2}=a+b-\sqrt{4ab}.$$ QED

(ii) Simplify $\sqrt{5-\sqrt{16}}$ and $\sqrt{6-\sqrt{20}}$ .

5 = 4 + 1 and 16 = 4 × 4 × 1, so $\sqrt{5-\sqrt{16}}=\sqrt{4}-\sqrt{1}=1$ .

Actually, there is a simpler solution

$$\sqrt{5-\sqrt{16}}=\sqrt{5-4}=\sqrt{1}=1.$$

6 = 5 + 1 and 20 = 4 × 5 × 1, so $\sqrt{6-\sqrt{20}}=\sqrt{5}-\sqrt{1}=\sqrt{5}-1$ .

**124.**

(a) Let $\alpha =1+\sqrt{2}.$ . Then ${\alpha}^{2}=3+2\sqrt{2}$ . Hence ${\alpha}^{2}-2\alpha =1$ , so α satisfies the quadratic polynomial equation
*x*^{2} − 2x − 1 = 0.

**Note:** Observe that the resulting polynomial is equal to

$$(x-(1+\sqrt{2}))(x-(1-\sqrt{2})).$$

In other words, to rationalize the coefficients, we need a polynomial which has both $\alpha =1+\sqrt{2}$ and its “conjugate” $1-\sqrt{2}$ as roots.

(b) Let $\alpha =1+\sqrt{3}$ . Then ${\alpha}^{2}=4+2\sqrt{3}$ . Hence ${\alpha}^{2}-2\alpha =2$ , so α satisfies the quadratic polynomial equation ${x}^{2}-2x-2=0$ .

**Note:** Observe that the resulting polynomial is equal to

$$(x-(1+\sqrt{3}))(x-(1-\sqrt{3})).$$

In other words, to rationalize the coefficients, we need a polynomial which has both $\alpha =1+\sqrt{3}$ and its “conjugate” $1-\sqrt{3}$ as roots.

(c) Let $\alpha =\sqrt{2}+\sqrt{3}$ . Then ${\alpha}^{2}=5+2\sqrt{6}$ , so ${\alpha}^{2}-5=2\sqrt{6}$ , and ${\left({\alpha}^{2}-5\right)}^{2}=24$ . Hence α satisfies the quartic polynomial equation ${x}^{4}-10{x}^{2}+1=0$ .

**Note:** Observe that the resulting polynomial is equal to

$$(x-(\sqrt{2}+\sqrt{3}))(x-(\sqrt{2}-\sqrt{3}))(x-(-\sqrt{2}+\sqrt{3}))(x-(-\sqrt{2}-\sqrt{3})).$$

In other words, the roots are: $\sqrt{2}+\sqrt{3}$ (as required), and also $\sqrt{2}-\sqrt{3},-\sqrt{2}-\sqrt{3}$ , and $-\sqrt{2}+\sqrt{3}$.

(d) Let $\alpha =\sqrt{2}+\frac{1}{\sqrt{3}}$ . Then

$${\alpha}^{2}=\frac{7}{3}+2\sqrt{\frac{2}{3}},$$

so

$${\left({\alpha}^{2}-\frac{7}{3}\right)}^{2}=\frac{8}{3},$$

and α satisfies the quartic polynomial equation

$${x}^{4}-\frac{14}{3}\xb7{x}^{2}+\frac{25}{9}=0.$$

**Note:**

$$\begin{array}{l}{x}^{4}-\frac{14}{3}\xb7{x}^{2}+\frac{25}{9}=\left(x-\left[\sqrt{2}+\frac{1}{\sqrt{3}}\right]\right)\left(x-\left[\sqrt{2}-\frac{1}{\sqrt{3}}\right]\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\xb7\left(x+\left[\sqrt{2}+\frac{1}{\sqrt{3}}\right]\right)\left(x+\left[\sqrt{2}-\frac{1}{\sqrt{3}}\right]\right),\end{array}$$

so the roots are:

$$x=\sqrt{2}+\frac{1}{\sqrt{3}},\sqrt{2}-\frac{1}{\sqrt{3}},-\sqrt{2}-\frac{1}{\sqrt{3}},-\sqrt{2}+\frac{1}{\sqrt{3}}.$$

**125.** A direct approach can be made to work in both cases (but see the **Notes**).

(a) Suppose to the contrary that $\sqrt{2}+\sqrt{3}=\frac{p}{q}$ , for some integers *p, q* with
*HCF*(p, q) = 1. Then $(5+2\sqrt{6}){q}^{2}={p}^{2}$ , so $\sqrt{6}$ is rational, and we can write $\sqrt{6}=\frac{r}{s}$ with HCF(*r,s*) = 1. But then 6s^{2} = r^{2} ; hence r
= *2t* must be even; so 3s^{2} = 2t^{2}, but then s must be even - contradicting *HCF(r,s*) = 1. Hence $\sqrt{2}+\sqrt{3}$
cannot be rational.

**Note:** It is slightly easier to rewrite the initial equation in the form

$$\sqrt{3}=\frac{p}{q}-\sqrt{2}$$

before squaring to get

$${\left(\frac{p}{q}\right)}^{2}-1=\frac{2p}{q}\sqrt{2},$$

which would imply that $\sqrt{2}$ is rational.

(b) Suppose to the contrary that $\sqrt{2}+\sqrt{3}+\sqrt{5}=\frac{p}{q}$ , for
some integers *p, q* with *HCF*(*p,q*) − 1. Then

$$10+2(\sqrt{6}+\sqrt{10}+\sqrt{15})={\left(\frac{p}{q}\right)}^{2},$$

so $\sqrt{6}+\sqrt{10}+\sqrt{15}$ is rational. Squaring $\sqrt{6}+\sqrt{10}+\sqrt{15}$ then gives that

$$\sqrt{60}+\sqrt{90}+\sqrt{150}=5\sqrt{6}+3\sqrt{10}+2\sqrt{15}$$

is rational. Subtracting $2\text{\hspace{0.17em}}(\sqrt{6}+\sqrt{10}+\sqrt{15})$ then shows that $3\sqrt{6}+\sqrt{10}$ is rational, and we can proceed as in part (a) to obtain a contradiction. Hence $\sqrt{2}+\sqrt{3}+\sqrt{5}$ cannot be rational.

**Note:** It is simpler to rewrite the original equation in the form

$$\sqrt{2}+\sqrt{3}=\frac{p}{q}-\sqrt{5}$$

before squaring to obtain

$$5+2\sqrt{6}=\left(5+{\left(\frac{p}{q}\right)}^{2}\right)-\frac{2p}{q}\sqrt{5},$$

whence $2\sqrt{6}+\frac{2p}{q}\sqrt{5}$ is rational, and we may proceed as in part (a).

**126.**

(i) We just have to fill in the missing bits of the partial factorisation

$${x}^{10}+1=\left({x}^{3}-1\right)\left({x}^{7}+{x}^{4}+\cdots \right)+\text{\hspace{0.17em}}remainder.$$

To produce the required term *x*^{10} we first insert *x*^{7}. This then creates an unwanted term “−
*x* ^{7}”, so we add +x^{4} to cancel this out. This in turn creates an unwanted term “− *x*
^{4}”, so we add +x to cancel this out. Hence the quotient is *x*^{7} + *x*^{4} + x, and the remainder is
“x + 1”:

$${x}^{10}+1=\left({x}^{3}-1\right)\left({x}^{7}+{x}^{4}+x\right)+(x+1).$$

**Note:** It is worth noting a short cut. The factorised term of the form (*x*^{3} − 1) (*x*^{7} + …) is
equal to zero when *x*^{3} = 1.

So one way to get the remainder is to “treat *x*^{3} as if it were equal to 1”. Then

$${x}^{10}={\left({x}^{3}\right)}^{3}\xb7x$$

is just like 1 · *x*, and *x*^{10} + 1 behaves as if it were equal to remainder.

(ii)

$${x}^{2013+1}=\left({x}^{2}-1\right)\left({x}^{2011}+{x}^{2009}+{x}^{2007}+\cdots +x\right)+(x+1),$$

so the remainder = *x* + 1.

**Note**: If we treat *x*^{2} “as if it were equal to 1”, then

$${x}^{2013}+1={\left({x}^{2}\right)}^{1006}\xb7x+1$$

behaves as if it were equal to 1 · *x* + 1

(iii) Apply the Euclidean algorithm to *m* and *n* in order to write *m = qn* + *r*, where $0\u2a7dr<n$
:

$${x}^{m}={x}^{qn+r}={\left({x}^{n}\right)}^{q}\xb7{x}^{r}.$$

Then

$$\begin{array}{ccc}{x}^{m}+1\hfill & =\hfill & {x}^{qn+r}+1\hfill \\ \hfill & =\hfill & ({x}^{n}-1)({x}^{n(q-1)+r}+{x}^{n(q-2)+r}+{x}^{n(q-3)+r}+\cdots +{x}^{r})+{x}^{r}+1.\hfill \end{array}$$

So the remainder is *x ^{r}* + 1.

**Note:** If we treat *x ^{n}* - 1 as if were 0 - that is, if we treat

*x*as if it were equal to 1 − then

^{n}$${x}^{m}+1={x}^{qn+r}+1={\left({x}^{n}\right)}^{q}\xb7{x}^{r}+1$$

which behaves like ${1}^{q}\xb7{x}^{r}+1$.

**127**. Suppose ${x}^{2013}+1=\left({x}^{2}+x+1\right)q(x)+r(x)$ , where deg(r(x)) < 2. Then

$$\begin{array}{ccc}({x}^{2013}+1)(x-1)\hfill & =\hfill & {x}^{2014}-{x}^{2013}+x-1\hfill \\ \hfill & =\hfill & ({x}^{3}-1)q\left(x\right)+(x-1)r\left(x\right).\hfill \end{array}$$

Now

$${x}^{2014}-{x}^{2013}+x-1=\left({x}^{3}-1\right)({x}^{2011}-{x}^{2008}+{x}^{2008}-{x}^{2007}+{x}^{2004}-{x}^{2003}+\cdots +x+2x-2$$

so the remainder *r*(*x*) = 2.

**Note:** If *x* satisfies *x*^{2} + *x* + 1 = 0, then *x*^{3} − 1 = 0 and $x\ne 1$ .

$\therefore {x}^{2013}+1={\left({x}^{3}\right)}^{671}+1$ behaves just like 1^{671} + 1 = 2, so r(x) = 2.

**128.**

(a)

$${(a+bi)}^{-1}=\frac{a}{{a}^{2}+{b}^{2}}-\left[\frac{b}{{a}^{2}+{b}^{2}}\right]i.$$

(b)

$$p(x)={x}^{2}-2ax+\left({a}^{2}+{b}^{2}\right).$$

(Suppose that the quadratic equation

$$p(x)={x}^{2}+cx+d=0,$$

with real coefficients *c*, *d*, has *x = a + ib* as a root. Then take the complex conjugate of the equation *p*(x) = 0 to see
that *x* = *a* − *ib* is also a rooti of

$$p(x)={x}^{2}+cx+d=0.$$

Therefore

$$\begin{array}{l}p(x)={x}^{2}+cx+d\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=(x-(a+ib))(x-(a-ib)),\end{array}$$

so *c* = −2*a*, and *d* = *a ^{2}* +

*b*

^{2}.)

**129**. Let the two unknown numbers be *α* and *β*. Then 10 = *α* + *β*,
and 40 = *αβ*, so *α* and *β* are roots of the quadratic equation *x*^{2} − 10*x*
+ 40 = 0. Hence

$$\alpha ,\beta =\frac{10\pm \sqrt{100-160}}{2}=5\pm \sqrt{-15}.$$

**130.**

(a) Applying a simple rearrangement:

$$\begin{array}{l}wz=r(\mathrm{cos}\theta +i\mathrm{sin}\theta )\xb7s(\mathrm{cos}\varphi +i\mathrm{sin}\varphi )\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=rs[(\mathrm{cos}\theta \xb7\mathrm{cos}\varphi -\mathrm{sin}\theta \mathrm{sin}\varphi )+i(\mathrm{cos}\theta \xb7\mathrm{sin}\varphi +\mathrm{sin}\theta \xb7\mathrm{cos}\varphi )]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=rs[\mathrm{cos}(\theta +\varphi )+i\mathrm{sin}(\theta +\varphi )]\end{array}$$

(by the usual addition formula: Problem **35**)

(b) By part (a),

$${(\mathrm{cos}\theta +i\mathrm{sin}\theta )}^{2}=\mathrm{cos}(2\theta )+i\mathrm{sin}(2\theta ).$$

Hence

$$\begin{array}{ccc}{(\mathrm{cos}\theta +i\mathrm{sin}\theta )}^{3}\hfill & =\hfill & {(\mathrm{cos}\theta +i\mathrm{sin}\theta )}^{2}(\mathrm{cos}\theta +i\mathrm{sin}\theta )\hfill \\ \hfill & =\hfill & [\mathrm{cos}\left(2\theta \right)+i\mathrm{sin}\left(2\theta \right)]\cdot (\mathrm{cos}\theta +i\mathrm{sin}\theta )\hfill \\ \hfill & =\hfill & \mathrm{cos}\left(3\theta \right)+i\mathrm{sin}\left(3\theta \right).\hfill \end{array}$$

Etc.

**Note:** This should really be presented as a “proof by mathematical induction”, where (having established the initial cases) we
“suppose the result holds for powers *n* = 1, 2, 3,…, *k*”, and then conclude that

$$\begin{array}{ccc}{(\mathrm{cos}\theta +i\mathrm{sin}\theta )}^{k+1}\hfill & =\hfill & {(\mathrm{cos}\theta +i\mathrm{sin}\theta )}^{k}(\mathrm{cos}\theta +i\mathrm{sin}\theta )\hfill \\ \hfill & =\hfill & [\mathrm{cos}\left(k\theta \right)+i\mathrm{sin}\left(k\theta \right)](\mathrm{cos}\theta +i\mathrm{sin}\theta )\hfill \\ \hfill & =\hfill & \mathrm{cos}\left(\right(k+1\left)\theta \right)+i\mathrm{sin}\left(\right(k+1\left)\theta \right).\hfill \end{array}$$

(c) ${z}^{n}={r}^{n}(\mathrm{cos}(n\theta )+i\mathrm{sin}(n\theta ))$ . Hence if ${z}^{n}=1$ , then $\left|{z}^{n}\right|={r}^{n}=1$ So r=1 (since $r\u2a7e0$ ).

**131.**

(a) We factorise: *x*^{3} − 1 = (*x* − 1) (*x*^{2} + *x* + 1, so the roots are *x* = 1;
and

$$x=\frac{-1\pm \sqrt{1-4}}{2}=\frac{-1\pm \sqrt{-3}}{2}=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$$

that is, the other two roots are

$$x=\mathrm{cos}\left(\frac{2\pi}{3}\right)+i\mathrm{sin}\left(\frac{2\pi}{3}\right)$$

and

$$x=\mathrm{cos}\left(-\frac{2\pi}{3}\right)+i\mathrm{sin}\left(-\frac{2\pi}{3}\right).$$

(b) We factorise:

$${x}^{4}-1=\left({x}^{2}-1\right)\left({x}^{2}+1\right)=(x-1)(x+1)\left({x}^{2}+1\right),$$

so the roots are *x* = 1, *x* = −1, *x* = *i*, *x* = −*i*.

(c) We factorise:

$$\begin{array}{ccc}{x}^{6}-1\hfill & =\hfill & [{\left({x}^{2}\right)}^{3}-1]\hfill \\ \hfill & =\hfill & ({x}^{2}-1)({x}^{4}+{x}^{2}+1)\hfill \\ \hfill & =\hfill & (x-1)(x+1)[{\left({x}^{2}\right)}^{2}+{x}^{2}+1],\hfill \end{array}$$

so the roots are

− *x* = 1, *x* = − 1, and

− four further values of *x* satisfying ${x}^{2}=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$: that is,

$$x=\mathrm{cos}\left(\frac{\pi}{3}\right)+i\mathrm{sin}\left(\frac{\pi}{3}\right)=\frac{1}{2}+\frac{\sqrt{3}}{2}i$$

and

$$x=\mathrm{cos}\left(\frac{2\pi}{3}\right)+i\mathrm{sin}\left(\frac{2\pi}{3}\right)=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$$

and

$$x=\mathrm{cos}\left(\frac{-\pi}{3}\right)+i\mathrm{sin}\left(\frac{-\pi}{3}\right)=\frac{1}{2}-\frac{\sqrt{3}}{2}i$$

and

$$x=\mathrm{cos}\left(\frac{-2\pi}{3}\right)+i\mathrm{sin}\left(\frac{-2\pi}{3}\right)=-\frac{1}{2}-\frac{\sqrt{3}}{2}i.$$

(d) We factorise:

$$\begin{array}{ccc}{x}^{8}-1\hfill & =\hfill & ({x}^{4}-1)({x}^{4}+1)\hfill \\ \hfill & =\hfill & ({x}^{2}-1)({x}^{2}+1)({x}^{2}+\sqrt{2}\cdot x+1)({x}^{2}-\sqrt{2}\cdot x+1)\hfill \end{array}$$

so the roots are

− *x* = 1, *x* = −1;

− *x* = *i*, *x* = − *i*, and

− the roots of ${x}^{2}+\sqrt{2}\xb7x+1=0$ and ${x}^{2}-\sqrt{2}\xb7x+1=0$ , which happen to be

$$x=\mathrm{cos}\left(\frac{\pi}{4}\right)+i\mathrm{sin}\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$$

and

$$x=\mathrm{cos}\left(-\frac{\pi}{4}\right)+i\mathrm{sin}\left(-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$$

and

$$x=\mathrm{cos}\left(\frac{3\pi}{4}\right)+i\mathrm{sin}\left(\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$$

and

$$x=\mathrm{cos}\left(-\frac{3\pi}{4}\right)+i\mathrm{sin}\left(-\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}.$$

**132**. [In Problem **114** you were left to work out the required factorisation with your bare
hands - and a bit of inspired guesswork. The suggested approach here is more systematic.]

The roots of *x*^{4} + 1 = 0 are complex numbers whose fourth power is equal to that is,

$$x=\mathrm{cos}\left(\frac{\pi}{4}\right)+i\mathrm{sin}\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$$

and

$$x=\mathrm{cos}\left(-\frac{\pi}{4}\right)+i\mathrm{sin}\left(-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$$

and

$$x=\mathrm{cos}\left(\frac{3\pi}{4}\right)+i\mathrm{sin}\left(\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$$

and

$$x=\mathrm{cos}\left(-\frac{3\pi}{4}\right))+i\mathrm{sin}\left(-\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i.$$

The first two are complex conjugates and give rise to two linear factors whose product is ${x}^{2}+\sqrt{2}\xb7x+1$ ; the other two are complex conjugates and give rise to two linear factors whose product is ${x}^{2}-\sqrt{2}\xb7x+1$ . Hence

$${x}^{4}+1=\left({x}^{2}+\sqrt{2}\xb7x+1\right)\left({x}^{2}-\sqrt{2}\xb7x+1\right).$$

**133**.

(a) The roots of *x*^{5} − 1 = 0 are precisely the five complex numbers of the form

$$\mathrm{cos}\left(\frac{2k\pi}{5}\right)+i\mathrm{sin}\left(\frac{2k\pi}{5}\right),\$for\$k=0,1,2,3,4:$$

that is,

$$\begin{array}{l}x=1\\ x=\mathrm{cos}\left(\frac{2\pi}{5}\right)+i\mathrm{sin}\left(\frac{2\pi}{5}\right)\\ x=\mathrm{cos}\left(\frac{4\pi}{5}\right)+i\mathrm{sin}\left(\frac{4\pi}{5}\right)\\ x=\mathrm{cos}\left(\frac{6\pi}{5}\right)+i\mathrm{sin}\left(\frac{6\pi}{5}\right)\\ x=\mathrm{cos}\left(\frac{8\pi}{5}\right)+i\mathrm{sin}\left(\frac{8\pi}{5}\right).\end{array}$$

From Problem **3**(c) we know that

$$\begin{array}{l}\mathrm{cos}\left(\frac{2\pi}{5}\right)=\frac{\sqrt{5}-1}{4}\text{\hspace{1em}}=\mathrm{cos}\left(\frac{8\pi}{5}\right)\\ \mathrm{sin}\left(\frac{2\pi}{5}\right)=\frac{\sqrt{10+2\sqrt{5}}}{4}=-\mathrm{sin}\left(\frac{8\pi}{5}\right)\\ \mathrm{cos}\left(\frac{4\pi}{5}\right)=-\mathrm{cos}\left(\frac{\pi}{5}\right)=-\frac{\sqrt{5}+1}{4}=\mathrm{cos}\left(\frac{6\pi}{5}\right)\\ \mathrm{sin}\left(\frac{4\pi}{5}\right)=\frac{\sqrt{10-2\sqrt{5}}}{4}=-\mathrm{sin}\left(\frac{6\pi}{5}\right).\end{array}$$

(b) The linear factor is clearly (x − 1). Each quadratic factor arises as the product of two of two conjugate linear factors. We saw in
Problem **128**(b) that two linear factors corresponding to roots a + bi and a - bi produce the quadratic factor ${x}^{2}-2ax+\left({a}^{2}+{b}^{2}\right)$ . Hence the two quadratic factors are:

$${x}^{2}-\frac{\sqrt{5}-1}{2}\xb7x+1,\text{\hspace{1em}}and\text{\hspace{1em}}{x}^{2}+\frac{\sqrt{5}+1}{2}\xb7x+1$$

(whose product is equal to *x*^{4} + *x*^{3} + *x*^{2} + *x* + 1).

**134.**

(a) Put *a =* 1, *y = x* + 1: then *x*^{3} + 3*x*^{2} − 4 = 0 becomes *y*^{3} −
3*y* = 2.

(b) Divide through by *a* (which we may assume is non zero, since otherwise it would not be a *cubic* equation), to obtain a cubic
equation

$${x}^{3}+p{x}^{2}+qx+r=0.$$

If we now put $y=x+\frac{p}{3}$ , then y^{3} incorporates both the *x*^{3} and the *x*^{2} terms, and the
equation reduces to:

$${y}^{3}+\left[q-3{\left(\frac{p}{3}\right)}^{2}\right]y+\left[r+2{\left(\frac{p}{3}\right)}^{3}-q\left(\frac{p}{3}\right)\right]=0.$$

**135.** Given the equation *x*^{3} + 3*x*^{2} − 4 = 0. Let *y* = *x* + 1.

(i) Then *y*^{3} = *x*^{3} + 3*x*^{2} + 3*x* + 1, so 0 = *x*^{3} + 3*x*^{2}
− 4 = *y*^{3} − 3*y* − 2.

(ii) Set *y* = *u + v* and use the fact that

$${(u+v)}^{3}={u}^{3}+3uv(u+v)+{v}^{3}$$

is an identity, and so holds for all *u* and *v*.

(iii) Solve “3uv = 3”, “u^{3} + v^{3} = 2”. Substitute $v=\frac{1}{u}$ from the first equation into the second to get the
quadratic equation in ${\left({u}^{3}\right)}^{2}-2{u}^{3}+1=0$ : that is, ${\left({u}^{3}-1\right)}^{2}=0$ , So *u*^{3} = 1.

(iv) **Hence u = 1 is certainly a solution.** (We know there are also complex cube roots of 1; these lead to the other two solutions of
the original cubic, but to “solve the equation” it is enough to find one solution.)

**Hence**

*v*= 1, so*y*=*u + v*= 2, and*x*= 2.**136**. The Euclidean algorithm for ordinary integers arises by repeating the division algorithm:

given integers $m,n(\ne 0)$ , there exists unique integersq, rsuch thatm = qn + rwhere $0\u2a7dr<n$ .

Here *q* is the *quotient* (the integer part of the division $m\xf7n$ ), and *r* is the *remainder. *If we then replace the initial pair (*m, n*) by
the new pair (*n, r*) and repeat until we obtain the remainder 0, then the last non-zero remainder is equal to *HCF*(*m,n*) (see
Problem **6**). The same idea also works for polynomials with integer coefficients (see Problem **126**).

We start by clarifying what we mean by *divisibility* for Gaussian integers. Given two Gaussian integers, *m* = *a* + *bi* and
*n* = *c* + *di*, we say that *n = c + di* divides *m = a + bi* (exactly) precisely

when there exists some other Gaussian integer

q = e + fisuch thatm = qn: that is,a + bi= (e + fi)(c + di).

For example, 2 + 3*i* divides −4 + 7*i* because (1 + 2*i*)(2 + 3*i*) = −4 + 7*i*.

If *m = a + bi* and *n = c + di* are any old Gaussian integers, then it will not in general be true that “n divides m”, but
we can imitate the division algorithm. The important idea here when carrying out particular calculations is to realize that “divide by *c +
di*” is the same as “multiply by $\frac{c-di}{{c}^{2}+{d}^{2}}$ ”

- first carry out the division $$m\xf7n=\frac{(a+bi)(c-di)}{{c}^{2}+{d}^{2}};$$
- then take the “nearest” Gaussian integer
*q = e + fi*, and let the difference*m*−*qn = r*be the*remainder.*

As for ordinary integers, any Gaussian integer that is a “common factor of *m* and n” is then automatically a common factor of
*n* and of *r* = *m* − *qn*, and conversely. That is, the common factors of m and *n* are precisely the same as the
common factors of *n* and *r*. So we can repeat the process replacing *m*, *n* by *n*, *r*. Provided the
“remainder” *r* is in some sense “smaller” than *n*, we can continue until we reach a stage where the remainder
*r* = 0 - at which point, the last non-zero remainder is equal to the *HCF*(*m,n*) (that is, the Gaussian integer which is the HCF of
the two initial Gaussian integers *m*, *n*).

The feature of the remainders that gets progressively smaller is their *norm* (see Problem **25**, and Problem **54**). As so often, this becomes clearer when we look at an example.

Let us try to find the HCF of the two Gaussian integers *m* = 14−42*i* and *n* = 4 −7*i*.

- First do the division $$m\xf7n=\frac{(14-42i)(4+7i)}{{4}^{2}+{7}^{2}}=\frac{350}{65}-\frac{70}{65}i.$$
- What is meant by the
*nearest*Gaussian integer may require an element of judgment; but it is clear that the answer is fairly close to 5 −*i = q*, where*qn*= 13 − 39*i*, with remainder*r = m*−*qn*= 1 − 3*i*. - Now repeat the process with
*n*,*r*: $$n\xf7r=\frac{(4-7i)(1+3i)}{{1}^{2}+{3}^{2}}=\frac{5}{2}+\frac{1}{2}i.$$ - The
*nearest*Gaussian integer is not well-defined, but the answer is fairly close to $3={q}^{\prime}$ . So ${q}^{\prime}r=3-9i$ , with remainder ${r}^{\prime}=n-{q}^{\prime}r=1+2i$ . - Now repeat the step with the pair
*r*= 1 − 3*i*and ${r}^{\prime}=1+2i$ , to discover that $$1-3i=-(1+i)(1+2i)$$ with remainder 0. Hence $$1+2i=HCF(14-42i,4-7i).$$

**Note:** One way to picture the process is to learn to “see” the Gaussian integers *geometrically.* Every Gaussian integer
(such as *a + bi*) can be written as an integer combination of the two basic Gaussian integers “1” and “*i*” -
namely

$$a+bi=a\times 1+b\times i.$$

Since 1 and *i* are both of length 1 and perpendicular to each other, this represents the set of all Gaussian integers as the dots in a
“square dot lattice” generated by translations in the *x*- and *y*- directions of the basic unit square spanned by 0, 1,
*i*, and 1 + *i*.

Any other given Gaussian integer, such as *n* = *c* + *di*, then generates a “stretched and rotated” square lattice,
which consists of all “Gaussian multiples” of *c* + *di* - generated by the basic square which is spanned by

$$0,\text{\hspace{1em}}(c+di)\times 1,\text{\hspace{1em}}(c+di)\times i,\text{\hspace{1em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}(c+di)\times (1+i).$$

Every Gaussian integer (or rather the point, or dot, which corresponds to it) lies either on the boundary, or inside, one of these larger
“stretched and rotated” squares: if the diagonal of one of these larger squares has length 2*k*, then any other Gaussian integer
*m = a* + *bi* lies inside one of these larger squares, and so lies *within distance k* (that is, half a diagonal) of some (Gaussian)
multiple *qn* of *n = c* + *di.* And the difference *m − qn* is precisely the required *remainder r.*

**Extra:** We interpret $\sqrt[3]{8}={8}^{\frac{1}{3}}=2$ . Prove that

$$\sqrt[1]{-\sqrt[1]{-1}}\approx 23\frac{1}{7}$$

(where ≈ denotes “approximately equal to”).