Problem 27

Problem 27: Relation between coefficients of quartic for real roots ( $✓$ $✓$ ) 1997 Paper III

In this question you may assume that, if $k_{1}, \dots, k_{n}$ are distinct positive real numbers, then

\frac{1}{n} \sum_{r = 1}^{n} k_{r} > {(\prod_{r = 1}^{n} k_{r})}^{\frac{1}{n}},

i.e. their arithmetic mean is greater than their geometric mean.

Suppose that $a$ , $b$ , $c$ and $d$ are positive real numbers such that the polynomial

f (x) = x^{4} - 4 a x^{3} + 6 b^{2} x^{2} - 4 c^{3} x + d^{4}

has four distinct positive roots.

(i): By considering the relationship between the coefficients of f and its roots, show that $c > d$ .
(ii): By differentiating f, show that $b > c$ .
(iii): Show that $a > b$ .

Comments

This result is both surprising and pleasing.

The question looks difficult, but you don’t have to go very far before you come across something to substitute into the given arithmetic mean/geometric mean (AM/GM) inequality.

To obtain the relationship between the coefficients and the roots for part (i), you need to write the quartic equation in the form $(x - p) (x - q) (x - r) (x - s) = 0$ .

Watch out for the condition in the given form of the arithmetic/geometric inequality that the numbers are distinct: you will have to show that any numbers (or algebraic expressions) you use in the inequality are distinct.

I puzzled over this question for ages, not understanding the idea behind the question²⁰

It will be clear from the proof that it can be generalised to any equation of the form

x^{N} + \sum_{k = 0}^{N - 1} (\binom{N}{k}) {(- a_{k})}^{N - k} x^{k} = 0,

where the numbers $a_{k}$ are distinct and positive, which has positive distinct roots.

Eventually, I realised that the result of the question has really nothing to do with quartic equations, though quartic equations provide a neat method of proof. The inequalities, which apply to any positive numbers, form a sequence interpolating between the AM and the GM. These inequalities were ﬁrst derived by Newton and Maclaurin in the 17th century but don’t seem to be very well known now.

Solution to problem 27

(i) First write

f (x) \equiv (x - p) (x - q) (x - r) (x - s)

where $p$ , $q$ , $r$ and $s$ are the four roots of the equation (known to be real, positive and distinct). Multiplying out the brackets and comparing with $x^{4} - 4 a x^{3} + 6 b^{2} x^{2} - 4 c^{3} x + d^{4}$ shows that $p q r s = d^{4}$ and $p q r + q r s + r s p + s p q = 4 c^{3}$ .

The required result, $c > d$ , follows immediately by applying the AM/GM inequality to the positive real numbers $p q r$ , $q r s$ , $r s p$ and $s p q$ :

c^{3} = \frac{p q r + q r s + r s p + s p q}{4} > {[(p q r) (q r s) (r s p) (s p q)]}^{1 ∕ 4} = {[p^{3} q^{3} r^{3} s^{3}]}^{1 ∕ 4} = {[d^{12}]}^{1 ∕ 4} .

Taking the cube root ( $c$ and $d$ are positive) preserves the inequality.

The AM/GM inequality at the beginning of the question is stated only for the case when the numbers are distinct (though in fact it holds provided at least two of the numbers are distinct). To use the inequality as above, we must therefore show that no two of $p q r$ , $q r s$ , $r s p$ and $s p q$ are equal. This follows immediately from the fact that the roots are distinct and non-zero. For if, for example, $p q r = q r s$ then $q r (p - s) = 0$ which means that $q = 0$ , $r = 0$ or $p = s$ all of which are ruled out.

(ii) The polynomial $f^{'} (x)$ is cubic so it has three zeros (roots). These are at the turning points of $f (x)$ , which lie between the zeros of $f (x)$ and are therefore distinct and positive.

Now

f^{'} (x) = 4 x^{3} - 12 a x^{2} + 12 b^{2} x - 4 c^{3},

so at the turning points

x^{3} - 3 a x^{2} + 3 b^{2} x - c^{3} = 0 .

Suppose that the roots of this cubic equation are $u$ , $v$ and $w$ all real, distinct and positive. Then comparing $x^{3} - 3 a x^{2} + 3 b^{2} x - c^{3} = 0$ and $(x - u) (x - v) (x - w)$ shows that $c^{3} = u v w$ and $3 b^{2} = v w + w u + u v$ .

The required result, $b > c$ , follows immediately by applying the AM/GM inequality to the positive distinct numbers $v w$ , $w u$ and $u v$ .

(iii) Apply similar arguments to $f^{″} (x) ∕ 12$ .

Post-mortem

The general inequalities mentioned in the comments above are interesting. For example, for four numbers $a_{1}$ , $a_{2}$ , $a_{3}$ and $a_{4}$ , we deﬁne

S_{1} = \frac{a_{1} + a_{2} + a_{3} + a_{4}}{4}, S_{2} = \frac{a_{2} a_{3} + a_{3} a_{1} + a_{1} a_{2} + a_{4} a_{1} + a_{4} a_{2} + a_{4} a_{3}}{6},

S_{3} = \frac{a_{2} a_{3} a_{4} + a_{1} a_{3} a_{4} + a_{1} a_{2} a_{4} + a_{1} a_{2} a_{3}}{4} and S_{4} = a_{1} a_{2} a_{3} a_{4} .

Except for the denominators and alternating signs, $S_{i}$ is the coefficient of $x^{4 - i}$ in the equation
$(x - a_{1}) (x - a_{2}) (x - a_{3}) (x - a_{4}) = 0$ , but that doesn’t matter. Maclaurin’s inequalities are

S_{1} \geq S_{2}^{\frac{1}{2}} \geq S_{3}^{\frac{1}{3}} \geq S_{4}^{\frac{1}{4}} .

What is really surprising is that we proved the three stronger inequalities from the apparently weaker AM $>$ GM inequality ( $S_{1} \geq S_{4}^{\frac{1}{4}}$ ): a rare example of pulling yourself up by your bootstraps.

Problem 28: Fermat numbers ( $✓$ $✓$ ) 2002 Paper II

The $n$ th Fermat number, $F_{n}$ , is deﬁned by

F_{n} = 2^{2^{n}} + 1, n = 0, 1, 2, \dots,

where $2^{2^{n}}$ means 2 raised to the power $2^{n}$ . Calculate $F_{0}$ , $F_{1}$ , $F_{2}$ and $F_{3}$ . Show that, for $k = 1$ , $k = 2$ and $k = 3$ ,

F_{0} F_{1} \dots F_{k - 1} = F_{k} - 2 .

(

*

)

Prove, by induction, or otherwise, that $(*)$ holds for all $k \geq 1$ . Deduce that no two Fermat numbers have a common factor greater than 1.

Hence show that there are inﬁnitely many prime numbers.

Comments

Fermat (1601–1665) conjectured that every number of the form $2^{2^{n}} + 1$ is a prime number. $F_{0}$ to $F_{4}$ are indeed prime, but Euler showed in 1732 that $F_{5}$ (4294967297) is divisible by 641. As can be seen, Fermat numbers get very big and not many more have been investigated; but those that have been investigated have been found not to be prime numbers. It is now conjectured that in fact only a ﬁnite number of Fermat numbers are prime numbers.

The Fermat numbers have a geometrical signiﬁcance as well: Gauss proved that a regular polygon of $n$ sides can be inscribed in a circle using a Euclidean construction (i.e. only a straight edge and a compass) if and only if $n$ is a power of 2 times a product of distinct Fermat primes.

The very last part of this question, showing that the number of primes is inﬁnite, is completely unexpected and delightful. It comes from Proofs from THE BOOK²¹, a set of proofs thought by Paul Erdös to be heaven-sent. Most of them are much less elementary this one. Erdös was an extraordinarily proliﬁc mathematician. He had almost no personal belongings and no home²². He spent his life visiting other mathematicians and proving theorems with them. He collaborated with so many people that every mathematician is (jokingly) assigned an Erdös number; for example, if you wrote a paper with someone who wrote a paper with Erdös, you are given the Erdös number 2. Most mathematicians seem to have an Erdös number less than 8.

²⁰I mentioned this in a talk I gave to sixth formers, and later saw irritatingly on http://www.thestudentroom.co.ukwww.thestudentroom.co.uk a comment to the effect that ‘even Dr Siklos can’t do this STEP question’. Of course I could do it; I wanted to understand it, as I hope you do too.