Problem 11:  Egyptian fractions ($✓$) 2000 Paper II

A number of the form $\frac{1}{N}$, where $N$ is an integer greater than 1, is called a unit fraction.

Noting that

guess a general result of the form

 $\frac{1}{N}=\frac{1}{a}+\frac{1}{b}\phantom{\rule{0.3em}{0ex}}$ ($\ast$)

and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions.

By writing $\left(\ast \right)$ in the form

$\left(a-N\right)\left(b-N\right)={N}^{2}$

and by considering the factors of ${N}^{2}$, show that if $N$ is prime, then there is only one way of expressing $\frac{1}{N}$ as the sum of two distinct unit fractions.

Prove similarly that any fraction of the form $\frac{2}{N}$, where $N$ is prime number greater than 2, can be expressed uniquely as the sum of two distinct unit fractions.

Fractions written as the sum of unit fractions are called Egyptian fractions: they were used by Egyptians. The earliest record of such use is 1900BC. The Rhind papyrus in the British Museum gives a table of representations of fractions of the form $\frac{2}{n}$ as sums of unit fractions for all odd integers $n$ between 5 and 101 — a remarkable achievement when you consider that algebra was 3,500 years in the future.

It is not clear why Egyptians represented fractions this way; maybe it just seemed a good idea at the time. Certainly the notation they used, in which for example $\frac{1}{n}$ was denoted by $n$ with an oval on top, does not lend itself to generalisation to fractions that are not unit. One of the rules for expressing non-unit fractions in terms of unit fractions was that all the unit fractions in should be distinct, so $\frac{2}{7}$ had to be expressed as $\frac{1}{4}+\frac{1}{28}\phantom{\rule{2.77695pt}{0ex}}$ instead of as $\frac{1}{7}+\frac{1}{7}\phantom{\rule{2.77695pt}{0ex}}$, which seems pretty daft.

It is not obvious that every fraction can be express in Egyptian form; this was proved by Fibonacci in 1202. There are however still many unsolved problems relating to Egyptian fractions.

Egyptian fractions have been called a wrong turn in the history of mathematics; if so, it was a wrong turn that favoured style over utility; no bad thing, in my opinion.

Solution to problem 11

A good guess would be that the ﬁrst term of the decomposition of $\frac{1}{N}$ is $\frac{1}{N+1}\phantom{\rule{0.3em}{0ex}}$, i.e. $a=N+1$. In that case, the other term is $\frac{1}{N}-\frac{1}{N+1}\phantom{\rule{0.3em}{0ex}}$ i.e. $b=N\left(N+1\right)\phantom{\rule{0.3em}{0ex}}$. That proves the result that every unit fraction can be expressed as the sum of two unit fractions.

The only factors of ${N}^{2}$ (since $N$ is prime) are $1\phantom{\rule{0.3em}{0ex}}$, $N\phantom{\rule{0.3em}{0ex}}$ and ${N}^{2}\phantom{\rule{0.3em}{0ex}}$. The possible factorisations of ${N}^{2}$ are therefore ${N}^{2}=1×{N}^{2}$ or ${N}^{2}=N×N$. We discount $N×N$, since this would lead to $a=b$, and this is ruled out in the question. That leaves only $a-N=1$ and $b-N={N}^{2}$ (or the other way round). Thus $N+1$ and ${N}^{2}+N$ are the only possible values for $a$ and $b$ and the decomposition is unique.

For the second half, set

$\frac{2}{N}=\frac{1}{a}+\frac{1}{b}$

where $a\ne b\phantom{\rule{0.3em}{0ex}}$. We need to aim for an equation with ${N}^{2}$ on one side, so that we can use the method of the ﬁrst part. We have $ab-\frac{1}{2}\left(a+b\right)N=0$ which we write as

$\left(a-\frac{N}{2}\right)\left(b-\frac{N}{2}\right)=\frac{{N}^{2}}{4}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{i.e.}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(2a-N\right)\left(2b-N\right)={N}^{2}\phantom{\rule{2.77695pt}{0ex}}.$

Thus $2a-N={N}^{2}$ and $2b-N=1$ (or the other way round). The decomposition is therefore unique and given by

$\frac{2}{N}=\frac{1}{\frac{1}{2}\left({N}^{2}+N\right)}+\frac{1}{\frac{1}{2}\left(N+1\right)}\phantom{\rule{2.77695pt}{0ex}}.$

The only possible ﬂy in the ointment is the $\frac{1}{2}$ in the denominators: $a$ and $b$ are supposed to be integers. However, all prime numbers greater than 2 are odd, so $N+1$ and ${N}^{2}+N$ are both even and the denominators are indeed integers.

Post-mortem

Another way of getting the last part (less systematically) would have been to notice that

$\frac{1}{N}=\frac{1}{\left(N+1\right)}+\frac{1}{N\left(N+1\right)}⇒\frac{2}{N}=\frac{1}{\frac{1}{2}\left(N+1\right)}+\frac{1}{\frac{1}{2}N\left(N+1\right)}\phantom{\rule{0.3em}{0ex}}$

which gives the result immediately. How might you have noticed this? Well, I noticed it by trying to work out some examples, starting with what is given at the very beginning of the question. If $\frac{1}{3}=\frac{1}{4}+\frac{1}{12}$ then $\frac{2}{3}=\frac{2}{4}+\frac{2}{12}$ which works.

Of course, the advantage of the systematic approach is that it allows generalisation: what happens if $N$ is odd but not prime; what happens if the numerator is 3 not 2?