A number of the form \(1/N\), where \(N\) is an integer greater than \(1\), is called a *unit fraction*.

Noting that \[\frac{1}{2} = \frac{1}{3} + \frac{1}{6}\quad\mbox{and}\quad\frac{1}{3}=\frac{1}{4}+\frac{1}{12},\] guess a general result of the form \[\frac{1}{N} = \frac{1}{a}+\frac{1}{b}\] and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions.

Based on the two examples given, we guess \[\frac{1}{N}=\frac{1}{N + 1} + \frac{1}{N(N+1)},\] i.e. \(a = N+1\) and \(b=N(N+1)\). To justify this, it suffices to note that \[\frac{1}{N + 1} + \frac{1}{N(N+1)} = \frac{N+1}{N(N + 1)} = \frac{1}{N}.\] In this expression, we can choose \(N\) to be an arbitrary integer greater than \(1\). Since \(N(N+1) > N+1\) if \(N>1\), it follows that any unit fraction can be expressed as the sum of two distinct unit fractions.

By writing the previous equation in the form \[(a − N)(b − N) = N^2\] and by considering the factors of \(N^2\), show that if \(N\) is prime, then there is only one way of expressing \(1/N\) as the sum of two distinct unit fractions.

We can convert the original equation into this form by first multiplying both sides by \(abN\) to get \[ab=Nb+Na.\] Now subtracting \(Na+Nb\) from both sides gives \(ab-Na-Nb=0\). If we add \(N^2\) to both sides, then the resulting expression can be factorised to get \((a-N)(b-N)=N^2\), which is what we wanted.

Without loss of generality we may assume \(a\leq b\). If \(N\) is prime then there are only two possible ways of factorising \(N^2\), namely \(N^2=N\times N\) and \(N^2=1\times N^2\). Comparing this with \[(a-N)(b-N)=N^2\] gives \(a-N=b-N=N\) in the first case and \(a-N = 1\) and \(b-N=N^2\) in the latter case. If we have \(a-N=b-N=N\) then \(a=b\) whereas \(a-N = 1\) and \(b-N=N^2\) yields \(b=N(N+1)>N+1=a\). Therefore, there is indeed only one way of expressing \(1/N\) as the sum of two **distinct** unit fractions.

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

#### Method 1

We begin by writing \[\frac{2}{N}=\frac{1}{a}+\frac{1}{b}\] and again assume that \(a < b\) (as we wish \(a\) and \(b\) to be distinct). Multiplying both sides by \(abN\) and subtracting terms gives \[2ab-Na-Nb=0.\] Unfortunately, this does not appear to factorise nicely because of the \(2\), so we try halving to get \[ab-\tfrac{1}{2}Na-\tfrac{1}{2}Nb=0.\] This is now almost what we had before, and we can factorise to get \[(a-\tfrac{1}{2}N)(b-\tfrac{1}{2}N)=\tfrac{1}{4}N^2.\] Multiplying through by \(4\) removes the fractions, giving \[(2a-N)(2b-N)=N^2,\] so as before we must have \(2a-N=1\), \(2b-N=N^2\) (as \(a\ne b\)), so \(a=\frac{N+1}{2}\), \(b=\frac{N^2+N}{2}\). These are both integers as \(N\) is an odd prime (making \(N+1\) and \(N^2+N=N(N+1)\) both even), so there is a unique way to express \(2/N\) as the sum of two distinct unit fractions.

#### Method 2

In this final part to the question, we are asked to solve a problem that is a modified version of an earlier part.

Sometimes we can do that using a similar approach, and that’s what we did in Method 1.

But sometimes we can also do it by using the answer to the earlier part to help with the later part, and that’s what we do in Method 2.

For the last part, we want to prove that if \(N>2\) is prime then we can uniquely write \[\frac{2}{N}=\frac{1}{c}+\frac{1}{d}\] with \(c\not= d\). If we have such a way then \[\frac{1}{N}=\frac{1}{2c} + \frac{1}{2d}\] where both \(2c\) and \(2d\) are integers. On the other hand, since \(N\) is a prime we already know that \[\frac{1}{N}=\frac{1}{N + 1} + \frac{1}{N(N+1)}\] is the unique way of expressing \(1/N\) as the sum of two distinct unit fractions. Moreover, as \(N\) is a prime greater than \(2\) it must be odd and therefore, both \(N+1\) and \(N(N+1)\) are even. Thus, \((N+1)/2\) and \(N(N+1)/2\) are integers and therefore, there is indeed a unique way of expressing \(2/N\) as the sum of two distinct unit fractions, namely \[\frac{2}{N}=\frac{1}{\frac{N+1}{2}}+\frac{1}{\frac{N(N+1)}{2}}.\]