Problem source

A number of the form $1/N$, where $N$ is  an integer greater
than 1, is called a \textit{unit fraction}.
Noting that 
\[
 \frac1 2 =\frac13 + \frac16\\\ 
\mbox{ and  }
\frac13 
= \frac14
+ \frac1{12},
\]
guess a general result of the form
$$
\frac1N
=\frac1a +\frac1b
\tag{*}
$$
and hence
 prove that any unit fraction can be expressed as the sum of 
two distinct unit fractions.
By writing $(*)$ 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.
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.

Solution source

Notice that $\frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)}$, so any unit fraction can be expressed as the sum of two distinct unit fractions.

\begin{align*}
&& \frac{1}N &= \frac1a + \frac1b \\
\Leftrightarrow && ab&= Nb+Na \\
\Leftrightarrow && 0 &= (a-N)(b-N)-N^2 \\
\Leftrightarrow && N^2 &= (a-N)(b-N)
\end{align*}

If $N$ is prime then the only factors of $N^2$ are $1,N$ and $N^2$. if $a-N = b-N = N$ then $a=b$ and we don't have distinct fractions. Therefore $a-N = 1$ and $b-N = N^2$ and we obtain the decomposition earlier (and it must be the only solution).


\begin{align*}
&& \frac2N &= \frac1a+\frac1b \\
\Leftrightarrow && 2ab &= Nb+Na \\
\Leftrightarrow && 4ab &= 2Na+2Nb \\
\Leftrightarrow && N^2 &= (2a-N)(2b-N)
\end{align*}

Therefore for $a,b$ to be distinct we must have $2a = N+1$ and $2b = N+N^2$ as the only possible factorisation. Both of the right hand sides are even so we can write

\[ \frac{1}{N} = \frac{1}{\frac{N+1}{2}} + \frac{1}{\frac{N(N+1)}{2}} \]

and this is unique