How do I prove that any unit fraction can be represented as the sum of two other distinct unit fractions? A number of the form $\frac{1}{n}$, where $n$ is an integer greater than $1$, is called a unit fraction.
Noting that
$\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$
and
$\frac{1}{3} = \frac{1}{4} + \frac{1}{12}$,
find 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 other distinct unit
fractions.

In reply to DavidH's question "is the above decomposition of a unit fraction into a pair of distinct unit fractions unique?":
If $n$ is not prime, then we can write $n = n_1n_2$ with $n_1 \neq 1 \neq n_2$, and then we have the decomposition
$$\frac{1}{n} = \frac{1}{n_1}\frac{1}{n_2} = \frac{1}{n_1}\Big(\frac{1}{n_2+1} + \frac{1}{n_2(n_2+1)}\Big) = \frac{1}{n_1(n_2+1)} + \frac{1}{n_1n_2(n_2+1)} \\ = \frac{1}{n+n_1} + \frac{1}{n(n_2+1)}$$
which is different to the decomposition
$$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)}$$
Now suppose $n$ is prime. Suppose
$$\frac{1}{n} = \frac{1}{n+a} + \frac{1}{c}$$
Assume also, wlog, that $n + a < c$, which tells us that $a < n$. Then
$$\frac{1}{c} = \frac{1}{n} - \frac{1}{n+a} = \frac{a}{n(n+a)}$$
But $n$ is prime, so $a$ and $n$ are coprime, as are $a$ and $a + n$, so we must have $a = 1$ and $c = n(n+1)$, i.e. the decomposition
$$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)}$$
is unique when $n$ is prime.
