Consider fractions such as $\frac{2}{5}$ and $\frac{2}{7}$ expressed as the sum of two unit fractions. Respectively, they can be expressed as $\frac{1}{3}+\frac{1}{15}$ and $\frac{1}{4}+\frac{1}{28}$.
Generally, $\frac{2}{n}$ can be expressed as $\frac{1}{a}+\frac{1}{b}$ where $\frac{b}{a}=n$ and $a=\frac{n+1}{2}$.
Substituting $a=\frac{n+1}{2}$ and $\frac{b}{a}=n$, it can clearly be seen why this must always be true.
However, I am wondering if there is only one unique way to express an irreducible fraction of the form $\frac{2}{n}$ as the sum of two distinct unit fractions. We have shown $\frac{2}{5}=\frac{1}{3}+\frac{1}{15}$, but can this be done any other way.
My attempt at proving this, makes me think there is only one way. However, I cannot complete the proof.
Here is my attempt.
$$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$$
$$2ab=n(a+b)$$
$$b(2a-n)=na$$
$$\frac{b(2a-n)}{a}=n$$
As $n$ and $a$ must both be positive integers $2a-n$ must also be a positive integer and must also be odd, as $ n$ must be odd, by definition of the irreducible fraction, must be coprime with $2$.
So obviously $2a-n=2k+1$ for some integer $k$. When $k=0$ we get our intial solution stated at the beginning. However, I cannot seem prove that for other values of $k$ this fails to work, and that the denominators are no longer integers etc. How can I proceed next?
Best Answer
For an odd integer $n$, consider the equation $\frac{2}{n} = \frac{1}{a}+\frac{1}{b}$. Then $2ab = na+nb$. This means that $(2a-n)(2b-n)=n^2$.
Notice that $\frac{2}{n} = \frac{1}{a}+\frac{1}{b}>\frac{1}{a}$ implies $2a-n>0$, and similarly $2b-n>0$.
This means that the $(2a-n)(2b-n)$ is a factorization of $n^2$ as two positive integers.
Now, given $uv=n^2$ any factorization of $n^2$ as two positive integers we can take $2a-n = u$ and $2b-n = v$, so $a=\frac{u+n}{2}$ and $b = \frac{v+n}{2}$ which are positive integers because $u$, $v$ and $n$ are odd.
This means that the solutions to your equation are in bijection with the factorizations of $n^2$, or equivalently, with the positive divisors $u$ of $n^2$.
For example, take $n=15$.
The divisors of $n^2$ are $1, 3, 5, 9, 15, 25, 45, 75, 225$.
This give us the factorizations:
$u\cdot v = 1\cdot 225, 3\cdot 75, 5\cdot 45, 9\cdot 25, 15\cdot 15, 25\cdot 9, 45\cdot 5, 75\cdot 3, 225\cdot 1$
which in turn give us the solutions:
$(a,b) = (8,120),(9,45),(10,30),(12,20),(15,15),(20,12),(30,10),(45,9),(120,8)$
I'm guessing that you are considering the solutions $(a,b)$ and $(b,a)$ as the same one, and you want $a\ne b$ so we end for this example with the solutions
$(a,b) = (8,120),(9,45),(10,30),(12,20)$
In general, we'll have $\frac{\tau(n^2)-1}{2}$ solutions ($\tau(k)$ is the numbers of positive divisors of $k$).