The diophantine equation $\frac{1}{a}+\frac{1}{b}=\frac{n}{a+b}$

Question

How do I solve This diophantine equation $\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{n}{a+b}$
of unknown $(a,b,n)\in \mathbb{(N^*)^3}$.

Answer 1

This is equivalent to $\frac{(a+b)^2}{ab}=n$. So a solution satisfies $ab | (a+b)^2$ which means $a|b^2$ and $b|a^2$. Further, this implies that $a$ and $b$ have the same prime divisors. So let $p_1,\dots,p_k$ be such that $a=\prod_{i=1}^k p_i^{\alpha_1}$ and $b=\prod_{i=1}^k p_i^{\beta_i}$.

Computing $\frac{(a+b)^2}{ab}$ now gives $2+\prod_{i=1}^k p_i^{\alpha_i-\beta_i}+\prod_{i=1}^k p_i^{\beta_i-\alpha_i}$. Since this is supposed to be an integer, we must have $\alpha_i=\beta_i$ for all $1\leq i\leq k$, so $a=b$.

So the solution set is $\{(a,a,4):a\in \mathbb{N}\}$.