Positive integer solutions to $\frac{1}{a} + \frac{1}{b} = \frac{c}{d}$ I was looking at the equation $$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$$ where $c$ and $d$ are positive integers such that $\gcd(c,d) = 1$.

I was trying to find positive integer solutions to this equation for $a, b$, given any $c$ and $d$ that satisfy the above conditions. I was also trying to find whether there are additional requirements on $c$ and $d$ so that positive integer solutions for $a$ and $b$ can even exist.

I found that this equation simplifies to $abc - ad - bd = 0$ so that $abc = d(a+b)$.
Also, since the equation is equivalent to $a+b = ab(\frac{c}{d})$, this means $a$ and $b$ are the roots of the quadratic $dx^2-abcx+abd = 0$ since their product is $ab$ and their sum is $a+b = ab(\frac{c}{d})$.
However, after I analyzed the quadratic I just ended up with $a = a$ and $b = b$.
Any ideas on how to solve this further?
Again, I need to find all the conditions on the positive integers $c$ and $d$ (where $\gcd(c,d) = 1$) such that positive integer solutions for $a, b$ can exist. And then also find the positive integer solutions for $a$ and $b$ given that those conditions are satisfied.

$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$
Above has solution:
$a=(3k-2)$
$b=(k-2)(3k-2)$
$c=(k-1)(3k-2)$
$d=(k-2)(3k-2)^2$
For k=5 we get:
$(a,b,c,d)=(13,39,52,507)$
