If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ then show that $a+b$ is a square. 
If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $$\frac{1}{a} + \frac{1}{b}= \frac{1}{c}$$ then show that $a+b$ is a perfect square.

This can be simplified to:
$$a+b = \frac{ab}{c}$$
Also, first few such examples of $(a,b,c)$ are $(12, 4, 3)$ and $(20, 5, 4)$. So, I have a feeling that $b$ and $c$ are consecutive.
I don't think I have made much progress.
Any help would be appreciated.

Rewrite as $(a-c)(b-c)=c^2$. First we show that $a-c$ and $b-c$ are relatively prime. Suppose to the contrary that the prime $p$ divides $a-c$ and $b-c$. Then $p$ divides $c$ and therefore $a$ and $b$, contradicting the fact that $\gcd(a,b,c)=1$.
Since $a-c$ and $b-c$ are relatively prime, it follows that $a-c=s^2$ and $b-c=t^2$, where $st=c$.
We conclude that $a=s^2+st$ and $b=t^2+st$, so $a+b=(s+t)^2$.
