Given primitive solution to $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, show $a+b$ is a perfect square If $a,b,c$ are positive integers and 
$\gcd(a,b,c)$ is $1$. Given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ then prove that $a+b$ is a perfect square.
I was trying to get something useful from the information given in the question but was unable to get that.

$$
\frac1a+\frac1b=\frac1c\implies ac+bc=ab
$$
$a\mid bc$. Therefore, each prime that divides $a$ must divide one and only one of $b$ or $c$ (since $(a,b,c)=1$). Therefore, $a=(a,b)(a,c)$. Similarly, $b=(a,b)(b,c)$ and $c=(a,c)(b,c)$.
Thus,
$$
a+b=\frac{ab}c=\frac{(a,b)(a,c)(a,b)(b,c)}{(a,c)(b,c)}=(a,b)^2
$$
