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.

Let $d=(a,b)$ be the g.c.d. of $a$ and $b$. We will show that $a+b=d^2$. Write $a=a_1d$ and $b=b_1d$. You also have $(a_1,b_1)=1$. We thus have $$\frac{1}{a}+\frac{1}{b}=\frac{1}{d}\left(\frac{1}{a_1}+\frac{1}{b_1}\right)=\frac{1}{c}$$$$\frac{a_1+b_1}{a_1b_1}=\frac{d}{c}$$ Observing that $(d,c)=1$ and $(a_1+b_1,a_1b_1)=1$, we get that $a_1+b_1=d$ and $a_1b_1=c$. But then $$a+b=d(a_1+b_1)=d^2$$
