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.

Ah well, since Jack deleted his answer, I'll fix it for him.
The equation is equivalent to $ac+bc=ab$ or $(a-c)(b-c)=c^2$.
Claim: $\gcd(a-c,b-c)=1$. 
Proof: $\gcd(a,b,c)=1$ implies $\gcd(a-c,b-c,c)=1$ which implies $\gcd(a-c,b-c,c^2)=1$. But $a-c\mid c^2$, so $\gcd(a-c,b-c,c^2)=\gcd(a-c,b-c)$.

When the product of two relatively prime positive integers is a perfect square, the two numbers are perfect squares. So:
$$a-c=d^2$$ for some $d\mid c$, and $b-c=\frac{c^2}{d^2}$.
So $$a+b = c+d^2 + c+\frac{c^2}{d^2} = \left(d+\frac{c}{d}\right)^2$$
[ In particular, Jack's assumption that $d=1$ or $d=c$ was wrong because in the case:
$$\frac{1}{10} + \frac{1}{15}=\frac{1}{6}$$
we have $d=2$.]
