<s>[INST] 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.
 [/INST] 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 $d=2$.]
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