"If $1/a + 1/b = 1 /c$ where $a, b, c$ are positive integers with no common factor, $(a + b)$ is the square of an integer" If $1/a  +  1/b  =  1 /c$   where $a, b, c$ are positive integers with no common factor, $(a + b)$ is the square of an integer.
I found this question in RMO 1992 paper !
Can anyone help me to prove this ?
May be it's too easy and simple . But , I'm just novice in number theory !

From the hypothesis we have
$$
(*)\quad (a-c)(b-c)=c^2.
$$
Let $d=gcd(a-c,b-c)$. Then $d^2|c^2$ and so  $d|c$. Since $d|(a-c), d|(b-c)$, we have $d|a, d|b, d|c$. Since $gcd(a,b,c)=1$, we get $d=1$. It follows from (*) that both $a-c$ and $b-c$ are perfect square numbers. Let $p^2=a-c, q^2=b-c$ with $p,q\in\mathbb{N}$. Then $c=pq$ and so
$$
a+b=(a-c)+(b-c)+2c=p^2+q^2+2pq=(p+q)^2.
$$ 
