"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 !

We have, $ab=(a+b)c$. Then $$a+b=(a/m)(b/n)=a^\prime b^\prime$$ where $a=ma^\prime ,b=nb^\prime$ and $c=mn$.
Now, let a prime $p\mid a^\prime$. Then, $p\mid a$ and hence $p\mid b$. If $p\mid n$, then $p\mid c$ but this is a contradiction. Hence $p\mid b^\prime$. Similarly, if a prime $p\mid b^\prime$ then $p\mid a^\prime$.
Hence, $a^\prime$ and $b^\prime$ has same prime factors. Now suppose for some prime $p$, let $r,s$ (with $r<s$) be the highest exponents such that $p^r\mid a^\prime$ and $p^s\mid b^\prime$. Then $p^s\mid a^\prime b^\prime, p^s\mid b$. Now if $p^s \mid  a$ then $p\mid m$, but then $p\mid c$ but this contradicts $\gcd (a,b,c)=1$. Hence prime factorizations of $a^\prime$ and $b^\prime$ are same and hence $a+b$ is square.
