If $\frac {1} {a} + \frac {1} {b}= \frac {1} {c}$ ; $a,b,c$ are natural numbers with no common factors. Prove that $(a+b)$ is a perfect square.
Source: Indian National Mathematics Olympiad by Rajeev Manocha.
What I have tried.
$\frac {1} {a} + \frac {1} {b}= \frac {1} {c}$
$\implies (a+b)c=ab$
$\implies a|(a+b)c$
But, $g.c.d(a,c)=1$
$\implies a|(a+b)$
$\implies a|b$
But, $g.c.d(a,b)=1$
$\therefore a=1$
In a similar way I have found $b=1$
But then $c=\frac {1} {2}$ which is contradicting that $c$ is a natural number.
The question asked to prove that $a+b$ is a square but I have proved that there is no integral solution of the given equation. Does my solution contain any error?
Assuming that $gcd(a,b,c)=1$.
Let $d=gcd(a,b)$. Then $a=da', b=db'$ and $gcd(d,c)=gcd(a',b')=1$.
$$ (a+b)c=ab \implies d(a'+b')c=d^2a'b' \implies (a'+b')c=da'b'$$
Now $gcd(a',b')=1$ implies that $$a'b'|c$$
So let $c=a'b'c'$, the equation becomes $$(a'+b')c'=d$$
Since $gcd(c,d)=1$ you get $gcd(c',d)=1$ and hence, the above implies $$a'+b'=d \\ c'=1$$
Then $$a+b=da'+db'=d(a'+b')=d^2$$
P.S. Baring an embarassing mistake, if we set $n:=a', m:=b'$ we get $d=m+n$ and hence $$a=(m+n)m \\ b=(m+n)n \\ c=mn$$ This shows that the general solution to $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ with $gcd(a,b,c)=1$ is given by the above with $gcd(m,n)=1$.
It is not hard to see that $gcd(a,b,c)=1 \Leftrightarrow gcd(m,n)=1$.