2021/039) if $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ where a,b,c are positive integers with no common factor then prove that a+b is a perfect square

 We have 

$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$

or $bc + ac = ab$

or $(a-c)(b-c)= c^2$

Now let a prime divide c 

There are two cases 

p divides a-c and b-c in which case p divides a,b,c so they have a common factor p which cannot be true

or $p^2$ divides a-c or b-c so there exists m and n(either of them can be 1) such that

$c= mn$ and $(a-c) = m^2$ and $(b-c) = n^2$   

so $a = m^2 + c$  and $b = n^2 +c$

so $a+b= m^2 + n^2 + 2c = m^2 + n^2 + 2mn = (m+n)^2$

or $a+b$ is a perfect square