If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $$\frac{1}{a} + \frac{1}{b}= \frac{1}{c}$$ then show that $a+b$ is a perfect square.
This can be simplified to: $$a+b = \frac{ab}{c}$$
Also, first few such examples of $(a,b,c)$ are $(12, 4, 3)$ and $(20, 5, 4)$. So, I have a feeling that $b$ and $c$ are consecutive.
I don't think I have made much progress.
Any help would be appreciated.
On
To show this, we note that $c(a+b)=ab$. Now let $g$ be the gcd of $a$ and $b$, which need not necessarily be $1$. Denote $a=a'g$ and $b=b'g$ so that we get $c(a'+b') = a'b'g$.
Because $a' + b'$ is relatively prime to both $a'$ and $b'$, it follows that it divides $g$. But g also divides $c(a'+b')$. Further, note that $g$ is coprime to $c$, because it was the gcd of $a$ and $b$, so that gcd($g$,$c$)=gcd($a$,$b$,$c$)=1. It follows that $a'+b'=g$, and therefore that $a+b = (a'+b')g = g^2$.
For example, $\frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ and $(3,6)=3$ and $3+6=9=3^2$.
Rewrite as $(a-c)(b-c)=c^2$. First we show that $a-c$ and $b-c$ are relatively prime. Suppose to the contrary that the prime $p$ divides $a-c$ and $b-c$. Then $p$ divides $c$ and therefore $a$ and $b$, contradicting the fact that $\gcd(a,b,c)=1$.
Since $a-c$ and $b-c$ are relatively prime, it follows that $a-c=s^2$ and $b-c=t^2$, where $st=c$.
We conclude that $a=s^2+st$ and $b=t^2+st$, so $a+b=(s+t)^2$.