Saudi Arabia Mathematical Competitions 2012

We will use the following result:

Lemma. If $x$ and $y$ are positive integers such that $\frac{x^2}{x+y}$ is an integer and prime, then $y\ge x$.

Proof of Lemma. Assume that $\frac{x^2}{x+y}=p$, where $p$ is a prime. We have $py=x(x-p)$, so $p\mid x(x-p)$, i.e. $p\mid x$. Let $x=up$, for some positive integer $u$. We get $$y=x(up-p)=x(u-1)p\ge x,$$ and we are done, since $u\ge 2$.

Applying Lemma in our situation, it follows $b\ge a$, $c\ge b$, $a\ge c$, meaning that $a=b=c$.