Saudi Arabia Mathematical Competitions

Let $d=c+1$. The equality $a-b=a^{2}c-b^2(c+a)$ implies $$(a-b)[c(a+b)-1]=b^2.$$ But $c(a+b)-1$ and $a-b$ are relatively prime. Indeed, if $p$ is a prime dividing $a-b$, then the above equality shows that $p$ also divides $b$, so $p$ will divide $a+b=(a-b)+2b$. Hence $p$ cannot divide $c(a+b)-1$. It follows that $|a-b|$ is a perfect square as well. A first nontrivial example is $18-22=18^2(-3)-22^2(-2)$.