Saudi Arabian 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 $18-22 = 18^{2}(-3) - 22^{2}(-2)$.