South-Afrika, 2011

We know that
$$(a-b)^2 + (ab+1)^2 = a^2b^2 + a^2 + b^2 + 1 = (a^2+1)(b^2+1),$$
so it will be enough to show that $a^2+1$ and $b^2+1$ are relatively prime. Suppose that there is a prime number $p$ that divides both $a^2+1$ and $b^2+1$. Then $p$ divides $a^2-b^2$, and hence $p$ divides one of $a+b$ or $a-b$.
If $p$ divides $a-b$, then $p$ divides $ab-b^2$, and since $p$ divides $b^2+1$, it divides $ab-b^2+b^2+1 = ab+1$, which is impossible, since it is assumed that $a-b$ and $ab+1$ are relatively prime. If $p$ divides $a+b$, a similar contradiction is derived.