Austria 2014

We have the trivial estimate $a^2=b\cdot (b+7)\ge b^2$ resulting in $a\ge b$ due to the non-negativity of $a$ and $b$. On the other hand, the inequality between the arithmetic and the geometric mean implies that $$a=\sqrt{b(b+7)}\le \frac{b+(b+7)}{2}=b+\frac{7}{2}.$$ Combining these inequalities shows that $a\in \{b,b+1,b+2,b+3\}$. Inserting these cases into the original equation only yields solutions for $a=b$ or $a=b+3$. These are $(a,b)\in \{(0,0),(12,9)\}$. $\square$