Selection tests for the Gulf Mathematical Olympiad 2013

We have $$0\equiv (a^3+1)-(b^3+1)\equiv (a-b)(a^2+ab+b^2)\equiv (a-b)ab(\mathrm{mod}{a}^2+b^2).$$ Let $d$ be a common divisor of $a$ and $a^2+b^2$. Then $d$ divides $a^3+1$ and $a^3$, so it divides $1$. Hence $a$ and $a^2+b^2$ are coprime. In a similar way $b$ and $a^2+b^2$ are coprime. Thus $a-b\equiv 0(\mathrm{mod}{a}^2+b^2)$.

If $a\ne b$ then $a^2+b^2\le |a-b|\le (a-b{)}^2<a^2+b^2$, since $ab\ge 1$, which is a contradiction. Hence $a=b=1$, since $a$ and $a^2+b^2$ are coprime.