, 2013

We have
$$0 \equiv (a^{3} + 1) - (b^{3} + 1) \equiv (a - b)(a^{2} + a b + b^{2}) \equiv (a - b) a b \pmod{a^{2} + b^{2}}.$$
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 $a - b \equiv 0 \pmod{a^{2} + b^{2}}$.

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