China Southeastern Mathematical Olympiad

By the condition of the problem, we have $a^2\mid (a^3+b^3+c^3)$, $b^2\mid (a^3+b^3+c^3)$ and $c^2\mid (a^3+b^3+c^3)$. Since $a$, $b$ and $c$ are coprime, we see that $a^2{b}^2{c}^2\mid (a^3+b^3+c^3)$.

Without loss of generality, suppose that $a\ge b\ge c$, so $$3a^3\ge a^3+b^3+c^3\ge a^2{b}^2{c}^2\Rightarrow a\ge \frac{b^2{c}^2}{3},$$ and $$2b^3\ge b^3+c^3\ge a^2\Rightarrow 2b^3\ge \frac{b^4{c}^4}{9}\Rightarrow b\le \frac{18}{c^4}.$$ We see that if $c\ge 2\Rightarrow b\le 1$, the result contradicts $b\ge c$. Thus, $c=1$.

If $c=1$ and $b=1$, then $a=1$, so $(a,b,c)=(1,1,1)$ is a solution.

If $c=1$, $b\ge 2$ and $a=b$, then $b^2\mid b^3+1$, which is a contradiction!

If $b\ge 2$ and $a>b>c=1$, then $$a^2{b}^2\mid (a^3+b^3+1)\Rightarrow 2a^3\ge a^3+b^3+1\ge a^2{b}^2\Rightarrow a\ge \frac{b^2}{2},$$ and by $c=1$, $$a^2\mid (b^3+1)\Rightarrow b^3+1\ge a^2\ge \frac{b^4}{4}\Rightarrow 4b^3+4\ge b^4.$$