China Mathematical Olympiad

Since $n$ is even, we have $p\ne 2$. As $p\nmid n$, we have $p\nmid k$. We may assume without loss of generality $0<k<p$. Set $a=k$, $b=p-k$, then $c=\frac{n-k(p-k)}{p}=\frac{n+k^2}{p}-k$.

By assumption, $c$ is an integer, and $a,b$ are distinct positive integers. It remains to be shown that $c>0$, and $c\ne a$, $b$. By the AM-GM inequality, we have $\frac{n}{k}+k\ge 2\sqrt{n}>p$, thus $n+k^2>pk$, hence $c>0$. If $c=a$, then $\frac{n+k^2}{p}-k=k$, thus $n=k(2p-k)$. Since $n$ is even, $k$ is also even, as a consequence $n$ is divisible by $4$, which contradicts the fact that $n$ is square-free. If $c=b$, then $n=p^2-k^2$. Since $n$ is even, $k$ is odd, implying that $n$ is again divisible by $4$, which is a contradiction.

We conclude that $a,b,c$ satisfy all the requirements, completing the proof. □