Since $n$ is even, we have $p \neq 2$. 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 $c \neq a$, $b$. 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. □