Croatian Mathematical Olympiad

Let $m^2 + n^2 + p^2 - 2mn - 2mp - 2np = k^2$ for some positive integer $k$. Since
$$k^2 + 4mp = m^2 + n^2 + p^2 - 2mn + 2mp - 2np = (m - n + p)^2$$
is a perfect square, there exists a positive integer $l$ such that $(k+l)^2 = k^2 + 4mp$, i.e. $2kl + l^2 = 4mp$, and hence $l$ is even. Thus $l = 2a$ and $a^2 < a(k+a) = mp < p^2$, hence $p > a$.
Since $p$ is prime, it follows that $p \mid k+a$, i.e. $k+a = bp$ for some positive integer $b$. Thus $ab = m < p$,
$$(bp + a)^2 = (k + 2a)^2 = (k + l)^2 = (m - n + p)^2$$
and $n = p + ab \pm (bp + a)$.
However, $p + ab - (bp + a) = (1-b)(p-a) \le 0$, so $n = p + m + bp + a$ meaning that every choice of $(a, b)$ yields valid $n$.

Now assume that two different choices $(a, b)$ and $(a', b')$ yield the same $n$, i.e. $p+m+bp+a = p+m+b'p+a'$. It follows that $-p(b-b') = a-a'$ and hence $p \mid a-a'$, which is impossible because $0 < |a-a'| < m < p$.
Thus, the number of valid positive integers $n$ is equal to the number of divisors of $m$, which does not depend on $p$.