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