31st Junior Turkish Mathematical Olympiad

Assume the contrary. The sum of the numbers is $$\frac{2n^{2}m(n^{2}m-1)}{(m^2+n^2)(m^2-n^2)}$$ Since $m$ and $n$ are relatively prime, we have $$\gcd (n^{2}m,m^2+n^2)=\gcd (n^{2}m,m^2-n^2)=1.$$ Then we obtain $(m^2+n^2)(m^2-n^2)\mid 2(n^{2}m-1)$. Note that $m,n\ge 1$ and we can not have $m=n=1$, and hence $n^{2}m-1>0$. Therefore, $|(m^2+n^2)(m^2-n^2)|\le 2(n^{2}m-1)<2n^{2}m$. As $m^2+n^2>n^2$, we get $|m^2-n^2|<2m$, and thus, $(m-1{)}^2-1<n^2<(m+1{)}^2-1$. So we have $n^2=m^2$ or $n^2=(m-1{)}^2$. It is clear that $n^2=m^2$ is impossible. Hence, $n^2=(m-1{)}^2$ which yields $n=m-1$. But in that case we obtain that one of the given numbers is not an integer, contradiction.