\section*{Problem 5 - 331035}

Determine all pairs \((m ; n)\) of positive integers \(m\) and \(n\) for which

\[
\frac{m^{2}}{m+1}+\frac{n^{2}}{n+1}
\]

is an integer.
}

It is

\[
\begin{aligned}
z & :=\frac{m^{2}}{m+1}+\frac{n^{2}}{n+1}=\frac{m^{2}-1}{m+1}+\frac{n^{2}-1}{n+1}+\frac{1}{m+1}+\frac{1}{n+1} \\
& =\frac{(m-1)(m+1)}{m+1}+\frac{(n-1)(n+1)}{n+1}+\frac{1}{m+1}+\frac{1}{n+1}=n-1+m-1+\frac{1}{m+1}+\frac{1}{n+1}
\end{aligned}
\]

exactly an integer if and only if \(z-(n-1)-(m-1)=\frac{1}{m+1}+\frac{1}{n+1}=: x\) is also an integer. Clearly, \(0 < x\) and since \(m \geq 1\) and \(n \geq 1\), we have

\[
0 < x = \frac{1}{m+1} + \frac{1}{n+1} \leq \frac{1}{1+1} + \frac{1}{1+1} = \frac{1}{2} + \frac{1}{2} = 1
\]

so \(x = 1\) must hold, as it is the only integer in the interval \((0 ; 1]\). Therefore, equality must also have held in the previous estimates, i.e., \(m = n = 1\).

Indeed, for the pair \((m ; n) = (1 ; 1)\), the number \(z = \frac{1^{2}}{1+1} + \frac{1^{2}}{1+1} = \frac{1}{2} + \frac{1}{2} = 1\) is an integer, so there is exactly one pair of positive integers, namely the one just considered, that satisfies the condition of the problem.