\section*{Problem 1 - 091041}

To be determined are all pairs of natural numbers such that each pair, together with the number 41, forms a triplet for which both the sum of the three numbers of the triplet and the sum of any two numbers arbitrarily selected from the triplet are squares of natural numbers.
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Natural numbers \(a, b \in \mathbb{N}\) and \(p, q, r, s \in \mathbb{N}\) are sought, such that the equations

\[
a+b+41=p^{2} a+b=q^{2} a+41=r^{2} b+41=s^{2}
\]

are satisfied. The difference of the first two equations yields \(41=p^{2}-q^{2}=(p+q)(p-q)\). Since \(p, q\) are natural numbers, the only solution is \(p+q=41, p-q=1 \Longleftrightarrow p=21, q=20\) and thus \(a+b=400\).

The last two equations directly yield the estimate \(r, s \geq 7\). By adding these, we obtain \(r^{2}+s^{2}=a+b+82=482\), from which \(r, s \leq 20\) follows. The last digit of a square number can only take the values \(0,1,4,5,6,9\).

For the sum of two square numbers to have the last digit 2, the last digits of \(r^{2}, s^{2}\) can only be 1,6, and thus the last digits of \(r, s\) can only be 1,4,6,9. Therefore, \(r, s \in\{9,11,14,16,19\}\) and it is found that \((r, s)=(11,19)\) or \((r, s)=(19,11)\) must hold. This results in \((a, b)=(80,320)\) and \((a, b)=\) \((320,80)\) as the only solutions.