\section*{Problem 2 - 141022}

Give all (ordered) triples \((x, y, z)\) that satisfy the following conditions!

(1) \(x-y=96\),

(2) \(y-z=96\),

(3) \(x, y\) and \(z\) are squares of natural numbers.
}

We first determine all pairs \((u, v) \in \mathbb{N}^{2}\) for which \(u^{2}-v^{2}=96\). Since \((u+v)(u-v)=96\), we simply need to go through all factorizations \(96=a \cdot b\) of 96 with \(a>b>0\), where the two complementary factors \(a, b\) of 96 must also have the same parity. This ensures that

\[
u:=\frac{a+b}{2}, v:=\frac{a-b}{2}
\]

is a pair \((u, v) \in \mathbb{N}^{2}\) with the above properties. This leads to the following 4 cases:

- \((a, b)=(48,2) \mapsto(u, v)=(25,23) \quad ; \quad \bullet(a, b)=(24,4) \mapsto(u, v)=(14,10)\)

- \((a, b)=(16,6) \mapsto(u, v)=(11,5) \quad ; \quad \bullet(a, b)=(12,8) \mapsto(u, v)=(10,2)\)

The only way to satisfy equations (1)-(3) in the problem statement for \(x, y, z \in \mathbb{N}\) is to set \(x=14^{2}=196, y=10^{2}=100, z=2^{2}=4\).