\section*{Exercise 1 - 011121}

\(3,4,5\) is a so-called Pythagorean triplet, as \(3^{2}+4^{2}=5^{2}\).

It is the only such triplet whose elements differ by only 1. Are there other triplets for the equation \(a^{2}+b^{2}=c^{2}\) where \(c=b+1\)?

What pattern can you recognize here? Try to find an expression that allows you to quickly find such triplets!
}

There are other number triples that satisfy the conditions \(a^{2}+b^{2}=c^{2}\) and \(c=b+1\):

\[
\begin{gathered}
a^{2}+b^{2}=(b+1)^{2}=b^{2}+2 b+1 \Rightarrow a^{2}=2 b+1 \\
\Rightarrow b=\frac{a^{2}-1}{2}
\end{gathered}
\]

From (1), it can be easily seen that \(a^{2}\) and thus \(a\) must be an odd number.

A triple \((a, b, c)\) with the required properties can thus be quickly found by assigning \(a\) an odd number and calculating \(b\) using (2).

\(c\) is then 1 greater than \(b\).

Thus, such a triple can be determined for any arbitrary odd natural number \(a\).