\section*{Problem 4 - 261034}

Determine among all the values that

\[
z=x^{2}+y^{2}+2 x-22
\]

can take for integer \(x\) and \(y\), the smallest value of \(z\) that is a natural number!

Give all pairs \((x ; y)\) of integers for which this value of \(z\) is obtained in (1)!
}

Transforming (1) yields

\[
z=x^{2}+y^{2}+2 x-22=(x+1)^{2}+y^{2}-23
\]

For \(z\) to be a natural number, \((x+1)^{2}+y^{2}-23 \geq 0\) must hold. The smallest natural number is 0, meaning \((x+1)^{2}+y^{2}=23\) must be true.

However, 23 cannot be represented as the sum of two squares, nor can 24. The number 25 is the first that can be represented as the sum of two integer squares, meaning \(z=2\) is the smallest value that is a natural number.

To determine the corresponding pairs \((x, y)\), 25 is decomposed as follows:

\[
(x+1)^{2}+y^{2}=25=0+25=25+0=9+16=16+9
\]

Thus, the following possibilities and resulting pairs are possible:

\begin{tabular}{ccccc|c}
\((x+1)^{2}\) & \(x+1\) & \(x_{i}\) & \(y^{2}\) & \(y_{i}\) & Pairs \((x, y)\) \\
\hline 0 & 0 & -1 & 25 & \(\pm 5\) & \((-1,5) ;(-1,-5)\) \\
25 & \(\pm 5\) & \(-6 ; 4\) & 0 & 0 & \((-6,0) ;(4,0)\) \\
9 & \(\pm 3\) & \(-4 ; 2\) & 16 & \(\pm 4\) & \((-4,-4) ;(-4,4) ;(2,-4) ;(2,4)\) \\
16 & \(\pm 4\) & \(-5 ; 3\) & 9 & \(\pm 3\) & \((-5,-3) ;(-5,3) ;(3,-3) ;(3,3)\)
\end{tabular}

Thus, there are 12 pairs \((x, y)\) for which \(z\) can take the smallest possible value of 2.