\section*{Problem 1 - 261021}

Determine all ordered pairs \((x ; y)\) of integers \(x\) and \(y\) for which

\[
3 x^{2}+3 y^{2}-18 x+12 y+27=0
\]

holds!
}

By the method of completing the square, from (1)

\[
(x-3)^{2}-9+(y+2)^{2}-4+9=0 \quad \Rightarrow \quad(x-3)^{2}+(y+2)^{2}=4
\]

Since \(x, y\) are integers, \((x-3),(y+2)\) must also be integers. Moreover, the summands in (2) are guaranteed to be \(>0\). The number 4 can be expressed as the sum of two integer squares as \(4=0+4=4+0\). This leads to two cases.

1. Case: \((x-3)^{2}=0\) and \((y+2)^{2}=4\)

This gives \(x=3\) and \(y+2= \pm 2\), resulting in the solution pairs \((3, 0)\) and \((3, -4)\) for \((x, y)\).

2. Case: \((x-3)^{2}=4\) and \((y+2)^{2}=0\)

This gives \(y=-2\) and \(x-3= \pm 2\), resulting in the solution pairs \((1, -2)\) and \((5, -2)\) for \((x, y)\).

A check confirms the four found pairs as solutions.