\section*{Problem 1 - 331021}

Determine whether there exists a four-digit square number \(q\) with the following properties (1), (2)! If such numbers exist, find all such square numbers!

(1) All four digits of \(q\) are less than 7.

(2) If each digit of \(q\) is increased by 3, the resulting four-digit number is also a square number.
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Let the square number \(m^{2}\) be the one resulting from (2) and let it be given that \(n^{2}=q\), where \(m\) and \(n\) are positive integers.

Then it follows that \(m^{2}=n^{2}+3333\) or \((m-n)(m+n)=m^{2}-n^{2}=3333=3 \cdot 11 \cdot 101\). Since \(m^{2}\) and \(n^{2}\) are four-digit numbers, it holds that \(30m+n\), which is a contradiction to \(n>0\). Therefore, \(m+n=101\) and \(m-n=33\), so \(m=67\) and \(n=34\) follow.

Indeed, \(q^{2}=34^{2}=1156\) and \(m^{2}=67^{2}=4489=1156+3333\) and all digit conditions are also satisfied, so 1156 is the only four-digit square number \(q\) that satisfies (1) and (2).