\section*{Problem 2 - 281012}

Antje wants to determine all four-digit natural numbers \(z\) that satisfy the following conditions (1), (2), (3):

(1) The first and second digits of \(z\) are the same.

(2) The third and fourth digits of \(z\) are the same.

(3) The number \(z\) is a perfect square.

Antje wants to solve this problem without using a table of numbers, a calculator, or any other computing device. How can she proceed?
}

I If \(z\) satisfies the required conditions and \(a, b\) are the first and third digit of \(z\) respectively, then from (1), (2): \(a\) and \(b\) are natural numbers with

\[
\begin{gathered}
1 \leq a \leq 9 \quad, \quad 0 \leq b \leq 9 \\
z=1000 a+100 a+10 b+b=11(100 a+b)
\end{gathered}
\]

Furthermore, according to (3), there exists a natural number \(n\) such that \(n^{2}=z\), thus \(n^{2}=11(100 a+b)\). The prime number 11 is therefore a divisor of \(n^{2}\) and consequently also a divisor of \(n\); hence there exists a natural number \(m\) with \(n=11 m\), thus

\[
11^{2} m^{2}=11(100 a+b) \quad ; \quad 11 m^{2}=100 a+b \quad ; \quad 11\left(m^{2}-9 a\right)=a+b
\]
i.e., \(a+b\) is divisible by 11. Since according to (4) \(1 \leq a+b \leq 18\) holds, this is only possible with

\[
a+b=11
\]

Thus, (6) leads to

\[
11 m^{2}=99 a+11 \quad ; \quad m^{2}=9 a+1
\]

For \(a=1, \ldots, 9\), \(9 a+1\) has the values \(10,19,28,37,46,55,64,73,82\); of these, only the value 64 generated for \(a=7\) is a perfect square. Therefore, and due to (7), (5), the conditions of the problem can only be satisfied with \(a=7, b=4, z=7744\).

II The number \(z\) satisfies (1), (2) and, since \(88^{2}=7744\), also (3).

With I. and II., it is proven that exactly the number \(z=7744\) satisfies the conditions of the problem.