## Task 1 - 070931

All four-digit square numbers are to be determined without using the number table, whose first two and last two basic digits are each equal.
Let $x$ be the first and $y$ the last digit of such a square number $z$. Then we have

$$
z=1000 x+100 x+10 y+y=1100 x+11 y=11(100 x+y)
$$

with $1 \leq x \leq 9$ and $0 \leq y \leq 9$.

This implies $11 \mid z$ and, since $z$ is a square number and 11 is a prime number, even $11^2 \mid z$. Therefore, $11 \mid 100 x+y$ must hold. However, $100 x+y=99 x+x+y$, and thus $11 \mid x+y$.

Due to the restrictions on $x$ and $y$, only $1 \leq x+y \leq 18$ and thus $x+y=11$ is possible.

This implies $100 x+y=99 x+x+y=11(9 x+1)$ and thus $z=11^2(9 x+1)$.

Since $z$ is a square number, $9 x+1$ must also be a square number. Given $1 \leq x \leq 9$, this is (as can be determined, for example, by calculating the numbers $9 x+1$ for $x=1, \ldots, 9$) only the case for $x=7$.

Conversely, $x=7$ indeed leads to $9 \cdot 7+1=64$, resulting in the square number $z=121 \cdot 64=7744=88^2$. This is thus the only four-digit square number with the required conditions.

## Solution adopted from $[5]$