XX OM - II - Task 2

Find all four-digit numbers in which the thousands digit is equal to the hundreds digit, and the tens digit is equal to the units digit, and which are squares of integers.
Suppose the number $ x $ satisfies the conditions of the problem and denote its consecutive digits by the letters $ a, a, b, b $. Then






The number $ x $ is divisible by $ 11 $, so as a square of an integer, it is divisible by $ 11^2 $, i.e., $ x = 11^2 \cdot k^2 $ ($ k $ - an integer), hence






Therefore,






The number $ a+b $ is thus divisible by $ 11 $. Since $ 0 < a \leq 9 $, $ 0 \leq b \leq 9 $, then $ 0 < a+b \leq 18 $, hence






Therefore, we conclude that $ b \ne 0 $, $ b \ne 1 $; since $ b $ is the last digit of the square of an integer, it cannot be any of the digits $ 2, 3, 7, 8 $. Thus, $ b $ is one of the digits $ 4, 5, 6, 9 $. The corresponding values of $ a $ are $ 7, 6, 5, 2 $, so the possible values of $ x $ are only the numbers $ 7744 $, $ 6655 $, $ 5566 $, $ 2299 $. Only the first one is a square of an integer.
The problem has one solution, which is the number $ 7744 $.