VIII OM - I - Task 6

Find a four-digit number, whose first two digits are the same, the last two digits are the same, and which is a square of an integer.
If $ x $ is the number sought, then






where $ a $ and $ b $ are integers satisfying the inequalities $ 0 < a \leq 9 $, $ 0 \leq b \leq 9 $. The number $ x $ is divisible by $ 11 $, since






Since $ x $ is a perfect square, being divisible by $ 11 $ it must be divisible by $ 11^2 $, so the number






is divisible by $ 11 $. It follows that $ a + b $ is divisible by $ 11 $, and since $ 0 < a + b \leq 18 $, then $ a + b = 11 $. Therefore,






from which we infer that $ 9a + 1 $ is the square of some natural number $ m $:






Since $ 9a + 1 \leq 82 $, then $ m \leq 9 $.
From the above,






It follows from this equality that the product $ (m + 1) (m - 1) $ is divisible by $ 9 $, and since at most one of the numbers $ m + 1 $ and $ m - 1 $ is divisible by $ 3 $, then one of them is divisible by $ 9 $. Considering that the natural number $ m $ is less than $ 10 $, we conclude from this that $ m + 1 = 9 $, so $ m = 8 $. In this case, $ a = 7 $, $ b = 4 $, and the sought number is $ 7744 = (88)^2 $.