IMO 1988 SL 25

Origin: GBR

Problem

A positive integer is called a double number if its decimal rep- resentation consists of a block of digits, not commencing with 0, followed immediately by an identical block. For instance, 360360 is a double num- ber, but 36036 is not. Show that there are infinitely many double numbers that are perfect squares.

Solution

Observe that 1001 = 7 \cdot 143, i.e., 103 = −1 + 7a, a = 143. Then by the binomial theorem, 1021 = (−1 + 7a)7 = −1 + 72b for some integer b, so that we also have 1021n \equiv−1 (mod 49) for any odd integer n > 0. Hence N = 49(1021n + 1) is an integer of 21n digits, and N(1021n + 1) =
3 7(1021n + 1) 2 is a double number that is a perfect square.