Problem

Source: 2006 Romania JBMO TST 5.4

Tags: Digits, number theory, Perfect Square



For a positive integer $n$ denote $r(n)$ the number having the digits of $n$ in reverse order- for example, $r(2006) = 6002$. Prove that for any positive integers a and b the numbers $4a^2 + r(b)$ and $4b^2 + r(a)$ can not be simultaneously squares.