Mod 4, $x^2$ is either 0 or 1. Mod 25, $x^2$ can take on $11$ values. A simple CRT argument shows that there are at most 22 quadratic residues $\pmod{100}$. So 23 numbers (instead of 43) should suffice, since Pigeonhole tells us that two of these numbers are such that their squares are congruent $\pmod{100}$.