Remainders after division by 100

$n$ is either odd $(2k+1)$ or even $(2k)$ , thus $n^2$ is either $(2k+1)^2 = 4k^2+4k+1$ or $(2k)^2 = 4k^2$ . Thus, $n^2 \equiv 0 \pmod{4}$ or $n^2 \equiv 1 \pmod {4}$ . Also, $n^2 \equiv 0, 1, 4, 6, 9, 11, 14, 16, 19, 21, 24 \pmod {25}$ . Since $4, 25$ are coprime, each possibility is attainable by using the Chinese Remainder Theorem. Hence $|S|=22$