6. (2004 Spanish Mathematical Olympiad) Find all two-digit positive integers $a, b$, such that $100a + b$ and $201a + b$ are both four-digit numbers and perfect squares.
6. Let $100a + b = m^2, 201a + b = n^2$, then $101a = n^2 - m^2 = (n - m)(n + m), m, n < 100$.
Therefore, $n - m < 100, n + m < 200, 101 \mid (m + n)$.
Thus, $m + n = 101$.
Substituting $a = n - m = 2n - 101$, we get
$201(2n - 101) + b = n^2$, which simplifies to $n^2 - 402n + 20301 = b \in (9, 100)$.
Upon verification, $n = 59, m = 101 - n = 42$.
Thus, $a = n - m = 17, b = n^2 - 402n + 20301 = 64$, i.e., $(a, b) = (17, 64)$.