3. Let positive integers $a, b$ be such that $15a + 16b$ and $16a - 15b$ are both squares of positive integers. Find the smallest value that the smaller of these two squares can take. ${ }^{[3]}$
(1996, China Mathematical Olympiad)
Given: $15a + 16b = r^2, 16a - 15b = s^2$ $\left(r, s \in \mathbf{N}_{+}\right)$.
Eliminating $a$ gives:
$481b = 16r^2 - 15s^2$
$\Rightarrow 481 \mid (16r^2 - 15s^2)$.
From equation (1):
$16r^2 \equiv 15s^2 \pmod{13}$.
If $13 \times r, 13 \times s$, then by Fermat's Little Theorem, $s^{12} \equiv 1 \pmod{13}$.
From equation (2):
$16r^2 s^{10} \equiv 15s^{12} \equiv 15 \equiv 2 \pmod{13}$.
Then $\left(4rs^5\right)^{12} = \left(16r^2 s^{10}\right)^6 \equiv 2^6 = 64$
$\equiv -1 \pmod{13}$.
This contradicts Fermat's Little Theorem.
Therefore, $13 \mid r, 13 \mid s$.
Similarly, it can be proven that $37 \mid r, 37 \mid s$ (omitted).
Since $(13, 37) = 1$, then $481 \mid r, 481 \mid s$.
Thus, $r^2 \geq 481^2, s^2 \geq 481^2$.
From $r^2 - s^2 = (15a + 16b) - (16a - 15b)$
$$
= -a + 31b,
$$

When $a = 481 \times 31, b = 481$,
$$
r^2 = s^2 = 481^2.
$$

Therefore, the minimum value sought is $481^2$.