6・65 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.
(37th International Mathematical Olympiad, 1996)
[Solution] Let positive integers $a, b$ satisfy that $15a + 16b$ and $16a - 15b$ are both squares of positive integers.
That is, $15a + 16b = r^2, 16a - 15b = s^2$.

Here $r, s \in \mathbb{N}$.
Eliminating $b$ yields $15^2a + 16^2a = 15r^2 + 16s^2$,

which is
$$
481a = 15r^2 + 16s^2.
$$

Eliminating $a$ yields $16^2b + 15^2b = 16r^2 - 15s^2$,
which is
$$
481b = 16r^2 - 15s^2.
$$

Therefore, $15r^2 + 16s^2$ and $16r^2 - 15s^2$ are both multiples of 481.
We will now prove that $r, s$ are also multiples of 481.
Since $481 = 13 \times 37$, it suffices to prove that $r, s$ are both multiples of 13 and 37.
First, we prove $13 \mid r, 13 \mid s$. We use proof by contradiction.
Assume $13 \nmid r, 13 \nmid s$. Since $13 \mid 16r^2 - 15s^2$,
then $16r^2 \equiv 15s^2 \pmod{13}$.
Because $13 \nmid r, 13 \nmid s$, and 13 is a prime, by Fermat's Little Theorem, $r^{12} \equiv 1 \pmod{13}, s^{12} \equiv 1 \pmod{13}$,
then (3) becomes $16r^2 \cdot s^{10} \equiv 15s^{12} \equiv 15 \equiv 2 \pmod{13}$,
and by Fermat's Little Theorem,
$\left(16r^2 \cdot s^{10}\right)^6 = \left(4rs^5\right)^{12} \equiv 1 \pmod{13}$,
and by (4), $2^6 \equiv 1 \pmod{13}$.
In fact, $2^6 = 64 \equiv -1 \pmod{13}$,
which is a contradiction, so $13 \mid r, 13 \mid s$.
Next, we prove $37 \mid r, 37 \mid s$. We still use proof by contradiction.
Assume $37 \nmid r, 37 \nmid s$.
Since $37 \mid 15r^2 + 16s^2, 37 \mid 16r^2 - 15s^2$,
then $37 \mid (16r^2 - 15s^2) - (15r^2 + 16s^2)$,
which is $37 \mid r^2 - 31s^2$,
so $r^2 \equiv 31s^2 \pmod{37}$.
And by $s^{36} \equiv 1 \pmod{37}$,
then $r^2s^{34} \equiv 31s^{36} \equiv 31 \pmod{37}$.
And $\left(r^2s^{34}\right)^{18} = \left(rs^{17}\right)^{36} \equiv 1 \pmod{37}$,
which is $31^{18} \equiv 1 \pmod{37}$.
However, $31^{18} \equiv (31^2)^9 \equiv (-6)^{18} \equiv 36^9 \equiv (-1)^9 \equiv -1 \pmod{37}$.

This contradicts the previous statement, so $37 \mid r, 37 \mid s$.
From the above, we know $481 \mid r, 481 \mid s$.
Thus, the perfect squares $15a + 16b$ and $16a - 15b$ are both at least $481^2$.
On the other hand, if we take $a = 481 \times 31, b = 481$, then
$15a + 16b = 481^2, 16a - 15b = 481^2$.
Therefore, the minimum value sought is $481^2$.