$\mathbf{N 4 2}$ (37-4, Russia) 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.
Let positive integers $a, b$ be such that $15a + 16b$ and $16a - 15b$ are both squares of positive integers. Let
$$
15a + 16b = r^2, \quad 16a - 15b = s^2,
$$

where $r, s$ are positive integers. Thus,

$$
\begin{array}{c}
15^2 a + 16^2 a = 15r^2 + 16s^2, \\
481a = 15r^2 + 16s^2. \\
16^2 b + 15^2 b = 16r^2 - 15s^2, \\
481b = 16r^2 - 15s^2.
\end{array}
$$

Therefore, $481 \mid 15r^2 + 16s^2, 481 \mid 16r^2 - 15s^2$. We will prove that $481 \mid r, 481 \mid s$. Since $481 = 13 \times 37$, it suffices to prove that $r, s$ are both multiples of 13 and 37.

First, we prove that $r, s$ are both multiples of 13. We use proof by contradiction. If not, at least one of $r, s$ is not divisible by 13, and since $13 \mid 16r^2 - 15s^2$, we have $13 \mid r, 13 \mid s$. Since
$$
16r^2 \equiv 15s^2 \quad (\bmod 13),
$$

we have
$$
16r^2 \cdot s^{10} \equiv 15s^{12} \equiv 15 \equiv 2 \quad (\bmod 13).
$$

Taking the 6th power on both sides, we get
$$
\begin{array}{c}
(4rs^5)^{12} \equiv 64 \quad (\bmod 13), \\
1 \equiv -1 \quad (\bmod 13),
\end{array}
$$

which is a contradiction.

Next, we prove that $37 \mid r, 37 \mid s$. If not, then $37 \nmid r, 37 \nmid s$. Since
$$
37 \mid 15r^2 + 16s^2, \quad 37 \mid 16r^2 - 15s^2,
$$

we have
$$
37 \mid r^2 - 31s^2,
$$

which means
$$
\begin{array}{c}
r^2 \equiv 31s^2 \quad (\bmod 37) \\
r^2 s^{34} \equiv 31s^{36} \equiv 31 \quad (\bmod 37).
\end{array}
$$

Taking the 18th power on both sides, we get
$$
(r s^{17})^{36} \equiv 31^{18} \quad (\bmod 37).
$$

Since $(r s^{17})^{36} \equiv 1 (\bmod 37)$ and $31^{18} \equiv (31^2)^9 \equiv ((-6)^2)^9 \equiv 36^9 \equiv (-1)^9 \equiv -1 \quad (\bmod 37)$, we have a contradiction.

Therefore, $481 \mid r, 481 \mid s$. Thus, these two squares are both at least $481^2$.
On the other hand, when $a = 481 \times 31, b = 481$, both perfect squares are $481^2$. Therefore, the minimum value is $481^2 = 231361$.