$2 \cdot 73$ Let positive integers $a, b$ make $15a + 16b$ and $16a - 15b$ both squares of positive integers. Find the smallest value that the smaller of these two squares can take.
[Solution] Let positive integers $a, b$ such that $15a + 16b$ and $16a - 15b$ are both squares of positive integers. Then we can set
$$
\left\{\begin{array}{l}
15a + 16b = r^2 \\
16a - 15b = s^2
\end{array}\right.
$$

where $r, s \in \mathbb{N}$.
Thus, $15r^2 + 16s^2 = 15^2 \cdot a + 16^2 \cdot a$,

which is
$$
15r^2 + 16s^2 = 481a \text{.}
$$

Also,
$$
16r^2 - 15s^2 = 16^2b + 15^2b \text{,}
$$

which is
$$
16r^2 - 15s^2 = 481b \text{.}
$$

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

If one of $r$ and $s$ is not a multiple of 13, then since $15r^2 + 16s^2$ is a multiple of 13, the other one is also not a multiple of 13.
Since $16r^2 - 15s^2$ is a multiple of 13, we have
$$
\begin{array}{l}
16r^2 \equiv 15s^2 \quad(\bmod 13), \\
16r^2 \cdot s^{10} \equiv 15s^{12}(\bmod 13),
\end{array}
$$

By Fermat's Little Theorem, since $(s, 13) = 1$, we have
$$
s^{12} \equiv 1 \quad(\bmod 13) \text{.}
$$

Thus,
$$
16r^2 \cdot s^{10} \equiv 15 \equiv 2 \quad(\bmod 13).
$$

That is,
$$
\left(4rs^5\right)^2 \equiv 2 \quad(\bmod 13).
$$

The left side of the above equation, $4rs^5$, is coprime with 13, so
$$
\left(\left(4rs^5\right)^2\right)^6 \equiv 1 \quad(\bmod 13).
$$

From (1), we get
$$
\begin{array}{cc}
2^6 \equiv 1 & (\bmod 13) \\
12 \equiv 1 & (\bmod 13), \text{ contradiction.}
\end{array}
$$

Therefore, $r$ and $s$ are both multiples of 13.
If one of $r$ and $s$ is not a multiple of 37, then the other one is also not a multiple of 37.
Since $15r^2 + 16s^2$ and $16r^2 - 15s^2$ are both multiples of 37, $r^2 - 31s^2$ is a multiple of 37, so we have
$$
\begin{array}{l}
r^2 \equiv 31s^2 \quad(\bmod 37) \\
r^2 \cdot s^{34} \equiv 31 \cdot s^{36} \equiv 31 \quad(\bmod 37)
\end{array}
$$

Taking the 18th power on both sides, we get
$$
\begin{array}{l}
\left(rs^{17}\right)^{36} \equiv 31^{18} \quad(\bmod 37) \\
1 \equiv (-6)^{18} \equiv \left((-6)^2\right)^9 \equiv (-1)^9 \equiv -1 (\bmod 37),
\end{array}
$$

contradiction.
Therefore, $r$ and $s$ are also multiples of 37. Thus, $r$ and $s$ are both multiples of 481, and $r^2$ and $s^2$ are both multiples of $481^2$.
On the other hand, let $a = 481 \times 31, b = 481$, then we have
$$
\begin{array}{l}
15a + 16b = 481 \times (15 \times 31 + 16) = 481^2, \\
16a - 15b = 481 \times (16 \times 31 - 15) = 481^2.
\end{array}
$$

In conclusion, the smallest value that the smaller of these two squares can take is $481^2$.