Example 16 Let positive integers $a, b$ be such that $15a + 16b$ and $16a - 15b$ are both squares of positive integers, find the smallest possible values of these two squares $\left(\mathrm{IMO}_{37-4}\right)$.
Solution: Let $15a + 16b = x^2$ and $16a - 15b = y^2$, where $x, y \in \mathbf{Z}^{+}$. Then, $15x^2 + 16y^2 = (13 \times 37)a$ and $16x^2 - 15y^2 = (13 \times 37)b$. If $(37, y) = 1$, let $yy_1 \equiv 1 \pmod{37}$. From $15x^2 + 16y^2 \equiv 0 \pmod{37}$ and $16x^2 - 15y^2 \equiv 0 \pmod{37}$, we get $15(x y_1)^2 \equiv -16 \pmod{37}$ and $16(x y_1)^2 \equiv 15 \pmod{37} \Rightarrow (x y_1)^2 \equiv -6 \pmod{37}$. This is impossible, as only $r^2 \equiv 0, \pm 1, \pm 3, \pm 4, \pm 7, \pm 9, \pm 10, \pm 12, \pm 16 \pmod{37}$. Therefore, $x = 37u$ and $y = 37v$. We have $0 \equiv 15u^2 + 16v^2 \equiv 2u^2 + 3v^2 \pmod{13}$ and $0 \equiv 16u^2 - 15v^2 \equiv 3u^2 - 2v^2 \pmod{13}$. If $(13, u) = 1$, let $uu_1 \equiv 1 \pmod{13}$. Then, we get $2(v u_1)^2 \equiv -3 \pmod{13}$ and $3(v u_1)^2 \equiv 2 \pmod{13} \Rightarrow (v u_1)^2 \equiv 5 \pmod{13}$. This is impossible, as only $r^2 \equiv 0, \pm 1, \pm 3, \pm 4 \pmod{13}$. Therefore, $u = 13s$ and $v = 13t$, so $x = 37 \times 13s$ and $y = 37 \times 13t$. Taking $s = t = 1$, we get the minimum values of $x^2$ and $y^2$ as $x^2 = y^2 = (13 \times 37)^2 = 231361$. At this point, $a = 13 \times 37 \times 31$ and $b = 13 \times 37$.