Example 20([37.4]) 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.
Let $15a + 16b = x^2$, $16a - 15b = y^2$ ($x, y \in \mathbb{N}$). Solving for $a, b$ gives:
$$15x^2 + 16y^2 = (13 \cdot 37)a, \quad 16x^2 - 15y^2 = (13 \cdot 37)b$$

We need to prove that $x, y$ are both multiples of $13 \cdot 37$. 13 and 37 are both prime numbers. The solution uses the concept of modular inverses (but only for specific moduli 37 and 13, without using general conclusions) and the special properties of the remainders of squares modulo specific numbers.
(i) If 37 does not divide $y$, then there must be $y'$ and $q$ such that $yy' + 37q = 1$ (this can be verified directly for $y \equiv 1, 2, \ldots, 36 \pmod{37}$), i.e., $yy' \equiv 1 \pmod{37}$ ($y'$ is called the inverse of $y$ modulo 37). Using $15x^2 + 16y^2 \equiv 0 \pmod{37}$, we get
$$15(x y')^2 \equiv -16 \pmod{37}, \quad (x y')^2 \equiv -6 \pmod{37}$$

This is impossible, because only
$$r^2 \equiv 0, \pm 1, \pm 3, \pm 4, \pm 7, \pm 9, \pm 10, \pm 11, \pm 12, \pm 16 \pmod{37}$$

So $x = 37u, y = 37v$, and we have
$$0 \equiv 15u^2 + 16v^2 \equiv 2u^2 + 3v^2 \pmod{13}$$
(ii) If 13 does not divide $u$, then there must be $u'$ and $r$ such that $uu' + 13r = 1$ (this can be verified directly for $u \equiv 1, 2, \ldots, 12 \pmod{13}$), i.e., $uu' \equiv 1 \pmod{13}$ ($u'$ is called the inverse of $u$ modulo 13). Using the above equation, we get
$$3(v u')^2 \equiv -2 \pmod{13}, \quad (v u')^2 \equiv -5 \pmod{13}$$

This is impossible, because only
$$r^2 \equiv 0, \pm 1, \pm 3, \pm 4 \pmod{13}$$

So $u = 13s, v = 13t$. Hence $x = 13 \cdot 37s, y = 13 \cdot 37t$, and thus
$$(13 \cdot 37)(15s^2 + 16t^2) = a, \quad (13 \cdot 37)(16s^2 - 15t^2) = b$$
(iii) Taking $s = t = 1$, we get $a = 13 \cdot 37 \cdot 31, b = 13 \cdot 37$, and the minimum value is $x^2 = y^2 = (13 \cdot 37)^2$