Example 6. (1MO-27 Preliminary Question) Let $A, B, C$ be three points on the edge of a circular pool, with $B$ due west of $C$, and $A B C$ forming an equilateral triangle with side lengths of 86 meters. A swimmer starts from $A$ and swims directly to $B$. After swimming $x$ meters, he reaches point $\boldsymbol{E}$, then turns and swims due west for $y$ meters, arriving at point $D$. If $x, y$ are both integers, find $y$.

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Given the figure, since $\triangle AEF$ is an equilateral triangle, we have
$$
AF = AE = x.
$$

By symmetry, $FG = DE = y$. Also, since
$$
AE \cdot EB = DE \cdot EG,
$$

we have
$$
\begin{aligned}
& x(86 - x) \\
= & y(x + y).
\end{aligned}
$$

Notice that if $x$ is odd, then $x(86 - x)$ is also odd, while $y(x + y)$ is always even (if $y$ is even, $y(x + y)$ is clearly even; if $y$ is odd, then $x + y$ is even, and $y(x + y)$ is still even).

Therefore, $x$ must be even. From (1), we know that $y(x + y)$ is even, so $y$ must also be even.

The above argument is essentially taking $\left(1; \mathbf{x}^{2}\right)$ modulo 2 to determine $x$. We have $\left(x + \frac{y}{2} - 43\right)^{2} + y^{2} = \left(43 - \frac{y}{2}\right)^{2}$.

This indicates that $\left|x + \frac{y}{2} - 43\right|, y, 43 - \frac{y}{2}$ form a Pythagorean triplet.

Let $d$ be the greatest common divisor of $\left|x + \frac{y}{2} - 43\right|$ and $y$. Then, by the theorem mentioned later, there exist positive integers $a, b, a > b$, such that
$y = 2abd$ (since $y$ is even),
$$
43 - \frac{y}{2} = (a^2 + b^2)d.
$$

From these two equations, we get
$$
(a^2 + ab + b^2)d = 43 \cdot (a > b > 0).
$$

Since 43 is a prime number, we have $d = 1$.
Finally, we solve the equation using the "bounding method"
$$
a^2 + ab + b^2 = 43.
$$

Since $a > b > 0$, we have $3b^2 < 43$, $b^2 < \frac{43}{3}$, $b^2 \leq 14$, so $b \leq 3$. It is easy to see that the only positive integer solutions are $b = 1$ and $a = 6$.

Thus, we know $y = 12$. Then, from $\left|x + \frac{y}{2} - 43\right| = a^2 - b^2$, we find $x = 2$ or $x = 72$.