32. (2005 National Training Team Training Question) Find all pairs of coprime positive integers $(x, y)$ such that: $x$ | $y^{2}+210, y \mid x^{2}+210$.
32. For positive integers $x, y$, we have
$x\left|y^{2}+210, y\right| x^{2}+210,(x, y)=1$
$\Leftrightarrow x y \mid x^{2}+y^{2}+210,(x, y)=1$
$\Leftrightarrow x^{2}+y^{2}+210=k x y,(x, y)=1, k$ is some positive integer.
It is easy to see that $(x, y)=1, (x, 210)=1, (y, 210)=1$ can be derived from any one of them, and $k \geqslant 3$.
If $x=y$, it is easy to see that $x=y=1, k=212$.
If $x=1$, then $k=y+\frac{211}{y}$, noting that 211 is a prime number, so $y=1$ or $211, k=212$.
If $1210$, then
$$
\begin{array}{l}
(y-x)^{2}+x^{2}>(y-x)^{2}+210=(k-2) x y \geqslant x y, \\
(y-x)^{2}>x(y-x) .
\end{array}
$$

Since $y-x>0$, we have $y-x>x$, i.e., $y>2 x$. Thus,
$$
y^{\prime}=\frac{x^{2}+210}{y}<\frac{x^{2}+210}{2 x}<\frac{x^{2}+x^{2}}{2 x}=x \text {. }
$$

Therefore, $\left(y^{\prime}, x\right)$ is a "smaller" solution, and the value of $k$ remains unchanged.
This means that every pair $(x, y)(1<x<y)$ can be derived from $(1,1)$ or $(1,211)$, and $k=212$. Since $(1,211)$ can also be derived from $(1,1)$, we can construct a sequence $\left\{a_{n}\right\}$ as follows: $a_{1}=a_{2}=1, a_{n+2}=212 a_{n+1}-a_{n}$, then all positive integer solutions to the original equation are $\left(a_{n}, a_{n+1}\right)$ and $\left(a_{n+1}, a_{n}\right), n \in \mathbf{Z}^{+}$.