18.3.61 ** Find all pairs of coprime positive integers $(x, y)$ such that $x\left|y^{2}+210, y\right| x^{2}+210$.
Parse $x^{2}+y^{2}+210$ is divisible by $x$, and also by $y$. Therefore, it is also divisible by $x y$. Let
$$
x^{2}+y^{2}+210=k x y, k \in \mathbf{N} .
$$

If $x=1$, then $y \mid 211, y=1$ or 211. If $y=1$, then $x=1$ or 221. Suppose $x>y>1$. From (1) and the fact that $x, y$ are coprime, we get $(x, 210)=(y, 210)=1$. Thus, $x$ and $y$ are both odd and $\geqslant 11$.

If $y=11$, then $x^{2}+331=11 k x$, so $x \mid 331, x=331$. But $x^{2}+331=331 \times 332$ is not divisible by 11; if $y=13$, then $x^{2}+379=13 k x$, so $x=379$. But $x^{2}+379=379 \times 380$ is not divisible by 13. Therefore, $y \geqslant 17$.
Since $k x y=x^{2}+y^{2}+210>2 x y$, we have $k \geqslant 3$.
$$
(x-y)^{2}+210=(k-2) x y \geqslant x y .
$$
$(x-y)^{2} \geqslant x y-210>x y-y^{2}=y(x-y)$, so $x>2 y$.
If $(x, y)=(a, b), a>b>1$ is a positive integer solution to (1), then by Vieta's formulas, $c=\frac{b^{2}+210}{a}=k b-a$ is also a solution to the equation $x^{2}-k b x+b^{2}+210=0$. Thus, $(x, y)=(b, c)$ is also a solution to (1).

From $c=k b-a$, we know $c$ is an integer. From $c=\frac{b^{2}+210}{a}$, we know $c$ is positive. Also, from (2) $a c=b^{2}+210b>1$, we can generate a new solution $(b, c), b>c$. If $c>1$, we can generate another new solution $(c, d), c>d$. This process can continue. Since a strictly decreasing sequence of natural numbers can only have a finite number of terms, we must eventually reach the solution $(211,1)$, i.e., $k=212$.

Thus, let $(x, y)=\left(a_{n+1}, a_{n}\right)$, then we have the recurrence relation $a_{n+1}=212 a_{n}-a_{n-1}, a_{1}=a_{2}=1$. Of course, $(x, y)-\left(a_{n}, a_{n+1}\right)$ is also a solution to this problem.