38 Find all integers $a$ such that the equation $x^{2}+a x y+y^{2}=1$ has infinitely many integer solutions.

Find all integers $a$ such that the equation $x^{2}+a x y+y^{2}=1$ has infinitely many integer solutions.
38. Discuss the number of integer solutions of the equation
$$x^{2}+a x y+y^{2}=1$$
for different values of $a$.
When $a=0$, the equation becomes $x^{2}+y^{2}=1$, which has only 4 sets of integer solutions.
If $a \neq 0$, then $(x, y)$ is a solution to equation (1) if and only if: $(x, -y)$ is a solution to the equation $x^{2}-a x y+y^{2}=1$. Therefore, we only need to discuss $ay$, then $(x, -a x+y)$ is also a solution to (1) (this solution can be obtained using Vieta's formulas when viewing (1) as a quadratic equation in $y$), and of course $( -a x+y, x)$ (satisfying $-a x+y>x>y$) is also a positive integer solution to (1), and by this recursion, we know that (1) has infinitely many positive integer solutions.

In summary, when $|a|>1$, equation (1) has infinitely many integer solutions, and when $|a| \leqslant 1$, (1) has only finitely many integer solutions.