Problem 1042

Let $\left(x_{0}, y_{0}\right)$ be a set of integer solutions to equation (1). Clearly, $x_{0} \neq 0, y_{0} \neq 0$, and
$$2017\left(x_{0} y_{0}+1\right)=x_{0}^{2}+y_{0}^{2}>0 \text {. }$$

From this, we know that $x_{0} y_{0}>0$, hence $x_{0}$ and $y_{0}$ have the same sign.
Without loss of generality, assume $x_{0}$ and $y_{0}$ are both positive integers (otherwise, replace $x_{0}$ and $y_{0}$ with $-x_{0}$ and $-y_{0}$ respectively), and $x_{0} \geqslant y_{0}$.
Consider equation (1) as a quadratic equation in $x$
$$x^{2}-2017 y_{0} x+y_{0}^{2}-2017=0 \text {. }$$

Thus, from a set of positive integer solutions $\left(x_{0}, y_{0}\right)$ $\left(x_{0} \geqslant y_{0}\right)$ of equation (1), we can derive another set of positive integer solutions $\left(y_{0}, x^{\prime}\right)$
$$\left(y_{0} \geqslant x^{\prime}\right) \text {, and } y_{0}+x^{\prime}<x_{0}+y_{0} \text {. }$$

The above process can be carried out infinitely, each time yielding a new set of solutions where the sum of the two numbers is strictly less than the sum of the two numbers in the previous set, and all these solutions are positive integer solutions, which is clearly impossible. Therefore, the original indeterminate equation has no integer solutions.