Problem 731

Assume there exists an integer solution pair $(x ; y)$ for the equation $2 x^{2}-7 y=1$.

Then, because $7 y \equiv 0(\bmod 7)$, it must hold that $2 x^{3} \equiv 1(\bmod 7)$, which means $x^{3} \equiv 4(\bmod 7)$. However, this is a contradiction to the fact that $x^{3} \equiv 0(\bmod 7)$ for $x \equiv 0(\bmod 7)$ and $x^{3} \equiv \pm 1(\bmod 7)$ for $x \neq 0$ $(\bmod 7)$, so $x^{3} \neq 4(\bmod 7)$ for any integer $x$.

Thus, the assumption is disproven.