Exercise 4. Find all positive integers $x$ and $y$ such that $x^{2}-2 \times y!=2021$.
Solution to Exercise 4: As is often the case with a Diophantine equation, it is necessary to start by examining the equation modulo small, well-chosen numbers.

Here, the equation viewed modulo 2 gives that $x^{2} \equiv 2021 \equiv 1(\bmod 2)$, so $x$ is odd. The equation modulo 3 requires a case distinction.

- If $y \geqslant 3$, then $y!=0(\bmod 3)$, so $x^{2} \equiv 2021 \equiv 2(\bmod 3)$, which is impossible.
- If $y=0$ or 1, then $x^{2}=2021+2=2023$ but 2023 is not a square, so there is no solution.
- If $y=2$, then $x^{2}=2021+2 \times 2=2025$ and since $x \geqslant 0, x=45$. Conversely, the pair $(45,2)$ is indeed a solution to the equation.

We conclude that the unique solution to the problem is the pair $(45,2)$.

Graders' Comment: The exercise is generally well done.