# Find all $x \in \mathbb Z$ and $y \in \mathbb N$ such that $x^2-y!=2001$ and prove your answer is correct

I have the following question:

Find all $x \in \mathbb Z$ and $y \in \mathbb N$ such that $x^2-y!=2001$. Prove that your answer is correct.

And I have absolutely no idea where to even start. I don't even know what my end going is going to be here. What do I have to do first?

For large enough $y$, $y!$ will be divisible by $9$ and then $x^2=y!+2001\equiv3\pmod 9$ which is impossible. How large, is large enough?

• Why is $x^2 \equiv 3$ (mod 9) impossible? – Mark Dodds Nov 3 '17 at 0:24
• @MarkDodds You could check all values of $x$ modulo $9$. – Lord Shark the Unknown Nov 3 '17 at 5:23

By trial of the first $5$ values for $y$

$$(y,x)=(4,\pm45)$$

For $x>5$, see previous answer.

• Actually, you missed $(y, x) = (4, -45)$ since we have $x \in \mathbb Z$ – Max Li Nov 7 '17 at 2:07
• @MaxLi Thank you for spotting that, I've corrected my typo and up-voted your comment. – Old Peter Nov 7 '17 at 18:57