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I would like to find all positive integers $n$ and $m$ such that $n^2 \equiv m! \ ( \text{mod } 2024)$. I see that for $m=1$ there is $n=45$ such that the relation holds.

I think that there is no other solutions. I would like to prove that. I know that $2024=2^3 \cdot 11 \cdot 23$. Hence, I cold investigate $n^2 \equiv m! \ ( \text{mod } 8)$, $n^2 \equiv m! \ ( \text{mod } 11)$, and $n^2 \equiv m! \ ( \text{mod } 23)$. Common quadratic residue of $8, 11$, and $23$ are $0,1,4$. Since $m$ must be positive, we reject $0$. Next, we have already covered $1$. Finally, there is no integer $m$ such that $m!=4$.

I think that this should prove the statement. Is this the right solution?

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  • $\begingroup$ There's also $n\equiv0\pmod{2024}$ and $m \geq 23$ $\endgroup$
    – Daniel Weber
    Commented Apr 28 at 19:06
  • $\begingroup$ Thank you for your comment. Actually, I was studying equation $n^2=m!+2024$. I think that it has only one solution in positive integers. I thought that I could show that the congruence modulo 2024 does not have a solution except $m=1, n=45$ and from this imply that the equation above does not have other solutions. $\endgroup$
    – Peter Johnson
    Commented Apr 28 at 19:18
  • $\begingroup$ Hint: look at this equation $\bmod 3$. $\endgroup$
    – Daniel Weber
    Commented Apr 28 at 19:41

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