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Does there exist a positive integer value of $n$ such that the sum of $1!^n+2!^n+3!^n+\cdots+2024!^n$ is a perfect power?

I know that $n\neq1$, as the last digit of the sum when $n\geq4$ is $3$ and no perfect square ends with $3$, and it cannot be a perfect power since $3^3$ does not divide the sum when $n\geq8$.

I don’t know about when $n=2$, since the only sure point when $1!^2+2!^2+3!^2+\cdots+n!^2$ cannot be a perfect power is when $n\geq1248828$,but I don’t have any idea if $1!^2+2!^2+3!^2+\cdots+2024!^2$ is a perfect power.

If $n=3$, then the sum isn’t a perfect square when $n\geq10$ as it is equal to $6\pmod{11}$, and it cannot be a perfect power since $3^3\nmid1!^3+2!^3+3!^3+\cdots+n!^3$ when $n\geq5$.

If $n=5,7$, then it is not a perfect power since $3^2$ dosen’t divide the sum.

Also, for all $n$ that is even, the sum can’t be a perfect square because the sum is equal to $2\pmod{3}$.

Note also that if $n$ is odd, then $1!^n+2!^n+3!^n+\cdots+2024!^n$ isn’t a perfect powers because there is some power of three that does not divide the sum after some point(in some cases, $9$ even dosen’t divide the sum), and forces the sum to be perfect square in order to be a perfect power. But also, the sum will “land” on a prime, where $1!^n+2!^n+3!^n+\cdots+2024!^n\equiv a\pmod{p}$, and $a$ isn’t a quadratic residue $\pmod{p}$.

So I suspect that $1!^n+2!^n+3!^n+\cdots+2024!^n$ is never a perfect square unless $n$ is divisible by $3$.

Now, it forces us that $n$ has to be even in order for $1!^n+2!^n+3!^n+\cdots+2024!^n$ to be a higher odd prime perfect power.

I used Pari GP and checked the all values of $n\leq10^{4}$ to see if there is a value of $n$ such that $1!^n+2!^n+3!^n+\cdots+2024!^n$ is a perfect power, but so far, none has been found.

Edit: This answer already ruled out about the sum being a perfect square.

Can $1!^n+2!^n+3!^n+\cdots+2024!^n$ be a perfect power?

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  • $\begingroup$ Considering the magnitude of the sum for $n=10^4$ , I am convinced enough that this can never be a perfect power. I think , you mean , you checked all values $n\le 10^4$. $\endgroup$
    – Peter
    Dec 31, 2023 at 6:26
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    $\begingroup$ "I don’t know about when n=2 , since the only sure point when 1!2+2!2+3!2+⋯+n!2 cannot be a perfect power is when n≥1248828" can you explain your reasoning here? also you probably really to avoid using the same letter $n$ for two different variables consider changing one of them. $\endgroup$ Dec 31, 2023 at 16:19
  • $\begingroup$ You don't know about $n=2$, yet you verified $n\leq 10^4$? Then you should know about $n=2$ case? $\endgroup$
    – Sil
    Feb 15 at 0:33
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    $\begingroup$ Using a similar approach to @OlderAmateur, I found that the number cannot be a perfect 2nd, 3rd, 5th, or 7th power for all $n \geq 84$. For a perfect 2nd power (i.e. a square) I used congruences mod $3^2, 13$. For 3rd power I used $3^2,7,19$. For 5th power I used $5^2,11$ and for a perfect 7th power I used congruences $3^2,7^2,29$. For all $n \geq 156$ I also ruled out it being a perfect power of 13 due to congruences mod $13^2,53,157$. $\endgroup$ Feb 18 at 12:25
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    $\begingroup$ Since you verified all $n \leq 10^4$ I can state that the number cannot be a perfect $e$-th power for all $n\in \mathbb Z$ and $e \in \{2,3,5,7,11,13,17,19,23,29\}$. $\endgroup$ Feb 18 at 12:55

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As you already demonstarted, the sum can only be a square $(\bmod 3)$ if $n$ = $1(\bmod 2) \implies n=1,3,5,7,9,11(\bmod 12)$

To be a square $(\bmod 5) \implies n = 3(\bmod 4) \implies n= \not1,3,\not5,7,\not9,11(\bmod 12)$

To be a square $(\bmod 9) \implies n=1,3(\bmod 6)\implies n=\not1,3,\not5,7,\not9,\not11(\bmod 12)$

To be a square $(\bmod 13) \implies n=1,5,11 (\bmod 12) \implies n = \not1,\not3,\not5,\not7,\not9,\not11 (\bmod 12)$ $$\implies \sum_N^{N\ge12} N!^n\not = k^2 $$

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  • $\begingroup$ But what about for the sum being a perfect cube or any higher odd prime perfect power? $\endgroup$ Feb 13 at 19:03

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