# 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?

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.

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

• 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 $n\le 10^4$. Dec 31, 2023 at 6:26
• "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. Dec 31, 2023 at 16:19
• You don't know about $n=2$, yet you verified $n\leq 10^4$? Then you should know about $n=2$ case?
– Sil
Feb 15 at 0:33
• 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 $13^2,53,157$. Feb 18 at 12:25
• 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 $e \in \{2,3,5,7,11,13,17,19,23,29\}$. Feb 18 at 12:55

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