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Is $7$ the largest possible value of $n$ for which $(1!+2!+3!+…n!)+16$ is a perfect power?

I noticed that $(7!+6!+5!+4!+3!+2!+1!)+16=77^2$ is a perfect power, and I don’t know if that is the largest possible value of $n$ for which $(1!+2!+3!+…n!)+16$ is a perfect power?

I can conclude that $(1!+2!+3!+…n!)+16$ is never a perfect cube if $n\geq6$, since it is equal to $4\pmod{9}$, but I don’t know if $(1!+2!+3!+…n!)+16$ can be a perfect square.

Since the last digit of $(1!+2!+3!+…n!)+16$ is $9$, I think it can be a perfect square if $n\geq8$, since the sum is $1\pmod{8}$, and $1\pmod{3}$.

I can also conclude that $(1!+2!+3!+…n!)+16$ cannot be a perfect 5th power if $n\geq10$, since $29$ is not a 5th power residue $\pmod{100}$.

Other than $3, 4, 5, 7$, are there any values of $n$ such that $(1!+2!+3!+…n!)+16$ is a perfect power?

Edit:

  • Since $3$ is not a 7th power residue $\pmod{29}$, $(1!+2!+3!+…n!)+16$ cannot be a perfect 7th power if $n\geq28$.
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  • $\begingroup$ I checked the first 20,000 (just using a simplistic brute force approach) and there is no other square apart from the first 4 solutions in your post. At this point, numbers are getting close to 100,000 digits so brute force becomes impractical. $\endgroup$
    – PC1
    Aug 4, 2023 at 20:06
  • $\begingroup$ Is $~6 = 2^1 \times 3^1~$ to be considered a perfect power? $\endgroup$ Aug 4, 2023 at 22:00
  • $\begingroup$ the power (exponent) on a perfect power must be 2 or more. So 6 isn't one. $\endgroup$
    – coffeemath
    Aug 7, 2023 at 3:22

2 Answers 2

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$$(1! + 2! + 3! + \ldots + n!) + 16 \equiv 14 \pmod{49}$$ for $n \ge 13$. So $(1! + 2! + 3! + \ldots + n!) + 16$ is not a perfect power if $n \ge 13$.

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  • $\begingroup$ Can you explain why $f(n)\equiv14 (\text{mod } 49)$ make it so that it's not a perfect power? More specifiacally, is $\text{mod }49$ more special than other $p^2$-type mods? And why does it verify, for instance, that $f(345)\neq 103^{19}$? $\endgroup$ Aug 7, 2023 at 7:32
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    $\begingroup$ This congruence implies that $f(n)$ is divisible by 7, but not divisible by $7^2$. But perfect power must be divisible by the square of all its prime divisors. $\mod 49$ is the first modulo for which there is a contradiction $\endgroup$ Aug 7, 2023 at 8:39
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    $\begingroup$ If $a^n\equiv 14\pmod{49}$ then $7|a$. But then $49|a^2$; so there's no such $a$. To your second point, $103^19\equiv 47\pmod{49}$, so isn't really relevant. $\endgroup$ Aug 7, 2023 at 8:43
  • $\begingroup$ Nice explanations, Denis and Chris, it all makes sense now. My question was dumb, I guess. $\endgroup$ Aug 7, 2023 at 10:05
  • $\begingroup$ @VeselinDimov not really, a lot of the implication is left for the reader. If you had the question, so did others. $\endgroup$ Aug 7, 2023 at 15:48
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A proof for perfect even power $m^{2k}$

Assume, on the contrary, that $n>7$, then $n! \equiv 0 \quad \pmod {32}$ and

$$ \begin{aligned} & 1 !+2 !+3 !+ 4!+5!+6! +7 !+8 !+\cdots+n !+16 \\ \equiv & 1+2+6+24+120+720+5030+16 \\ \equiv & 5919 \\ \equiv & -1(\bmod 32) \end{aligned} $$

which contradicts to the fact that $m^{2k}\not \equiv -1\quad \pmod {32}$ for any integers $m$ and $k$.

Therefore $1 !+2 !+3 !+4!+5!+6!+7 !+8 !+\cdots+n !+16$ is never a perfect even power for $n>8$. On the other hand, $1 !+2 !+3 !+\ldots+7 !+16=77^2$ reveals that $7$ is largest integer to make the sum a perfect even power.

Unfortunately, this method can’t be applied to perfect odd powers.

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