# Can the sum of $2025$ consecutive factorials be a perfect power?

Can the sum of $$2025$$ consecutive factorials be a perfect power?

My thoughts:

If the sum of $$2025$$ consecutive factorials is a perfect power, then the number of trailing zeroes in the sum must be a perfect (number of trailing zeroes)th power.

So this means that if the sum of $$2025$$ consecutive factorials is a perfect power, then the sum must be a perfect (number of trailing zeores)th residue $$\pmod{p}$$, where $$p$$ is a certain prime number.

For example, the sum of $$10!+11!+12!+\cdots+2034!$$ contains $$2$$ trailing zeroes as all factorials from $$10!$$ to $$14!$$ contains $$2$$ trailing zeroes, and since the sum contains $$2$$ trailing zeroes, then it must be a perfect square in order to be a perfect power.

By using Pari GP, I tried to check all factorials from $$1!$$ to $$10^{6}!$$, but I did not succeed.

• suggest you investigate five consecutive factorials that may not begin at $1$ Let's see, factor $2! + 3! + 4! + 5! + 6!$ and then factor $3! + 4! + 5! + 6! + 7!$ and so on. I can see that things change a little when beginning after 5 itself. The point, though, is that you will have no idea how to deal with 2025 factorials unless you know how to deal with 5 Feb 2 at 4:35
• @willjagy I did already tried that by using Pari GP, but I wasn’t successful. Feb 4 at 13:46
• @ThirdyYabata I think Will was suggesting to do a little algebra with pen and paper. For example, what can you do with $5! +6! +7! +8! +9!$. Running programs is good, but sometimes a little algebra can be more illuminating :) Feb 5 at 2:16
• One neat thing about the sum is that every subsequent term is a multiple of every preceding term. Perhaps you could use that to help with your question..? Feb 5 at 16:47

Suppose $$X=n!+(n+1)!+\cdots+(n+2024)!$$ is a perfect power. Let $$Y=\frac X{n!}=1+(n+1)+(n+1)(n+2)+\cdots+(n+1)\cdots(n+2024).$$ Note that $$Y<\frac{(n+2025)!}{n!}<(n+2025)^{2025}.$$ For $$X$$ to be a perfect power, each prime which divides $$n!$$ exactly once must also divide $$Y$$. In particular, $$(n/2)^{\pi(n)-\pi(n/2)}\leq \prod_{n/2 This implies that $$\pi(n)-\pi(n/2)\leq 2025\log_{n/2}(n+2025);$$ as long as $$n>200$$, say, $$n+2025<(n/2)^2$$, and so we have $$$$\tag{\star}\pi(n)-\pi(n/2)\leq 4050.$$$$ On the other hand, we have explicit bounds on $$\pi(n)$$ from explicit versions of the prime number theorem: see for example this article of Dusart, as described on Wikipedia, which gives $$\frac{x}{\log x-1}\leq \pi(x)\leq \frac{x}{\log x-1.1}$$ for $$x\geq 60184$$. This implies, for $$n\geq 130000$$, \begin{align} \pi(n)-\pi(n/2) &\geq \frac n{\log n-1}-\frac{n/2}{\log(n/2)-1.1}\\ &>\frac n{\log n-1}\left(1-\frac{\log n-1}{2(\log(n/2)-2)}\right)\\ &=\frac n{\log n-1}\cdot\frac{\log n-3}{2\log n-4}. \end{align} As long as $$\log n>10$$, we have $$\pi(n)-\pi(n/2)\geq \frac n{\log n-1}\cdot\frac 7{16}.$$ Since $$n/(\log n-1)$$ is an increasing function of $$n$$, we have that, when $$n\geq 130000$$, $$\pi(n)-\pi(n/2)\geq \frac 7{16}\frac{130000}{\log(130000)-1}\geq \frac{7}{16}\cdot\frac{130000}{11}\geq 5000.$$ This contradicts ($$\star$$). We conclude that $$X$$ can only be a perfect power for $$n<130000$$. Since you have already checked these cases, we conclude that the sum in question is never a perfect power.