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Prove/Disprove $\displaystyle \sum_{i=1}^{p-1} (i!)^2 \not\equiv 0 (\text{mod } p)$.

While I was trying this problem, I had to show that $\displaystyle p \not\Bigg|\sum_{i=1}^{p-1} (i!)^2$. So I tried to solve this but failed. Here's my attempt.

\begin{align} & \text{Approach 1. What's }[i!(p-1-i)!]^2 \text{ mod } p?\\ \ \\ & \text{Note. } {p-1 \choose i}^2 \equiv 1 (\text{mod } p). \\ \Rightarrow \; & \left[ \frac {(p-1)!} {i!(p-1-i)!} \right]^2 \equiv 1 (\text{mod }p). \\ & \text{By Wilson's theorem, } (p-1)! \equiv -1 (\text{mod } p). \\ \therefore \; & \left[i!(p-1-i)!\right]^2 \equiv 1 (\text{mod } p). \\ \ \\ & \text{Approach 2. How can } 2\sum_{i=1}^{p-1} (i!)^2 \text{ change its form?} \\ &2\sum_{i=1}^{p-1} (i!)^2 = \sum_{i=1}^{p-1} \left\{(i!)^2+[(p-i)!]^2\right\}\\ & = \sum_{i=1}^{p-1} \left\{ [i!+(p-i)!]^2 -2\cdot i!(p-i)!\right\} \end{align}

Uh... Any tips?

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  • $\begingroup$ Remember that when you want to write something like $\sum_{i=1}^n a+\sum_{i=1}^n b$, you have to write $\sum_{i=1}^n (a+b)$. Without the parenthesis, it becomes very hard to understand. I have already suggested an edit for you. $\endgroup$
    – IraeVid
    Jun 11, 2023 at 14:47
  • $\begingroup$ @IraeVid Ohh, I've missed that. I'll edit it...(I already voted for 'reject' in the suggested edits review... sorry for that.) $\endgroup$
    – RDK
    Jun 11, 2023 at 14:48

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It seems that the statement is wrong. According to an answer here the smallest counterexample is $p=1248829$

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  • $\begingroup$ Oh, I see. It was my mistake that I didn't look for the other posts. $\endgroup$
    – RDK
    Jun 12, 2023 at 14:18

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