Motivated by my study of determinants and permanents, here I present several conjectures on $p$-adic congruences involving permutations.
As usual, we let $S_n$ be the symmetric group consisting of all the permutations of $\{1,\ldots,n\}$.
Conjecture 1. Let $p$ be an odd prime. Then $$\sum_{\tau\in S_{p-1}}\prod_{j=1\atop \tau(j)\not=j}^{p-1}\frac1{j-\tau(j)}\equiv 1+(-1)^{(p-1)/2}\pmod p\tag{1}$$ and $$\sum_{\tau\in S_{p-1}}\mathrm{sign}(\tau)\prod_{j=1\atop \tau(j)\not=j}^{p-1}\frac1{j-\tau(j)}\equiv 2\pmod p.\tag{2}$$
Conjecture 2. For any prime $p$, we have $$\sum_{\tau\in S_{p-1}}\prod_{j=1\atop \tau(j)\not=j}^{p-1}\frac{j+\tau(j)}{j-\tau(j)}\equiv((p-2)!!)^2\pmod{p^2}.\tag{3}$$ If $p$ is an odd prime, then $$\sum_{\tau\in S_{p-1}}\mathrm{sign}(\tau)\prod^{p-1}_{j=1\atop \tau(j)\not=j}\frac{j+\tau(j)}{j-\tau(j)}\equiv\frac{(-1)^{(p+1)/2}}{p-2}((p-2)!!)^2\pmod{p^2}.\tag{4}$$
Conjecture 3. Let $p$ be a prime with $p\equiv3\pmod4$.
(i) We have $$\sum_{\tau\in S_{(p-1)/2}}\prod_{j=1\atop \tau(j)\not=j}^{(p-1)/2}\frac1{j^2-\tau(j)^2} \equiv \sum_{\tau\in S_{(p-1)/2}}\mathrm{sign}(\tau)\prod_{j=1\atop \tau(j)\not=j}^{(p-1)/2}\frac1{j^2-\tau(j)^2}\equiv1\pmod p.\tag{5}$$
(ii) If $p>3$, then $$\sum_{\tau\in S_{(p-1)/2}}\mathrm{sign}(\tau)\prod^{(p-1)/2}_{j=1\atop \tau(j)\not=j}\frac{j^2+\tau(j)^2}{j^2-\tau(j)^2}\equiv0\pmod{p^2}.\tag{6}$$ When $p\equiv 7\pmod 8$, we have $$\sum_{\tau\in S_{(p-1)/2}}\mathrm{sign}(\tau)\prod^{(p-1)/2}_{j=1\atop \tau(j)\not=j}\frac{j^2+\tau(j)^2}{j^2-\tau(j)^2}\equiv0\pmod{p^3}.\tag{7}$$
QUESTION. How to prove the congruences $(1)-(7)$?
In my opinion, $(1)$, $(2)$ and $(5)$ should be relatively easy, but the other conjectural congruences might be quite challenging.
Your comments are welcome!
