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For a matrix $[a_{j,k}]_{1\le j,k\le n}$, its permanent is given by $$\mathrm{per}[a_{j,k}]_{1\le j,k\le n}=\sum_{\tau\in S_n}\prod_{j=1}^na_{j,\tau(j)}.$$

Let $p$ be an odd prime. I have proved the congruences \begin{align}\mathrm{per}[j+k]_{1\le j,k\le p-1}\equiv&-2-4(p-1)!\pmod{p^2},\tag{1} \\\mathrm{per}[j+k]_{1\le j,k\le p}\equiv& p\pmod{p^2},\tag{2} \\\mathrm{per}[j+k]_{0\le j,k\le p-1}\equiv&-p\pmod{p^2}.\tag{3} \end{align} Motivated by this, I make the following conjecture.

Conjecture. For any odd prime $p$, we have $$\mathrm{per}[|j-k|]_{1\le j,k\le p}\equiv-\frac12\pmod p.\tag{4}$$

Remark. I have verified $(4)$ for $p=3,5,7,11,13,17,19,23$. My method to prove $(1)$-$(3)$ does not work for the congruence $(4)$.

QUESTION. How to prove the congruence $(4)$ for each odd prime $p$ ?

I only ask how to prove $(4)$. Please ignore $(1)$-$(3)$ since I have already proved them completely.

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  • $\begingroup$ The OEIS gives the the permanent of these matrices oeis.org/A085807 for $n\leq 37$. From this reference, we know that (4) holds for all odd primes $p\leq 37$. $\endgroup$
    – Joseph Van Name
    Commented Aug 29, 2021 at 17:12
  • $\begingroup$ It seems that $$ \mathrm{per}[|j^2-k^2|]_{1\le j,k\le p} \equiv \frac{p(p+1)}{2} \pmod {p^2}$$ for any odd prime $p$, $$\mathrm{per}[j^2+jk+k^2]_{1\le j,k\le p} \equiv \frac{p(p-1)}{2} \pmod {p^2}$$ for any odd prime $p>3$. $\endgroup$
    – Deyi Chen
    Commented Aug 30, 2021 at 15:07
  • $\begingroup$ For my proofs of $(1)$-$(3)$, one may consult my recent preprint Arithmetic properties of some permanents availabe from arxiv.org/abs/2108.07723 . $\endgroup$
    – Zhi-Wei Sun
    Commented Aug 31, 2021 at 0:57

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