# On the permanent $\text{per}[i^{j-1}]_{1\le i,j\le p-1}$ modulo $p^2$

Let $$p$$ be an odd prime. It is well-known that $$\det[i^{j-1}]_{1\le i,j\le p-1}=\prod_{1\le i I'm curious about the behavior of the permanent $$\text{per}[i^{j-1}]_{1\le i,j\le p-1}$$ modulo powers of $$p$$. This leads me to formulate the following conjecture on the basis of my computation.

Conjecture. If $$p$$ is a Fermat prime, then $$\text{per}[i^{j-1}]_{1\le i,j\le p-1}\equiv\frac{p-1}2!\ p\pmod{p^2}.$$ If $$n\not\equiv2\pmod4$$ and $$n$$ is not a Fermat prime, then $$\text{per}[i^{j-1}]_{1\le i,j\le n-1}\equiv 0\pmod{n^2}.$$ For $$n\equiv2\pmod4$$, we have $$\text{per}[i^{j-1}]_{1\le i,j\le n-1}\not\equiv 0\pmod{n}.$$

QUESTION: Is the above conjecture true?

Now I show that $$\text{per}[i^{j-1}]_{1\le i,j\le p-1}\equiv0\pmod p$$ for any odd prime $$p$$. (It is easy to see that this implies that $$n\mid \text{per}[i^{j-1}]_{1\le i,j\le n}$$ for all $$n=3,4,\ldots$$.) Let $$g$$ be a primitive root modulo $$p$$. Then $$\text{per}[i^{j-1}]_{1\le i,j\le p-1}\equiv\sum_{\sigma\in S_{p-1}}\prod_{i=1}^{p-1}g^{i(\sigma(i)-1)}=\prod_{i=1}^{p-1}g^{-i}\times\text{per}[g^{ij}]_{1\le i,j\le p-1}\pmod p.$$ Thus we may appy my argument in the related question 316836 to get that $$\text{per}[i^{j-1}]_{1\le i,j\le p-1}\equiv0\pmod p$$ since the order of $$g$$ mod $$p$$ is the even number $$p-1$$.