Let $p$ be an odd prime. It is well-known that $$\det[i^{j-1}]_{1\le i,j\le p-1}=\prod_{1\le i<j\le p-1}(j-i)\not\equiv0\pmod p.$$ 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 $p-1$.

Your comments are welcome!

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.