I am not sure if the following result is known or an equivalent result is known. I think, if the result holds then this could be used an elementary number theory exercise.

Let $k \in \mathbb{Z} \setminus \{0\}$ be fixed. For $n\in \mathbb{N} $ , let $A_{n,k}=[a^{nk}_{ij}]$ be a $n\times n$ matrix such that $$ a^{nk}_{ij}= \begin{cases} i & i= j \\ k & \text{otherwise} \end{cases} $$ As a matrix, $A_{n,k}$ has the following form $$ A_{n,k}=\left[ \begin{matrix} 1 & k & \ldots & k\\ k & 2 & \ldots & k\\ \vdots & \vdots & \ddots & \vdots\\ k & k &\ldots & n \end{matrix} \right] $$ Conjecture: Let $n\geq |k|$. If $k$ is odd , then $n$ is prime number if and only if $$|{\det(A_{n,k})}|\equiv -|{k}| \mod n $$ and if $k$ is even then $n$ is prime if and only if $$|{\det(A_{n,k})}|\equiv -k \mod n .$$

It is easy to figure out for $k\geq 1$ , $\det A_{n,k}=(k-1)!(n-k)!$ . Thus, for $k=1$, you basically get $\det A_{n,1}=(n-1)!$ which is equivalent to Wilson's theorem. For $k>1$, we get $n$ prime iff $(k-1)!(n-k)!\equiv (-1)^k \mod n.$ I wrote a Matlab code to check and it seems to be true. Does anyone have a counterexample or know how to prove it? The Gauss's generalization of Wilson's theorem seem to be related to this but that only considers(as far as I know) integers less than and that are co-prime to $n$.

P. S. For $k=-1$, I calculated the determinant in a non trivial way and got $\det A_{n,-1}={n!\big(n-(1+n)(\mathcal{H}_n-1)\big)}$ where $\mathcal{H}_n$ is the $n$th harmonic number so the determinant of this matrix seem to relate the $n$-harmonic number with prime numbers as well.


1 Answer 1


This is for the case $k>0$. If $n$ is prime, then it can be seen from this post that

$${{n-1}\choose {k-1}}=(-1)^{k-1} \pmod n.$$ Combining with $(n-1)!=-1 \pmod n$, one gets $(k-1)!(n-k)!=(-1)^k$ for all $1\leq k \leq n$.

For the converse, suppose $(k-1)!(n-k)!=(-1)^k \pmod n$ for some $1\leq k\leq n$. We claim that $n$ must be prime. On the contrary, suppose $n$ is composite and let $p$ be a prime factor of $n$. If $k>p$, then $(k-1)!$ is divisible by $p$. If $k\leq p$, then $n-k\geq n-p \geq p$ and so $(n-k)!$ is divisibly by $p$. In either case $(k-1)!(n-k)!=0 \pmod p$ contradicting the assumption.

For $k<0$, let $l=-k$ and use this post to get $$\det A=\prod_{i=1}^n(i+l)-l\sum_{i=1}^n\prod_{j\neq i}(j+l)\equiv -l \prod_{j\neq n-l}(j+l) \pmod n,$$ since all other terms have a factor of $n$. One has for both odd and even $k$ $$-l\prod_{j\neq n-l}(j+l)=-l(1+l)\cdots(n-1)(n+1)\cdots(n+l)\equiv -l(l+1)\cdots(n-1)(1)(2)\cdots (l)\equiv-l(n-1)! \equiv k(n-1)! \pmod n.$$

Therefore, $\det A \equiv -k \pmod n$ if and only if $n$ is prime by Wilson's theorem.

  • $\begingroup$ Great! I am surprised the statement holds !. $\endgroup$
    – BR Pahari
    Sep 23, 2018 at 17:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .