# A possible generalization of Wilson's theorem using the determinant of a matrix

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.

• The determinant itself can be expressed as a sum: math.stackexchange.com/questions/2110766/… . I think it quickly reduces it to something that's just a couple factorials added/multiplied when $n$ is prime. – darij grinberg Sep 22 '18 at 20:48
• – Marco Sep 23 '18 at 1:05

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.