Consider the all-familiar *Vandermonde determinant* $V\_n(x\_1,\dots,x\_n)$ of the matrix of $(i,j)$-entries $M\_n(i,j)=x\_j^{i-1}$ so that
$$V\_n(x\_1,\dots,x\_n)=\prod\_{1\leq i<j\leq n}(x\_j-x\_i).$$
Let's specialize the variables $x\_k=k\pi\_k$ where $\pi\in\mathfrak{S}\_n$ is a permutation of $n$ letters $\{1,2,\dots,n\}$.


Assume $n>3$. I like to ask:



> 
> **QUESTION.** Is this true? $V\_n(1\pi\_1,2\pi\_2,\dots,n\pi\_n)$ is congruent to $0$ moduluo $n$ for any permutation $\pi\in\mathfrak{S}\_n$.
> 
> 
> 



Vandermonde $V_n$ mod $n$
Per comments above, for a counterexample we have with necessity $\pi\_n=n$ and prime $n$. The case $n=2$ is trivial, so I assume that $n$ is an odd prime.


The elements $U:=\{ 1,2,\dots,n-1\}$ form the unit group of $GF(n)$ and the mapping $i\mapsto i\pi\_i$ has to be a permutation of $U$. Such mappings are called *complete* and [it's known](https://www.combinatorics.org/ojs/index.php/eljc/article/download/v19i1p34/pdf/) that they do not exists whenever the group has a nontrivial, cyclic Sylow 2-subgroup. In our case, $U$ has an even order and thus a nontrivial Sylow 2-subgroup, and at the same time [all its subgroups are cyclic](https://math.stackexchange.com/q/59903). Hence, no complete mappings exist, providing an affirmative answer to the question.


