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 $\pi\in\mathfrak{S}_n$.
Best Answer
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 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. Hence, no complete mappings exist, providing an affirmative answer to the question.