Usually, I like working with determinants related to the Vandermonde matrix, i.e.
$$\det(x\_j^{i-1})=\prod\_{i<j}(x\_j-x\_i).$$
However, I run into some unusual matrix and its determinant. Define the $(2n)\times (2n)$ matrix $\mathbb{M}\_{2n}$ with entries
$$\mathbb{M}\_{2n}(i,j)=
\begin{cases}
(x\_j-x\_i)(x\_i-x\_{2n}) \qquad \text{if $i<j$}, \\
(x\_j-x\_i)(x\_j-x\_{2n}) \qquad \text{if $i\geq j$}.
\end{cases}$$



> 
> **QUESTION.** Is this true?
> $$\det\mathbb{M}\_{2n}=(x\_1-x\_2)^2(x\_2-x\_3)^2\cdots(x\_{2n-1}-x\_{2n})^2(x\_{2n}-x\_1)^2.$$
> 
> 
> 



A determinant of perfect square polynomials
The answer is **yes** for any $n \geq 2$. (The $n=1$ case should be treated separately.)


The determinant, as a polynomial, contains $(x\_m−x\_{m+1})$ $(m=1,2,...,2n-1)$ and $(x\_{2n}-x\_1)$ as factors, because setting $x\_m=x\_{m+1}$ will make the $m$th and $(m+1)$th column of $\mathbb{M}\_{2n}$ equal, and setting $x\_{2n}=x\_1$ will make the $(2n)$th and first column of $\mathbb{M}\_{2n}$ equal, making the determinant zero.


Since [the determinant of a skew-symmetric matrix can always be written as the square of a polynomial in the matrix entries that only depend on the size of the matrix](https://en.wikipedia.org/wiki/Pfaffian), the determinant contains the factor $(x\_1−x\_2)^2(x\_2−x\_3)^2⋯(x\_{2n−1}−x\_{2n})^2(x\_{2n}−x\_1)^2$. As the determinant cannot have degree higher than that of this factor, the only thing left is to plug in values of $x\_k$ such that LHS=RHS≠0 in the last equation stated in the question.


If we let $x\_k=1$ if $k$ is odd and $0$ otherwise, then


$\mathbb{M}\_{2n}(i,j)=
\begin{cases}
\phantom{-}1 \qquad \text{if $i<j$},i \text{ odd}, j \text{ even} \\
-1 \qquad \text{if $i > j$}, i \text{ even}, j \text{ odd}\\
\phantom{-}0 \qquad \text{otherwise}
\end{cases}$.


The $(2n-1)$th row has only one nonzero element, namely $\mathbb M\_{2n-1,2n}=1$, and the $(2n-1)$th column has only one nonzero element, namely $\mathbb M\_{2n,2n-1}=-1$. By applying the [Leibniz formula for determinants](https://en.wikipedia.org/wiki/Leibniz_formula_for_determinants), we see that $\det \mathbb M\_{2n}=\det \mathbb M\_{2n-2}$. By mathematical induction, we have $\det \mathbb{M}\_{2n}=1$ for any $n$. As the RHS also evaluates to $1$, we claim that LHS=RHS for any $x\_k$s.


