Given real numbers $x_1,x_2,\ldots,x_{2n}$. Let $A$ be the skew-symmetric $2n\times 2n$ matrix with entries.... Given real numbers $x_1,\cdots,x_{2n}$, let $A$ be the skew-symmetric $2n \times 2n$ matrix with entries $a_{i,j}=(x_i-x_j)^2$ for $1 \le i < j \le 2n$. Prove
$$\det(A)=4^{n-1}\left((x_1-x_2)(x_2-x_3)\cdots(x_{2n-1}-x_{2n})(x_{2n}-x_{1})\right)^2$$
I have tried to use Induction but I am not able to move from step $n$ to $n+1$. How we can proceed.
 A: Let $p(x_1,\dots,x_{2n})$ be the desired determinant as a polynomial in $x_1,\dots,x_{2n}$; it is a homogeneous polynomial of total degree $4n$. Note that the matrix given by $(x_2,x_3,\dots,x_{2n},x_1)$ can be formed by the matrix given by $(x_1,\dots,x_{2n})$ by swapping the first and last row, swapping the first and last column, negating the last row, and negating the last column, so
$$p(x_1,x_2,\dots,x_{2n})=p(x_2,\dots,x_{2n},x_1).$$
We claim that $(x_1-x_2)^2\mid p$; from this and the above will follow that
$$(x_1-x_2)^2(x_2-x_3)^2\cdots (x_{2n-1}-x_{2n})^2(x_{2n}-x_1)^2\mid p,$$
and thus (noting the degree of $p$) that $p$ is a scalar multiple of the desired polynomial.
Fix arbitrary real $x_2,\dots,x_{2n}$, and let $x_1=x_2+\epsilon$, where $\epsilon$ may vary. It suffices to show that $p(x_2+\epsilon,x_2,x_3,\dots,x_{2n})=O(\epsilon^2)$. The matrix $A$ defined by these values looks like
$$\begin{pmatrix}
0&\epsilon^2&(x_2+\epsilon-x_3)^2&(x_2+\epsilon-x_4)^2&\cdots\\
-\epsilon^2&0&(x_2-x_3)^2&(x_2-x_4)^2&\cdots\\
-(x_2+\epsilon-x_3)^2&-(x_2-x_3)^2&*&*&\cdots\\
-(x_2+\epsilon-x_4)^2&-(x_2-x_4)^2&*&*&\cdots\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{pmatrix},$$
where $*$ denotes entries we will not be concerned with. Subtracting the second row from the first, and then the second column from the first (which does not change the determinant) gives
$$\begin{pmatrix}
0&\epsilon^2&2(x_2-x_3)\epsilon+\epsilon^2&2(x_2-x_4)\epsilon+\epsilon^2&\cdots\\
-\epsilon^2&0&(x_2-x_3)^2&(x_2-x_4)^2&\cdots\\
-2(x_2-x_3)\epsilon-\epsilon^2&-(x_2-x_3)^2&*&*&\cdots\\
-2(x_2-x_4)\epsilon-\epsilon^2&-(x_2-x_4)^2&*&*&\cdots\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{pmatrix},$$
so two factors of $\epsilon$ can be taken out, one from the first row and one from the first column. This shows that $p(x_2+\epsilon,x_2,\dots,x_{2n})=O(\epsilon^2)$, and thus that $(x_1-x_2)^2\mid p$.

So, now all we need to do is find the constant factor. We'll do this by find $p(\tfrac12,-\tfrac12,\tfrac12,-\tfrac12,\dots)$. Here,
$$(x_1-x_2)^2(x_2-x_3)^2\cdots (x_{2n-1}-x_{2n})^2(x_{2n}-x_1)^2=1,$$
so we need to show that $\det A = 4^{n-1}$. The matrix $A$ in this case looks like
$$\begin{pmatrix}0&1&0&1\\-1&0&1&0\\0&-1&0&1\\-1&0&-1&0\end{pmatrix}$$
(to give the example of $n=2$). By changing the order of the rows and columns to put the odd-indexed rows and columns in order before the even-indexed rows and columns (this does not change the sign of the determinant), we get
$$\det A=\begin{vmatrix}0&B_n\\-B_n^T&0\end{vmatrix}$$
where $B_n$ is the $n\times n$ matrix with $1$s on and above the diagonal and $-1$s below it, so $\det A=(\det B_n)^2$. Note that $\det B_1=1$; subtracting the second row of $B_n$ from the first gives the block matrix
$$\begin{pmatrix}2&0\\-1&B_{n-1}\end{pmatrix},$$
where $0$ denotes a row of $n-1$ zeros and $-1$ denotes a column of $n-1$ copies of $-1$. So, $\det B_n=2\det B_{n-1}$, and so by induction $\det B_n=2^{n-1}$. This means $\det A=4^{n-1}$ and
$$p(x_1,\dots,x_{2n})=4^{n-1}(x_1-x_2)^2(x_2-x_3)^2\cdots (x_{2n-1}-x_{2n})^2(x_{2n}-x_1)^2,$$
as desired.
