A binomial determinant fomula Is there an existing or elementary proof of the determinant identity
$
\det_{1\le i,j\le n}\left( \binom{i}{2j}+ \binom{-i}{2j}\right)=1
$?
 A: I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$. 
\begin{multline}
\scriptsize\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ \scriptsize\hskip1cm=  \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 4 & 9 & 16 & 25 & 36 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}
\end{multline}
The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.
A: This is true, and in fact you can show a slightly more general fact:
$$\det_{1\le i,j\le n}\left( \binom{x_i}{2j}+ \binom{-x_i}{2j}\right)=\prod_{i=1}^n x_i^2 \prod_{i<j} (x_j^2-x_i^2) \prod_{j=1}^n \frac{2}{(2j)!}.$$
It is easy to show that this evaluates to $1$ when you set $x_i=i$. To prove this, notice that every column is an even polynomial in the $x_i$ of degree $2j$ with leading coefficient $\frac{2}{(2j)!}$. So elementary column operations will bring the matrix in a familiar Vandermonde form. 
