# Does $(2n)!$ divide $\det[(i^2+j^2)^n]_{0\le i,j\le n-1}$ for each integer $n>2$?

For $n=1,2,3,\ldots$ let $a_n$ denote the determinant $\det[(i^2+j^2)^n]_{0\le i,j\le n-1}$. Then $$a_1=0,\ a_2=-1,\ a_3=-17280,\ a_4= 1168415539200.$$

QUESTION: Is it true that $(2n)!\mid a_n$ for all $n=3,4,\ldots$?

I even conjecture that $$a_n'=\frac{(-1)^{n(n-1)/2}a_n}{2\prod_{k=1}^n(k!(2k-1)!)}$$ is a positive integer for every integer $n>2$. Note that \begin{gather*}a_3'=1,\ a_4'=559,\ a_5'=10767500,\ a_6'=9372614611500. \end{gather*}

The question is similar to my previous question http://mathoverflow.net/questions/302130. But it seems that darij grinberg's method there does not work for the present question.

• My method easily generalizes to showing that $\det\left(\left(x_i+y_j\right)^n\right)_{0\leq i \leq n-1,\ 0 \leq j\leq n-1}$ is divisible by $n^2 V\left(x_0, x_1, \ldots, x_{n-1}\right) V\left(y_0, y_1, \ldots, y_{n-1}\right)$ whenever $x_0, x_1, \ldots, x_{n-1}, y_0, y_1, \ldots, y_{n-1}$ are $2n$ integers and $n \geq 3$. (Here, $V\left(a_0, a_1, \ldots, a_{n-1}\right)$ means the Vandermonde determinant $\prod\limits_{0 \leq i < j \leq n-1} \left(a_j - a_i\right)$.) Also, ... – darij grinberg Jun 6 '18 at 7:17
• ... we have $V\left(0^2, 1^2, \ldots, \left(n-1\right)^2\right) = \prod\limits_{k=1}^{n-1} \left(k!^2 \dbinom{2k-1}{k}\right) = \prod\limits_{k=1}^{n-1} \left(k \left(2k-1\right)!\right)$. Since you have two of these $V$ factors around, you can kill almost all of your denominator; only $2 n \left(2n-1\right)!$ remains. I'm not sure where to get the $\left(2n-1\right)!$ from, though, particularly if $2n-1$ is prime. – darij grinberg Jun 6 '18 at 7:23

This is not an answer but a claim regarding some generalized determinant: $$\det\left[(x_i+x_j)^n\right]_{i,j=0}^{n-1}$$ is divisible by the square of the Vandermonde $$\prod_{i<j}^{0,n-1}(x_i-x_j)^2.$$ Letting $x_i=i^2$ recovers your matrix.
• when we divide by the square of the Vandermonde we get a symmetric polynomial in $x_i$'s of degree $n$ which has degree 2 in each specific variable. Maybe it is explicitly computable? – Fedor Petrov Jun 6 '18 at 9:45
• @FedorPetrov: Let $x_0, x_1, \ldots, x_{n-1}, y_0, y_1, \ldots, y_{n-1}$ be $2n$ elements of a commutative ring $R$. Let $\overrightarrow{x} = \left(x_0, x_1, \ldots, x_{n-1}\right)$ and $\overrightarrow{y} = \left(y_0, y_1, \ldots, y_{n-1}\right)$. For any $\overrightarrow{z} = \left(z_0, z_1, \ldots, z_{n-1}\right) \in R^n$, we let $V \left( \overrightarrow{z} \right)$ be the Vandermonde determinant $\prod\limits_{0 \leq i < j \leq n-1}\left(z_j - z_i\right)$, and we let $e_i\left(\overrightarrow{z}\right)$ be the $i$-th elementary symmetric polynomial in ... – darij grinberg Jun 6 '18 at 13:23
• ... the $z_0, z_1, \ldots, z_{n-1}$ whenever $i$ is a nonnegative integer. (Note that $e_0\left(\overrightarrow{z}\right) = 1$.) Then, $\det \left( \left(x_i + y_j\right)^n \right)_{0 \leq i \leq n-1,\ 0 \leq j \leq n-1}$ $= V\left(\overrightarrow{x}\right) V\left(\overrightarrow{y}\right) \sum\limits_{k=0}^n \left(\prod\limits_{i \in \left\{0,1,\ldots,n\right\};\ i \neq k} \dbinom{n}{i}\right) e_{n-k}\left(\overrightarrow{x}\right) e_k\left(\overrightarrow{y}\right)$. This follows from the argument using Cauchy-Binet that I gave in the comments to mathoverflow.net/questions/302130 . – darij grinberg Jun 6 '18 at 13:25