# What’s the sign of $\det\left(\sqrt{i^2+j^2}\right)_{1\le i,j\le n}$?

Suppose $$A=(a_{ij})$$ is a $$n×n$$ matrix by $$a_{ij}=\sqrt{i^2+j^2}$$. I have tried to check its sign by matlab. l find that the determinant is positive when n is odd and negative when n is even. How to prove it？

Recall that $$e^{-|x|}=\int_0^\infty e^{-sx^2}\,d\mu(s)$$ for some non-negative probability measure $$\mu$$ on $$[0,+\infty)$$ (Bernstein, totally monotone functions, etc.). Thus $$e^{-t|x|}=\int_0^\infty e^{-st^2x^2}\,d\mu(s)$$. for $$t>0$$. In particular, since $$(e^{-st^2(i^2+j^2)})_{i,j}$$ are rank one non-negative matrices, their mixture $$(e^{-t\sqrt{i^2+j^2}})_{i,j}$$ is a non-negative matrix for all $$t>0$$. Hence, considering the first two terms in the Taylor expansion as $$t\to 0+$$, we conclude that $$-A$$ is non-negative definite on the $$n-1$$-dimensional subspace $$\sum_i x_i=0$$, so the signature of $$A$$ is $$1,n-1$$. To finish, we just need to exclude the zero eigenvalue, i.e., to show that the columns are linearly independent, which I leave to someone else (i.e., up to this point we have shown that the determinant has the conjectured sign or is $$0$$).