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\det\big[(i^2+cij+dj^2)^{n-2}\big]_{0\leq i,j\leq n-1}\equiv 0\pmod {n^2} \] with no condition on $c$ and $d$. If $n=p$ is prime, the same congruence holds whenever the Legendre symbol $\leg{d}{p}$ is $-1$. For composite $n$, a polynomial determinant is divisible by two Vandermonde factors; after specialisation, their product already yields the required square divisor. For prime $n=p$, we estimate the rank of the matrix modulo $p$. The required rank defect follows from a coefficient cancellation obtained from the involution $t\mapsto d/t$ on $\Fp^\times$ and the condition $\leg{d}{p}=-1$.
From: Yutong Zhang [view email]
[v1]
Tue, 19 May 2026 07:37:46 UTC (12 KB)
[v2]
Sat, 23 May 2026 20:03:22 UTC (12 KB)
[v3]
Wed, 27 May 2026 18:08:39 UTC (12 KB)
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