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I have made the followng conjecture on the basis of my computation.

Conjecture. For any prime $p\equiv3\pmod4$ with $p>3$, we have $$\det\left[x+\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}=-2^{(p-1)/2}x$$ and $$\det\left[x+\left(\frac{i-j}p\right)\right]_{1\le i,j\le(p-1)/2}=x,$$ where $(\frac{\cdot}p)$ is the Legendre symbol.

The case $x=0$ is easy and known. Any ideas towards the general solution?

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  • $\begingroup$ I have reduced the conjecture to the case $x=1$, but I have no idea how to handle the case $x=1$. $\endgroup$ – Zhi-Wei Sun Dec 12 '18 at 17:37
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The first part of the conjecture follows from a result of Robin Chapman. Letting $C_{p}(x)$ be the $\frac{p-1}{2}$ by $\frac{p-1}{2}$ matrix with $(i,j)$ entry $x+(\frac{i+j-1}{p})$, where $x$ is an indeterminate, he proved that when $p\equiv 3 \pmod 4$, $$\det C_{p}(x)=-2^{(p-1)/2}x.$$ If we let $D_{p}(x)$ be the reflection of $C_{p}(x)$ w.r.t. the antidiagonal, then we have $$\det C_{p}(x)=\det D_{p}(x)$$ and $$D_{p}(x)=\left[x+\left(\frac{(p+1)/2-j+(p+1)/2-i-1}p\right)\right]_{1\le i,j\le(p-1)/2}=\left[x-\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}$$ Hence $$\det \left[x-\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}=-2^{(p-1)/2}x.$$ So $$\det \left[x+\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}=(-1)^{(p-1)/2}\det \left[-x-\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}=(-1)(-2^{(p-1)/2}(-x))=-2^{(p-1)/2}x$$

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