On $\det[x+(\frac{i\pm j}p)]_{1\le i,j\le(p-1)/2}$ for primes $p\equiv 3\pmod 4$

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?

• I have reduced the conjecture to the case $x=1$, but I have no idea how to handle the case $x=1$. – Zhi-Wei Sun Dec 12 '18 at 17:37

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 $$\det \left[x+\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}=(-1)^{(p-1)/2}\det \left[-x-\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}=(-1)(-2^{(p-1)/2}(-x))=-2^{(p-1)/2}x$$