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Let $p$ be an odd prime. Motivated by my question A determinant problem for primes $p\equiv 1\pmod4$, here I introduce the matrices $A^+_p$ and $A^-_p$ whose definitions are as follows: $$A^+_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^+=\left(\frac jp\right) \ \text{and}\ a_{ij}^+=\left(\frac{i+j}p\right)\ \text{for}\ i>1,$$ $$A^-_p=[a_{ij}^-]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^-=\left(\frac jp\right) \ \text{and}\ a_{ij}^-=\left(\frac{i-j}p\right)\ \text{for}\ i>1.$$

QUESTION: Let $p\equiv3\pmod4$ be a prime. Is it true that $\det A_p^-=(-1)^{(p-3)/4}?$ When $p>3$, is it true that $\det A_p^+=-2^{(p-3)/2}$?

Based on my computation, I conjecture that the question has a positive answer.

Let $p$ be an odd prime, and let $(\frac{\cdot}p)$ denote the Legendre symbol. Motivated by my question A determinant problem for primes $p\equiv 1\pmod4$, here I introduce the matrices $A^+_p$ and $A^-_p$ whose definitions are as follows: $$A^+_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^+=\left(\frac jp\right) \ \text{and}\ a_{ij}^+=\left(\frac{i+j}p\right)\ \text{for}\ i>1,$$ $$A^-_p=[a_{ij}^-]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^-=\left(\frac jp\right) \ \text{and}\ a_{ij}^-=\left(\frac{i-j}p\right)\ \text{for}\ i>1.$$

QUESTION: Let $p\equiv3\pmod4$ be a prime. Is it true that $\det A_p^-=(-1)^{(p-3)/4}?$ When $p>3$, is it true that $\det A_p^+=-2^{(p-3)/2}$?

Based on my computation, I conjecture that the question has a positive answer.

Let $p$ be an odd prime. Motivated by my question A determinant problem for primes $p\equiv 1\pmod4$, here I introduce the matrices $A^+_p$ and $A^-_p$ whose definitions are as follows: $$A^+_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^+=\left(\frac jp\right) \ \text{and}\ a_{ij}^+=\left(\frac{i+j}p\right)\ \text{for}\ i>1,$$ $$A^-_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^-=\left(\frac jp\right) \ \text{and}\ a_{ij}^-=\left(\frac{i-j}p\right)\ \text{for}\ i>1.$$

QUESTION: Let $p\equiv3\pmod4$ be a prime. Is it true that $\det A_p^-=(-1)^{(p-3)/4}?$ When $p>3$, is it true that $\det A_p^+=-2^{(p-3)/2}$?

Based on my computation, I conjecture that the question has a positive answer.

Let $p$ be an odd prime. Motivated by my question A determinant problem for primes $p\equiv 1\pmod4$, here I introduce the matrices $A^+_p$ and $A^-_p$ whose definitions are as follows: $$A^+_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^+=\left(\frac jp\right) \ \text{and}\ a_{ij}^+=\left(\frac{i+j}p\right)\ \text{for}\ i>1,$$ $$A^-_p=[a_{ij}^-]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^-=\left(\frac jp\right) \ \text{and}\ a_{ij}^-=\left(\frac{i-j}p\right)\ \text{for}\ i>1.$$

QUESTION: Let $p\equiv3\pmod4$ be a prime. Is it true that $\det A_p^-=(-1)^{(p-3)/4}?$ When $p>3$, is it true that $\det A_p^+=-2^{(p-3)/2}$?

Based on my computation, I conjecture that the question has a positive answer.

Let $p$ be an odd prime. Motivated by my question A determinant problem for primes $p\equiv 1\pmod4$, here I introduce the matrices $A^+_p,A^-_p,B_p^+,B_p^-$ whose definitions are as follows: $$A^+_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^+=\left(\frac jp\right) \ \text{and}\ a_{ij}^+=\left(\frac{i+j}p\right)\ \text{for}\ i>1,$$ $$A^-_p=[a_{ij}^-]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^-=\left(\frac jp\right) \ \text{and}\ a_{ij}^-=\left(\frac{i-j}p\right)\ \text{for}\ i>1,$$ $$B^+_p=[b_{ij}^+]_{0\le i,j\le (p-1)/2}\ \text{with}\ b_{0j}^+=\left(\frac jp\right) \ \text{and}\ b_{ij}^+=\left(\frac{i+j}p\right)\ \text{for}\ i>0,$$ $$B^-_p=[b_{ij}^-]_{0\le i,j\le (p-1)/2}\ \text{with}\ b_{0j}^-=\left(\frac jp\right) \ \text{and}\ b_{ij}^-=\left(\frac{i-j}p\right)\ \text{for}\ i>0.$$

QUESTION: Let $p\equiv3\pmod4$ be a prime. Is it true that $$\det A_p^-=(-1)^{(p-3)/4}\ \ \text{and}\ \ \det B_p^-=-1?$$ When $p>3$, is it true that $$\det A_p^+=-2^{(p-3)/2}\ \ \text{and}\ \ \det B_p^+=2^{(p-1)/2}?$$

Based on my computation, I conjecture that the question has a positive answer.

Let $p$ be an odd prime. Motivated by my question A determinant problem for primes $p\equiv 1\pmod4$, here I introduce the matrices $A^+_p$ and $A^-_p$ whose definitions are as follows: $$A^+_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^+=\left(\frac jp\right) \ \text{and}\ a_{ij}^+=\left(\frac{i+j}p\right)\ \text{for}\ i>1,$$ $$A^-_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^-=\left(\frac jp\right) \ \text{and}\ a_{ij}^-=\left(\frac{i-j}p\right)\ \text{for}\ i>1.$$

QUESTION: Let $p\equiv3\pmod4$ be a prime. Is it true that $\det A_p^-=(-1)^{(p-3)/4}?$ When $p>3$, is it true that $\det A_p^+=-2^{(p-3)/2}$?

Based on my computation, I conjecture that the question has a positive answer.