On the determinant $\det[\sec2\pi\frac{jk}p]_{0\le j,k\le(p-1)/2}$

On the basis of my computation, I have the following conjecture involving the secant function.

Conjecture. Let $$p$$ be an odd prime and define $$S_p:=\det\left[\sec2\pi\frac{jk}p\right]_{0\le j,k\le (p-1)/2}.$$ If $$p\equiv 1\pmod4$$, then $$S_p=0$$. If $$p\equiv3\pmod4$$, then $$s_p=\frac{(\frac 2p)S_p}{2^{(p-3)/2}p^{(p+1)/4}}$$ is a positive odd integer.

Remark. Via Mathematica I find that $$s_3=s_7=s_{11}=1$$ and $$s_{19}=19$$.

Let $$p$$ be an odd prime. For $$a=1,\ldots,p-1$$ let $$\pi_a$$ be the permutation on $$\{0,1,\ldots,(p-1)/2\}$$ such that $$\pi_a(k)$$ is the unique $$r\in\{0,\ldots,(p-1)/2\}$$ with $$ak$$ congruent to $$r$$ or $$-r$$ modulo $$p$$. By a result of H. Pan in the paper arXiv:0601026, the sign of $$\pi_a$$ is $$(\frac ap)^{(p+1)/2}$$. If $$p\equiv3\pmod4$$, then by applying the automorphism $$\sigma_a$$ in the Galois group of $$\mathbb Q(e^{2\pi i/p})$$ with $$\sigma(e^{2\pi i/p})=e^{2\pi ia/p}$$, we see that $$\sigma_a(S_p)=\det\left[\sec 2\pi\frac{ajk}p\right]_{0\le j,k\le(p-1)/2}=\left(\frac ap\right)^{(p+1)/2}S_p$$ with the help of Pan's result. Thus, when $$p\equiv3\pmod4$$ we have $$\sigma_a(S_p)=S_p$$ for all $$a=1,\ldots,p-1$$, and hence $$S_p$$ is rational. If $$p\equiv1\pmod 4$$, then for each $$a=1,\ldots,p-1$$ we have $$\sigma_a(\sqrt p)=\sum_{x=0}^{p-1}\sigma_a(e^{2\pi ix^2/p})=\sum_{x=0}^{p-1}e^{2\pi iax^2/p}=\left(\frac ap\right)\sqrt p.$$ Thus, $$\sigma_a(S_p/\sqrt p)=S_p/\sqrt p$$ for all $$a=1,\ldots,p-1$$, and hence $$S_p/\sqrt p\in\mathbb Q$$.

Similarly, I can show that for any odd prime $$p$$ the numbers $$\frac1{2^{(p-1)/2}p^{(p-3)/4}}\det\left[\sec2\pi \frac{jk}p\right]_{1\le j,k\le(p-1)/2}$$ and $$\frac1{2^{(p-1)/2}p^{(p-5)/4}}\det\left[\csc2\pi \frac{jk}p\right]_{1\le j,k\le(p-1)/2}$$ are rational numbers. I conjecture that these two numbers are always integers.

• For the last sequence, one gets $[1, 1, 0, 1, 1, 2, 1, 0, 8, 0, 37, 242, 844, 0, \dots]$. – F. C. May 19 at 6:01
• And for the previous one, one gets $[1, 1, 2, 3, 9, 32, 95, 402, 9408, 40672, 1174257, 6844400, 41323172, 418892388, \dots]$. – F. C. May 19 at 6:15
• It seems that the squares of the matrices have coefficients in $\mathbb{Z}$. – F. C. May 19 at 6:17
• And for the original question, one gets $[1, 1, 1, 19, 67, 5084, 3756652, 34907699, 109337677693,\dots]$. – F. C. May 19 at 12:34