English

On some determinants involving the tangent function

Number Theory 2023-12-25 v6

Abstract

Let pp be an odd prime and let a,bZa,b\in\mathbb Z with pabp\nmid ab. In this paper we mainly evaluate Tp(δ)(a,b,x):=det[x+tanπaj2+bk2p]δj,k(p1)/2  (δ=0,1).T_p^{(\delta)}(a,b,x):=\det\left[x+\tan\pi\frac{aj^2+bk^2}p\right]_{\delta\le j,k\le (p-1)/2}\ \ (\delta=0,1). For example, in the case p3(mod4)p\equiv3\pmod4 we show that Tp(1)(a,b,0)=0T_p^{(1)}(a,b,0)=0 and Tp(0)(a,b,x)={2(p1)/2p(p+1)/4if (abp)=1,p(p+1)/4if (abp)=1,T_p^{(0)}(a,b,x)=\begin{cases} 2^{(p-1)/2}p^{(p+1)/4}&\text{if}\ (\frac{ab}p)=1, \\p^{(p+1)/4}&\text{if}\ (\frac{ab}p)=-1,\end{cases} where (p)(\frac{\cdot}p) is the Legendre symbol. When (abp)=1(\frac{-ab}p)=-1, we also evaluate the determinant det[x+cotπaj2+bk2p]1j,k(p1)/2.\det[x+\cot\pi\frac{aj^2+bk^2}p]_{1\le j,k\le(p-1)/2}. In addition, we pose several conjectures one of which states that for any prime p3(mod4)p\equiv3\pmod4 there is an integer xp1(modp)x_p\equiv1\pmod p such that det[sec2π(jk)2p]0j,kp1=p(p+3)/2xp2.\det\left[\sec2\pi\frac{(j-k)^2}p\right]_{0\le j,k\le p-1}=-p^{(p+3)/2}x_p^2.

Keywords

Cite

@article{arxiv.1901.04837,
  title  = {On some determinants involving the tangent function},
  author = {Zhi-Wei Sun},
  journal= {arXiv preprint arXiv:1901.04837},
  year   = {2023}
}

Comments

20 pages. To appear in Ramanujan J

R2 v1 2026-06-23T07:12:22.345Z