How prove $\left(\sum\cos{\frac{2k-1}{p}\pi}\right)\cdot\left(\sum\cos{\frac{2k-1}{p}\pi}\right)$

Question

Question:let $p$ be an odd prime number,let $A$ be the set of the (postive and less than $p$) quadratic residues modulo $p$,and $B$ be the set of the (positive and less than $p$ quadraric non-residues modulo $p$,

if we let $$p=4t+1$$ prove or disprove $$\left(\sum_{1\le k\le 2t,(2k-1)\in A}\cos{\dfrac{2k-1}{p}\pi}\right)\cdot\left(\sum_{1\le k\le 2t,(2k-1)\in B}\cos{\dfrac{2k-1}{p}\pi}\right)=-\dfrac{t}{4}$$

This problem is my found,because I know prove this follow

(1):$p=5,t=1,$ $$\left(\cos{\dfrac{\pi}{5}}\right)\cdot\left(\cos{\dfrac{3\pi}{5}}\right)=-\dfrac{1}{2}$$

(2):$p=13,t=3$ $$\left(\cos{\dfrac{3\pi}{13}}+\cos{\dfrac{9\pi}{13}}+\cos{\dfrac{\pi}{13}}\right) \cdot\left(\cos{\dfrac{7\pi}{13}}+\cos{\dfrac{11\pi}{13}}+\cos{\dfrac{5\pi}{13}}\right)= -\dfrac{3}{4}$$ and so on,and I use computer to have this:

maybe this is true,and How prove it? Thank you

Answer 1

Note that $-1$ is a square $\bmod\,p$ and $$2k-1 \equiv 2\left(k+\frac{p-1}{2}\right) \equiv -2\left(-k+\frac{p+1}{2}\right)\mod p.$$ Let us first assume that $2$ is also a square $\bmod\,p$ and $k \in \{1,\ldots,(p-1)/2\}$ in what follows. Then the following are equivalent:

• $2k-1$ is a quadratic residue
• $k+(p-1)/2$ is a quadratic residue
• $-k+(p+1)/2$ is a quadratic residue

Let $\zeta_n$ denote the root of unity $\exp(2\pi\mathrm{i}/n)$. Then we have

$$\begin{eqnarray} \sum_{2k-1 \textrm{ square}}\cos\left(\frac{2k-1}{p}\pi\right)&=&\tfrac{1}{2}\sum_{2k - 1 \textrm{ square}}\left(\zeta_{2p}^{2k-1}+\zeta_{2p}^{-2k+1}\right)\\ &=&-\tfrac{1}{2}\sum_{2k-1 \textrm{ square}}\left(\zeta_{p}^{k+(p-1)/2}+\zeta_{p}^{-k+(p+1)/2}\right)\\ &=&-\tfrac{1}{2}\sum_{k \textrm{ square}}\zeta_{p}^k\\ &=&\frac{1-\sqrt{p}}{4} \end{eqnarray}$$

where the last equality follows from the explicit expression for a quadratic Gauss sum. Then $$\left(\sum_{2k-1 \textrm{ square}}\cos\left(\frac{2k-1}{p}\pi\right)\right) \left(\sum_{2k-1 \textrm{ not square}}\cos\left(\frac{2k-1}{p}\pi\right)\right) = \frac{1-\sqrt{p}}{4} \left(\frac{1}{2}-\frac{1-\sqrt{p}}{4}\right) = \frac{1-p}{16}. $$

The case that $2$ is not a square $\bmod\,p$ can be shown similarly but now you have to use the equivalence of:

• $2k-1$ is a quadratic residue
• $k+(p-1)/2$ is a quadratic non residue
• $-k+(p+1)/2$ is a quadratic non residue