Exercise 5.34

Proof. Let $p$ be an odd prime number, $p\ne 3$ and $\zeta =e^{2\mathit{i\pi }∕p}$.

(As $\cos <!--\; FUNCTION\; APPLICATION\; -->(2\alpha )=1-2{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2\alpha$, so $3-4{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2\alpha =1+2\cos <!--\; FUNCTION\; APPLICATION\; -->\alpha$.)

Let

Since $f:{𝔽}_p^{\text{∗}}\to {𝔽}_p^{\text{∗}}$ defined by $\bar{j}\mapsto 2\bar{j}$ is a bijection,

Therefore

${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^{p-1}{\zeta }^{-k}={\left({\zeta }^p\right)}^{-(p-1)∕2}=1$. Moreover, ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^{p-1}(1-{\zeta }^{3k})={\prod }_{k=1}^{p-1}(1-{\zeta }^k)$, since $\bar{k}\mapsto 3\bar{k}$ is a bijection in ${𝔽}_p^{\text{∗}}$, thus $P=1$, and consequently

Thus ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^{(p-1)∕2}\left(3-4{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2\left(\frac{2\mathit{\pi j}}{p}\right)\right)=\pm 1$.

Let $\nu$ be the number of negative factors in this product.

If $1\le j\le (p-1)∕2$, then $0<4\mathit{\pi j}∕p<2\pi$.

Let $\mu$ be the number of integers $j,1\le j\le (p-1)∕2$ such that the least remainder of $3j$ is negative. Since $3\le 3j\le 3(p-1)∕2$, these $j$ are the integers such that $(p-1)∕2<3j\le p-1$, and since $3j\ne p∕2$, such that $\frac{p}{2}<3j<p$, so $\mu =\nu$. Therefore

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