# On the values of $\prod_{k=1}^{(p-1)/2}(e^{2\pi i/12}-e^{2\pi i k^2/p})$ for primes $p>3$

In a recent preprint, I investigated $$S_p(x):=\prod_{k=1}^{(p-1)/2}(x-e^{2\pi ik^2/p}),$$ where $$p$$ is an odd prime and $$x$$ is a root of unity.

Motivated by Question 337879 and Question 338325, here I pose my conjecture on $$s_p:=S_p(e^{2\pi i/12})=\prod_{k=1}^{(p-1)/2}(e^{2\pi i/12}-e^{2\pi i k^2/p})$$ for primes $$p>3$$.

Conjecture. Let $$p>3$$ be a prime.

(i) If $$p\equiv13\pmod{24}$$, then $$s_p=i(-1)^{\frac{p-5}8+|\{1\le k<\frac p4:\ (\frac kp)=-1\}|}(x_p\sqrt3-y_p\sqrt p),$$ where $$(\frac kp)$$ is the Legendre symbol, and $$(x_p,y_p)$$ is the least positive integer solution to the diophantine equation $$3x^2+1=py^2$$.

(ii) When $$p\equiv19\pmod{24}$$, we may write $$p=(4x)^2+3y^2$$ with $$x,y\in\mathbb Z$$, and we have $$s_p=(-1)^{(p-19)/24+x}(1+i)\frac{1+\sqrt3}2.$$

(iii) If $$p\equiv1,7\pmod{24}$$, then $$(-1)^{\lfloor \frac{h(-p)}2\rfloor+|\{1\le k<\frac p{12}:\ (\frac kp)=1\}|} e^{2\pi i(p-1)/48}s_p>0,$$ where $$h(-p)$$ is the class number of the imaginary quadratic field $$\mathbb Q(\sqrt {-p})$$.

(iv) If $$p\equiv5\pmod6$$ but $$p\not\equiv23\pmod{24}$$, then $$(-1)^{\lfloor \frac{h(-p)}2\rfloor+|\{1\le k<\frac p{12}:\ (\frac kp)=1\}|} e^{2\pi i5(p-1)/48}s_p>0.$$ When $$p\equiv 23\pmod{24}$$, we have $$(-1)^{\lfloor \frac{h(-p)}2\rfloor+|\{1\le k<\frac p{12}:\ (\frac kp)=1\}|} e^{2\pi i5(p-1)/48}s_p<0.$$

I have checked my above conjecture numerically.

QUESTION. How to prove the conjecture? How to determine the exact value of $$s_p$$ for a general prime $$p>3$$ with $$p\not\equiv13,19\pmod{24}$$?