# On the product $\prod_{k=1}^{(p-1)/2}(x-e^{2\pi i k^2/p})$ with $x$ a root of unity

Let $$p$$ be an odd prime. Dirichlet's class number formula for quadratic fields essentially determines the value of the product $$\prod_{k=1}^{(p-1)/2}(1-e^{2\pi ik^2/p})$$. I think it is interesting to investigate the product $$S_p(x)=\prod_{k=1}^{(p-1)/2}(x-e^{2\pi i k^2/p})$$ with $$x$$ a root of unity. In a recent preprint available from http://arxiv.org/abs/1908.02155, I determined the value of $$S_p(i)$$ for $$p\equiv 1\pmod4$$. For the cubic root $$\omega=(-1+\sqrt{-3})/2$$ of unity, I have proved in the same preprint that $$(-1)^{|\{1\le k\le\lfloor\frac{p+1}3\rfloor:\ (\frac kp)=-1\}|}S_p(\omega)=\begin{cases}1&\text{if}\ p\equiv1\pmod{12},\\\omega \varepsilon_p^{h(p)}&\text{if}\ p\equiv5\pmod{12},\end{cases}$$ where $$(\frac kp)$$ is the Legendre symbol, $$\varepsilon_p$$ and $$h(p)$$ are the fundamental unit and the class number of the real quadratic field $$\mathbb Q(\sqrt p)$$.

Question 1. How to determine the value of $$S_p(i)$$ for primes $$p\equiv3\pmod4$$? How to determine the value of $$S_p(\omega)$$ for primes $$p\equiv 7,11\pmod{12}$$?

Question 2. Let $$p>3$$ be a prime and let $$n>2$$ be an integer. Define $$f_n(p)=(-1)^{|\{1\le k<\frac p{2^n}:\ (\frac kp)=1\}|}S_p(e^{2\pi i/2^n})$$ Via numerical computation, I guess that $$e^{-2\pi i(p-1)/2^{n+2}}f_n(p)>0$$ if $$p\equiv1\pmod4$$, and $$(-1)^{(h(-p)+1)/2}f(p)e^{-2\pi i(p+2^n-1)/2^{n+2}}>0$$ if $$p\equiv3\pmod4$$, where $$h(-p)$$ is the class number of the imaginary quadratic field $$\mathbb Q(\sqrt{-p})$$. How to prove this observation? How to determine the exact values of $$S_p(e^{2\pi i/2^n})$$ $$(n=3,4,\ldots)$$?

New Addition (August 12, 2019). I have conjectures on the exact values of $$S_p(i)$$ and $$S_p(\omega)$$ for primes $$p\equiv 3\pmod4$$. For the conjectural value of $$S_p(i)$$ with $$p\equiv3\pmod4$$, see my posted answer. Here I state my conjecture on $$S_p(\omega)$$.

Conjecture. Let $$p>3$$ be a prime with $$p\equiv 3\pmod4$$, and let $$(x_p,y_p)$$ be the least positive integer solution to the diophantine equation $$3x^2+4\left(\frac p3\right)=py^2.$$ Then \begin{align}S_p(\omega)=&(-1)^{(h(-p)+1)/2}\left(\frac p3\right)\frac{x_p\sqrt3-y_p\sqrt{p}}2 \\&\times\begin{cases}i&\text{if}\ p\equiv7\pmod{12}, \\(-1)^{|\{1\le k<\frac p3:\ (\frac kp)=1\}|}i\omega&\text{if}\ p\equiv11\pmod{12}. \end{cases}\end{align}

For example, this conjecture predicts that $$S_{79}(\omega)=i\frac{\sqrt{79}-5\sqrt3}2\ \ \text{and}\ \ S_{227}(\omega)=i\omega(1338106\sqrt3-153829\sqrt{227}).$$

• A a supplement to my conjecture on $S_p(\omega)$, for any prime $p\equiv3\pmod4$, I conjecture that the equation $3x^2+4(\frac p3)=py^2$ always has integer solutions. – Zhi-Wei Sun Aug 14 '19 at 7:38
• I have just expanded Question 2. See also mathoverflow.net/questions/338876 for the values of $S_p(e^{2\pi i/12})$. – Zhi-Wei Sun Aug 22 '19 at 15:43

Let $$p>3$$ be a prime with $$p\equiv3\pmod 4$$. We first show that $$(i-(\frac{2}{p}))S_p(i)\in \mathbb{Q}(\sqrt{p}).$$

Clearly $${\rm Gal}(\mathbb{Q}(i,\zeta_p)/\mathbb{Q}(\sqrt{p}))=\{\sigma_a: a\in (\mathbb{Z}/4p\mathbb{Z})^{\times},(\frac{p}{a})=+1\}.$$ Here $$\sigma_a: \zeta_{4p}\mapsto\zeta_{4p}^a.$$

Then for each $$\sigma_a\in {\rm Gal}(\mathbb{Q}(i,\zeta_p)/\mathbb{Q}(\sqrt{p}))$$, if $$a\equiv 1\pmod 4$$ and $$(\frac{a}{p})=+1$$, then clearly $$\sigma_a$$ acts trivially on $$((i-(\frac{2}{p})))S_p(i)$$. If $$a\equiv 3\pmod 4$$ and $$(\frac{a}{p})=-1$$, then $$\sigma_a((i-(\frac{2}{p}))S_p(i))=(-i-(\frac{2}{p}))\prod_{1\le k\le \frac{p-1}{2}}(-i-\zeta_p^{-k^2}).$$ Noting that $$S_p(i)S_p(-i)=(\frac{2}{p})$$ and $$\prod_{1\le k\le \frac{p-1}{2}}(-i-\zeta_p^{k^2})\prod_{1\le k\le \frac{p-1}{2}}(-i-\zeta_p^{-k^2}) =\frac{(-i)^p-1}{-i-1}=-i,$$ one can easily verify that $$\sigma_a$$ fixes $$(i-(\frac{2}{p}))S_p(i)$$.

Next we let $$\varepsilon_p>1$$ and $$h(4p)$$ be the fundamental unit and class number of $$\mathbb{Q}(\sqrt{p})$$ respectively. By the class number formula we have (here we let $$(\frac{p}{\cdot})$$ be the character modulo $$4p$$ of field $$\mathbb{Q}(\sqrt{p})$$, and let $$e^{2\pi i/4p}=i^s\times e^{2\pi it/p}$$ with $$ps+4t=1.$$) \begin{align*} \varepsilon_p^{h(4p)}=&\frac{\prod_{1\le b\le 2p-1,(\frac{p}{b})=-1}\sin(\pi b/4p)}{\prod_{1\le c\le 2p-1,(\frac{p}{c})=+1}\sin(\pi c/4p)} \\=&\prod_{1\le b\le 2p-1, (\frac{p}{b})=+1}\frac{\sin(\pi(2p-b)/4p)}{\sin(\pi b/4p)} \\=&(-i)^{\frac{p-1}{2}}\prod_{1\le b\le 2p-1,(\frac{p}{b})=+1}\frac{1+e^{2\pi ib/4p}}{1-e^{2\pi ib/4p}} \\=&(-i)^{\frac{p-1}{2}}\prod_{1\le b\le 2p-1,(\frac{p}{b})=+1}\frac{1+i^{sb}\zeta_p^{tb}}{1-i^{sb}\zeta_p^{tb}} \\=&(-i)^{\frac{p-1}{2}}\prod_{1\le b\le p-1, b\equiv 1\pmod4, (\frac{b}{p})=1}\frac{1-i\zeta_p^{tb}}{1+i\zeta_p^{tb}}\prod_{1\le b\le p-1, b\equiv 3\pmod4, (\frac{b}{p})=-1}\frac{1+i\zeta_p^{tb}}{1-i\zeta_p^{tb}}\\&\times \prod_{1\le b\le p-1, b\equiv 1\pmod4, (\frac{b}{p})=-1}\frac{1-i\zeta_p^{-tb}}{1+i\zeta_p^{-tb}}\prod_{1\le b\le p-1, b\equiv 3\pmod4, (\frac{b}{p})=1}\frac{1+i\zeta_p^{-tb}}{1-i\zeta_p^{-tb}}. \end{align*}

Then we obtain \begin{align*} \varepsilon_p^{h(4p)}=&(-i)^{\frac{p-1}{2}}(-1)^{\#\{1\le b\le p-1: (\frac{p}{b})=-1\}} \prod_{1\le b\le p-1,2\nmid b, (\frac{b}{p})=1}\frac{1-i\zeta_p^{tb}}{1+i\zeta_p^{tb}} \prod_{1\le b\le p-1,2\nmid b,(\frac{b}{p})=-1}\frac{1+i\zeta_p^{tb}}{1-i\zeta_p^{tb}} \\=&(-i)^{\frac{p-1}{2}}(-1)^{\#\{1\le b\le p-1: (\frac{p}{b})=-1\}\cup\{1\le b\le p-1: 2\mid b, (\frac{b}{p})=1\}}\prod_{1\le k\le \frac{p-1}{2}}\frac{1-i\zeta_p^{k^2}}{1+i\zeta_p^{k^2}} \\=&(-i)^{\frac{p+3}{2}}(-1)^{\#\{1\le b\le p-1: (\frac{p}{b})=-1\}\cup\{1\le b\le p-1: 2\mid b, (\frac{b}{p})=1\}}\cdot\frac{S_p(-i)}{S_p(i)}. \end{align*}

Finally we get $$\varepsilon_p^{h(4p)}S_p(i)^2=(\frac{2}{p})(-i)^{\frac{p+3}{2}}(-1)^{\#\{1\le b\le p-1: (\frac{p}{b})=-1\}\cup\{1\le b\le p-1: 2\mid b, (\frac{b}{p})=1\}}.$$

Dr. Timothy Foo has kindly sent me his following observation (based on his numerical computation) about $$S_p(i)$$ for primes $$p\equiv3\pmod4$$: $$(i-(\frac 2p))S_p(i)$$ has the form $$a+b\sqrt p$$ with $$a,b\in\mathbb Z$$.

Now I report that I have found the exact value of $$S_p(i)$$ for primes $$p\equiv3\pmod4$$. Namely, I have formulated the following conjecture on the basis of my computation.

Conjecture. Let $$p>3$$ be a prime with $$p\equiv3\pmod4$$, and let $$h(-p)$$ be the calss number of the imaginary quadratic field $$\mathbb Q(\sqrt{-p})$$. Let $$\varepsilon_p$$ and $$h(p)$$ be the fundamental unit and the class number of the real quadratic field $$\mathbb Q(\sqrt p)$$ respectively. Write $$\varepsilon_p^{h(p)}=a_p+b_p\sqrt p$$ with $$a_p$$ and $$b_p$$ positive integers. Then $$\left(i-(-1)^{(p+1)/4}\right)S_p(i)=(-1)^{\frac{h(-p)+1}2\cdot\frac{p+1}4}(s_p-t_p\sqrt p),$$ where $$s_p=\sqrt{a_p+(-1)^{(p+1)/4}}\ \ \ \text{and}\ \ \ t_p=\frac{b_p}{s_p}$$ are positive integers.

Example. For $$p=79$$, we have $$h(-p)=5$$, $$h(p)=3$$ and $$\varepsilon_p=80+9\sqrt p$$. Note that $$\varepsilon_p^{h(p)}=(80+9\sqrt{79})^3=2047760 + 230391\sqrt{79},$$ and $$s_p=\sqrt{2047760+1}=1431\ \ \ \text{and}\ \ \ t_p= \frac{230391}{1431}=161.$$ Thus the conjecture for $$p=79$$ states that $$(i-1)S_{79}(i)=1431-161\sqrt{79}.$$

• It is easy to see that for any prime $p\equiv3\pmod4$ we have $$S_p(i)S_p(-i)=\prod_{k=1}^{(p-1)/2}\left(1+e^{2\pi i2k^2/p}\right)=\left(\frac 2p\right).$$ – Zhi-Wei Sun Aug 10 '19 at 8:46
• If we write $\varepsilon_p^{h(4p)}=a_p+b_p\sqrt{p}$ and write $(i-(\frac{2}{p}))S_p(i)=u_p+v_p\sqrt{p}$, then from the identity I got we have $$(u_p+v_p\sqrt{p})^2=2\times(-1)^{\frac{p+5}{4}+\delta_p}(a_p-b_p\sqrt{p}).$$ Here $$\delta_p=\#\{1\le b\le p-1: 2\nmid b,(\frac{p}{b})=-1\}\cup\{1\le b\le \frac{p-1}{2}: (\frac{b}{p})=\frac{2}{p}).$$ From this we obtain that $$a_p\equiv (\frac{2}{p})(-1)^{\frac{p+5}{4}+\delta_p}\equiv (-1)^{\delta_p-1}\pmod p$$ and $$u_p^2-pv_p^2=(\frac{2}{p})\times2,$$ i.e., $(u_p,v_p)$ is a solution of the equation $$x^2-py^2=(\frac{2}{p})\times2.$$ – Hai-Liang Wu Aug 17 '19 at 2:46
• From the above comment, we see that in fact $\delta_p\equiv \frac{p+5}{4}\pmod 2$. Hence we have $a_p\equiv (-1)^{\frac{p+1}{4}}\pmod p$. – Hai-Liang Wu Aug 17 '19 at 23:36