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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 $S_p(e^{2\pi i/2^n})$ $(n=3,4,\ldots)$?

Your comments are welcome!

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 $$S_{79}(\omega)=i\frac{\sqrt{79}-5\sqrt3}2\ \ \text{and}\ \ S_{227}(\omega)=i\omega(1338106\sqrt3-153829\sqrt{227}).$$

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  • $\begingroup$ 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. $\endgroup$ – Zhi-Wei Sun Aug 14 '19 at 7:38
  • $\begingroup$ I have just expanded Question 2. See also mathoverflow.net/questions/338876 for the values of $S_p(e^{2\pi i/12})$. $\endgroup$ – Zhi-Wei Sun Aug 22 '19 at 15:43
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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 $$\varepsilon_p^{h(4p)}S_p(i)^2=(\frac{2}{p})(-i)^{\frac{p+3}{2}}(-1)^{\#\{1\le b\le p-1: (\frac{p}{b})=-1\}\cup\{1\le b\le p-1: 2\mid b, (\frac{b}{p})=1\}}.$$

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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 $$(i-1)S_{79}(i)=1431-161\sqrt{79}.$$

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  • $\begingroup$ It is easy to see that for any prime $p\equiv3\pmod4$ we have $$S_p(i)S_p(-i)=\prod_{k=1}^{(p-1)/2}\left(1+e^{2\pi i2k^2/p}\right)=\left(\frac 2p\right).$$ $\endgroup$ – Zhi-Wei Sun Aug 10 '19 at 8:46
  • $\begingroup$ 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 $$x^2-py^2=(\frac{2}{p})\times2.$$ $\endgroup$ – Hai-Liang Wu Aug 17 '19 at 2:46
  • $\begingroup$ From the above comment, we see that in fact $\delta_p\equiv \frac{p+5}{4}\pmod 2$. Hence we have $a_p\equiv (-1)^{\frac{p+1}{4}}\pmod p$. $\endgroup$ – Hai-Liang Wu Aug 17 '19 at 23:36

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