9
$\begingroup$

Question. Is my following conjecture new? How to prove it?

Conjecture. Let $p>3$ be a prime with $p\equiv3\pmod 8$, and let $h(-p)$ denote the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. Then we have $$h(-p)=\frac1{2\sqrt p}\sum_{k=1}^{(p-1)/2}\csc\left(2\pi\frac{k^2}p\right).$$

I have checked the conjecture numerically for all primes $3<p<10^5$ with $p\equiv3\pmod8$. Your comments are welcome!

$\endgroup$
  • 2
    $\begingroup$ Via Galois theory and quadratic Gauss sums, we see that the right-hand side of the formula is a rational number. $\endgroup$ – Zhi-Wei Sun Aug 6 at 8:48
  • 7
    $\begingroup$ Assuming the analytic class number formula this should be provable using just trigonometry. E.g. one could expand $1/\sin(2\pi x/p) = -2i\zeta_p^x /(1-\zeta_p^{2x}) = (2i\zeta_p^x/p) \sum_{y=1}^{p-1} y \zeta_p^{2xy}$. $\endgroup$ – François Brunault Aug 6 at 11:55
13
$\begingroup$

Here is a possible approach based on a formula of Zhang (see Page 432 of Wenpeng Zhang, On the mean values of Dedekind sums. J. Théor. Nombres Bordeaux 8 (1996), no. 2, 429–442.) Recalling that $$\cot\Big(\frac{\pi c}{p}\Big)=\frac{2p}{\pi\varphi(p)}\sum_{\chi(-1)=-1}\overline{\chi}(c)L(1,\chi)$$ and $\csc(x)=\cot(x/2)-\cot(x),$ we find \begin{align*} \sum_{1\leqslant k\leqslant p/2}\csc\Big(\frac{2\pi k^2}{p}\Big) &=\sum_{1\leqslant k\leqslant p/2}\Big\{\cot\Big(\frac{\pi k^2}{p}\Big)-\cot\Big(\frac{2\pi k^2}{p}\Big)\Big\}\\ &=\frac{2p}{\pi\varphi(p)}\sum_{\chi(-1)=-1}\{1-\overline{\chi}(2)\}L(1,\chi)\sum_{1\leqslant k\leqslant p/2}\overline{\chi}(k^2)\\ &=\frac{p}{\pi\varphi(p)}\sum_{\chi(-1)=-1}\{1-\overline{\chi}(2)\}L(1,\chi)\sum_{k\bmod p}\overline{\chi}(k^2). \end{align*} By orthogonality, the above quantity becomes \begin{align*} \frac{p}{\pi}\{1-\overline{\chi}_2(2)\}L(1,\chi_2)=\frac{2p}{\pi}L(1,\chi_2), \end{align*} where $\chi_2$ denotes the quadratic character mod $p$ and we have used the fact that $\chi_2(2)=-1$ for $p\equiv3\bmod8$. The desired identity then follows from the class number formula of Dirichlet.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.