Let $p\equiv 8 \mod 9$ be a prime, I find the following equation:

$$2\sum_{\substack{0<x<p\\ 2|x}}\sum_{r|p^2-x^2}\left(\frac{-3}{r}\right)=p+1.$$ where $\left(\frac{-3}{r}\right)$ is the Kronecker symbol. I checked it for many $p$ using computer. Does anyone have ideal how to prove it?

  • $\begingroup$ It may help to note $(-3\mid r)=(r\mid3)$ is $1$ if $r\equiv1\bmod3$, $-1$ if $r\equiv2\bmod3$. $\endgroup$ – Gerry Myerson Nov 21 '19 at 22:12
  • $\begingroup$ @GerryMyerson, Thank you for help! $\endgroup$ – yhb Nov 22 '19 at 8:43

The identity can be rewritten as $$\sum_{\substack{|x|<p\\ 2|x}}\sum_{r|p^2-x^2}\left(\frac{-3}{r}\right)=p+2,$$ because for $x=0$ the inner sum is $1-1+1=1$. Writing $x=2c$, the identity becomes $$\sum_{|c|<p/2}\,\sum_{r|p^2-4c^2}\left(\frac{-3}{r}\right)=p+2.$$ The inner sum counts the number of integral representations $p^2-4c^2=a^2+ab+b^2$ divided by $6$, hence the identity is equivalent to the statement that the number of integral representations of $p^2$ by the quadratic form $a^2+ab+b^2+4c^2$ equals $6(p+2)$. We shall verify this by Siegel's mass formula, as it appears in Über die analytische Theorie der quadratischen Formen, Ann. of Math. 36 (1935), 527-606.

As the class of $a^2+ab+b^2+4c^2$ is alone in its genus, the number of representations of $p^2$ can be calculated as a product of local densities: $$r(p^2)=\alpha_\infty\alpha_2\alpha_3\alpha_p\prod_{q\nmid 6p}\alpha_q.$$ By Hilfssatz 26 and (71) and the line below (59) in Siegel's paper, $$\alpha_\infty=\frac{p}{\sqrt{3}}\cdot\frac{\pi^{3/2}}{\Gamma(3/2)}=\frac{2\pi }{\sqrt{3}}p.$$ By Hilfssatz 13 in Siegel's paper, $$\alpha_2=\frac{3}{2}\qquad\text{and}\qquad\alpha_3=\frac{4}{3}.$$ By Hilfssatz 16 in Siegel's paper, $$\alpha_p=\left(1-p^{-2}\right)\left(1+\frac{p^{-1}}{1+p^{-1}}\right)=(1-p^{-1})(1+2p^{-1}).$$ Finally, by Hilfssatz 12 in Siegel's paper, $$\prod_{q\nmid 6p}\alpha_q=\prod_{q\nmid 6p}(1+\chi(q)q^{-1})=\frac{2}{1-p^{-1}}\prod_{q\neq 3}(1+\chi(q)q^{-1}),$$ where $\chi$ denotes the nontrivial quadratic character modulo $3$. Therefore, $$r(p^2)=(p+2)\frac{8\pi}{\sqrt{3}}\prod_{q\neq 3}(1+\chi(q)q^{-1}).$$ We can identify the product over $q\neq 3$ as $$\prod_{q\neq 3}(1+\chi(q)q^{-1})=\prod_{q\neq 3}\frac{1-q^{-2}}{1-\chi(q)q^{-1}}=\frac{9}{8}\cdot\frac{6}{\pi^2}L(1,\chi)=\frac{3\sqrt{3}}{4\pi},$$ hence in the end $$r(p^2)=(p+2)\frac{8\pi}{\sqrt{3}}\cdot\frac{3\sqrt{3}}{4\pi}=6(p+2).$$ The proof is complete.

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  • $\begingroup$ It is very good, Thank you so much! I note that p+1 is the class number of the quadratic order with discriminant -3*4*p^2, is it possible to prove using class number formula? I try, but not success. $\endgroup$ – yhb Nov 22 '19 at 8:22
  • $\begingroup$ @yhb: I don't know, but I have now included a detailed proof using Siegel's mass formula. See my updated post. $\endgroup$ – GH from MO Nov 22 '19 at 19:34
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    $\begingroup$ All right! Thank you so much for the nice answer! $\endgroup$ – yhb Nov 22 '19 at 21:10
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    $\begingroup$ hi! what is your name? This equation is find during my reasearch and I want to mention you in my paper. $\endgroup$ – yhb Nov 26 '19 at 12:40
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    $\begingroup$ All right! Sounds good! $\endgroup$ – yhb Nov 27 '19 at 15:29

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