# how to prove an equation involving sums of Kronecker symbol

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

$$2\sum_{\substack{0 where $$\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?

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

The identity can be rewritten as $$\sum_{\substack{|x| because for $$x=0$$ the inner sum is $$1-1+1=1$$. Writing $$x=2c$$, the identity becomes $$\sum_{|c| 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.