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  • $\begingroup$ It is easy to see that for any prime $p\equiv3\pmod4$ we have $$S_p(i)S_p(-i)=\prod_{k=1}^{(p-1)/2}\left(1+e^{2\pi i2k^2/p}\right)=\left(\frac 2p\right).$$ $\endgroup$ Commented Aug 10, 2019 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 $$x^2-py^2=(\frac{2}{p})\times2.$$ $\endgroup$ Commented Aug 17, 2019 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 $a_p\equiv (-1)^{\frac{p+1}{4}}\pmod p$. $\endgroup$ Commented Aug 17, 2019 at 23:36