A question about the Legendre symbol $\left(\frac{1+a}{p}\right)$

Question

Let $p$ be a prime number such that $p\equiv-1\,(\text{mod}\,16)$ and let $a$ be an integer such that $2a^2\equiv 1\,(\text{mod}\,p)$. Is it true that $$ \left(\frac{1+a}{p}\right)=1\ ? $$ Thanks!

Answer 1

Over the complex numbers $$2\cos^2\frac\pi8-1=\frac1{\sqrt2}$$ and so $$\sqrt2 \cos\frac\pi8=\sqrt{1+\frac1{\sqrt2}}.$$ If we let $\eta=\exp(\pi i/8)$ then $$\sqrt{1+\frac1{\sqrt2}}=\frac14(\eta^2+\eta^{-2})(\eta+\eta^{-1}) =\frac{\eta^3+\eta+\eta^{-1}+\eta^{-3}}4.$$ Why is this relevant? Let now $\eta$ be a primitive $16$-th root of unity generating a finite extension $K$ of $\Bbb F_p$. Let $$b=\frac{\eta^3+\eta+\eta^{-1}+\eta^{-3}}4.$$ Then $$b^2=1+a$$ where $$a=\frac{\eta^2+\eta^{-2}}2$$ satisfies $2a^2=1$. We conclude that the Legendre symbol $\left(\frac{1+a}p\right)=1$ iff $b\in\Bbb F_p$.

Now $b\in F_{p}$ iff $b^p=b$. But since $p\equiv -1\pmod{16}$ then $\eta^p=\eta^{-1}$, and so $$b^p=\frac{\eta^{3p}+\eta^{p}+\eta^{-p}+\eta^{-3p}}{4^p} =\frac{\eta^{-3}+\eta^{-1}+\eta+\eta^3}{4} =b$$ in $K$. Therefore $\left(\frac{1+a}p\right)=1$.