If $p\equiv1\pmod{4}$, do we have $\sqrt{p}\in\mathbb{Q}(\zeta_p)$?

Question

Why is it true that if $p\equiv1\pmod{4}$ then $\sqrt{p}\in\mathbb{Q}(\zeta_p)$?

Answer 1

For $p \equiv 1 \bmod 4$ then $(-1)^{(p-1)/2} = 1$ so there is $c^2 \equiv -1 \bmod p$ and $(a,b) \to (a+cb,a-cb)$ is bijective $Z/(p)\times Z/(p) \to Z/(p)\times Z/(p)$ therefore $$(\sum_{a \bmod p} e^{2i \pi a^2/p})^2= \sum_{a,b \bmod p} e^{2i \pi (a^2+b^2)/p}=\sum_{a,b \bmod p}e^{2i \pi (a+cb)(a-cb)/p}=\sum_{u,v \bmod p}e^{2i \pi uv/p}\\= p+\sum_{u\bmod p,u \ne 0} \sum_{v \bmod p}e^{2i \pi uv/p}= p$$

The class field theory approach is to find the conductor of $\mathbb{Q}(\sqrt{p})/\mathbb{Q}$ is $(p)$, not sure how without using $\sum_{a \bmod p} e^{2i \pi a^2/p} = \pm \sqrt{p}$ (for $p\equiv 3 \bmod 4$ the conductor is $(4p)$, that is the Artin map is obtained from Dirichlet characters $\bmod 4p$)