Let $K= \mathbb{Q}(\omega)$ be the cyclotomic field of the $p^{th}$ roots of unity for an odd prime $p$. Let $g = \sum_{n=0}^{p-1} \omega^{n^2}.$ Here $\omega$ denote a primtive $p^{th}$ root of $1.$ I want to show that [$\mathbb{Q}(g): \mathbb{Q}] =2.$ I spent last 30 hours on this problem, but I don't see any progress. The good thing is that using this result, I could prove some cool facts about $g.$ Please note that the problem might require the definition of Lagrange resolvents.
The following is the definition of Lagrange resolvents. Let $K$ be any cyclic extension of degree $n$ over a field $F$ of characteristics not dividing $n$, and which contains the $n^{th}$ root of $1.$ Let $\sigma$ be a generator of the cyclic group Gal$(K/F).$ Then for $\alpha \in K, $ and any $n^{th}$ root of unity $\zeta$, the Lagrange resolvent $(\alpha, \zeta) = \sum_{j=0}^{n-1} \zeta^{j} \sigma^{j}(\alpha).$ By the way, this is an exercise from Dummit and Foote.Thanks so much.
The Galois group of $\Bbb Q(\omega)/\Bbb Q$ consists of maps $\sigma_a$ for $1\le a\le p-1$ characterised by $\sigma_a(\omega)=\omega^a$. Can you show that (i) $\sigma_a(g)=g$ when $a$ is a quadratic residue modulo $p$, (ii) $\sigma_a(g)=-g$ when $a$ is a quadratic non-residue modulo $p$?