Understanding/computing quadratic Gauss Sum This is an Exercise(Exercise #11 section 14.7)from Dummit and Foote. This exercise has six different parts. I was able to show the first three parts, but I'm stuck with the $4^{th}$ one. Any help would be highly appreciated. Here is the problem statement.
Let the classical Gauss sum, $g = \sum_{i=0}^{p-1} \zeta_{p}^{i^2}.$ I want to show that $g= \overline{g}$ if  $p \equiv 1$ mod $4,$ and $g= -\overline{g}$ if  $p \equiv 3$ mod $4.$
 Note that $p$ is an odd prime, and $\zeta_p$ is a primitive $p^{th}$ root of unity. Please note that I don't haave any algebraic number theory background. It would be awesome if you could provide some solution/suggestions using basic Galois Theory. Thank you very much. 
By the way, there is a hint in the book. It says to use the Lagrange resolvents of $\zeta_p.$
Edit:


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*$\sum_{\tau  \in H} \tau (\zeta_p)$ = $\sum_{a=squre} \tau (\zeta_p)^a$ where $H$ is a index 2 subgroup of Gal$(Q(\zeta_p)/Q),$ and $\sum_{\tau  \in \sigma (H)} \tau (\zeta_p)$ = $\sum_{b \neq  squre} \tau (\zeta_p)^b$ where $\sigma$ generates the group Gal$(Q(\zeta_p)/Q).$

*$\sum_{\tau  \in H} \tau (\zeta_p)$ + $\sum_{\tau  \in \sigma (H)} \tau (\zeta_p)$ = $(\zeta_p, 1)=-1,$ and $\sum_{\tau  \in H} \tau (\zeta_p)$-$\sum_{\tau  \in \sigma (H)} \tau (\zeta_p)$ = $(\zeta_p, -1).$

*$g= (\zeta_p, -1)= \sum_{i=0}^{p-2} (-1)^i \sigma^i(\zeta_p)$
 A: Since $\bar\zeta_p = \zeta_p^{-1}$ $$\bar g = \sum_{i=0}^{p-1} \zeta_{p}^{-i^2}.$$
From this you can see that if $-1$ is a square mod $p$ then sum is just a rearrangement of the sum for $g$ (think about the exponents reduced mod $p$, pick $m^2 = (-1) \pmod p$ and see that $-i^2 = (m i)^2$ hits every number $\pmod p$ that $i^2$ does), so $g = \bar g$. Otherwise everything is negated so $\bar g = - g$.

To elaborate a bit on the nonsquare case:
Consider the numbers $0$, $1$, ..., $p-1$ mod $p$.
Their squares are $0^2$, $1^2$, ..., $(p-1)^2$ mod $p$, every number except $0$ occurs twice here. Call this list (A).
If $-1$ is a square mod $p$ then multiplying that list of squares by it just rearranges it (because that value will be nonzero hence invertible).
If $-1$ is not a square then the resulting set of numbers are all the nonsquares twice (and one zero). Call this list (B).
Taking list (A) and (B) together gives you every number mod $p$ twice so we get $g + \bar g = \sum_{i=0}^{p-1} 2 \zeta^i = 2\cdot 0$

For a concrete example take $p = 7$, then $$g = \zeta^0 + \zeta^1 + \zeta^4 + \zeta^2 + \zeta^2 + \zeta^4 + \zeta^1$$ and $$\bar g = \zeta^0 + \zeta^6 + \zeta^3 + \zeta^5 + \zeta^5 + \zeta^3 + \zeta^6$$ when added you get the complete sum twice which comes out as zero.
