Let $p > 2$ be a prime and $q = p^r$ for some $r \in \mathbb{Z}^+$. I will assume that all roots of unity lie in $\mathbb{C}\_p^{\times}$. Let $\zeta$ a primitive $p$-th root of unity. Let $Tr : \mathbb{F}\_q \to \mathbb{F}\_p$ be the trace. Also, denote $\pi$ to be the maximal prime in $\mathbb{Z}\_p[\zeta]$ such that $\pi^{p-1} = -p$.


For a multiplicative character $\psi: \mathbb{F}\_q \to \mathbb{C}\_p^{\times}$ ($\psi(0) = 0$), the Gauss sum for $\psi$ is defined to be
\begin{align\*}
G(\psi) = \sum\_{c \in \mathbb{F}\_q} \psi(c) \zeta^{Tr\_{\mathbb{F}\_q/\mathbb{F}\_p}(c)}
\end{align\*}


I feel that the following should be true:


If $\psi$ is order 2 and $r =1$ then $G(\psi) = \pi^{(p-1)/2}$ or $-i\pi^{(p-1)/2}$, where $i \in \mathbb{C}\_p$ is a solution to $X^2 + 1 = 0$.


My thought is that $\pi^{(p-1)/2}$ is playing the role of $i\sqrt{p}$ in the traditional value for the Gauss sum (where we view everything taking place in $\mathbb{C}$). Is this correct? Or is there something more subtle going on that I'm missing?



Value of the quadratic Gauss sum viewed in $\mathbb{C}_p$
That $G(\psi,\zeta)^2 = \psi(-1)p$ is pure algebra, so it holds in $\mathbf C$ or $\mathbf C\_p$ or any other field not of characteristic $2$ that contains a nontrivial $p$th root of unity.


You could write down a $p$-adic formula for your quadratic Gauss sum using the Gross–Koblitz formula.


First let's normalize the link between your nontrivial $p$th root of unity and your choice of $\pi$ such that $\pi^{p-1} = -p$. To each $\pi$ there is a unique nontrivial $p$th root of unity $\zeta$ such that $\zeta \equiv 1 + \pi \bmod \pi^2$, where the congruence means $\lvert\zeta - (1 + \pi)\rvert\_p \leq \lvert\pi\rvert\_p^2$, or equivalently $\lvert\zeta - (1 + \pi)\rvert\_p < \lvert\pi\rvert\_p$ since $\mathbf Q\_p(\pi) = \mathbf Q\_p(\zeta)$. Write the $\zeta$ fitting that congruence mod $\pi^2$ as $\zeta\_{\pi}$.


Every character of $\mathbf F\_q^\times$ with values in $\mathbf C\_p$ is a power of the Teichmüller character $\omega\_q$ (interpret $\mathbf F\_q$ as $\mathbf Z\_p[\zeta\_{q-1}]/(p)$). For the Gross–Koblitz formula it is convenient to write characters of $\mathbf F\_q^\times$ as powers of $\omega\_q^{-1}$, say as $\omega\_q^{-k}$ for $0 \leq k < q-1$. The quadratic character $\psi$ of $\mathbf F\_q^\times$ is $\omega\_q^{(q-1)/2} = \omega\_q^{-(q-1)/2}$, so $k = (q-1)/2$.
Let the base $p$ expansion of $k$ be $d\_0 + d\_1p + \cdots + d\_{f-1}p^{f-1}$. When $k = (q-1)/2 = (p^f-1)/2$, all of its base $p$ digits are $(p-1)/2$, so the sum of the base $p$ digits is $f(p-1)/2$.


The Gross–Koblitz formula for the quadratic character $\psi$ says
$$
-G(\psi,\zeta\_\pi) = \pi^{f(p-1)/2}\Gamma\_p\left(\frac{(p-1)/2}{q-1}\right)^f, 
$$
where $\Gamma\_p$ is Morita's $p$-adic Gamma-function. Note the minus sign on the left side: normalizing Gauss sums with an overall minus sign is reasonable for various purposes, like here and in the Hasse–Davenport relation. On the right side of the formula above,
$\pi^{f(p-1)/2}$ is a square root of $\pi^{f(p-1)} = (-p)^f = (-1)^fq$.


$\newcommand\sgn{\genfrac(){}{}}$In the special case $q = p$ (so $f = 1$), you're working with the classical quadratic Gauss sum for $\mathbf F\_p$ and the Legendre symbol. In this case
$$
-G\left(\sgn\cdot p,\zeta\_\pi\right) = \pi^{(p-1)/2}\Gamma\_p\left(\frac{1}{2}\right),
$$
where $\pi^{(p-1)/2}$ is a square root of $\pi^{p-1} = -p$. For $p > 2$ it is known that $\Gamma\_p(1/2)^2 = -\sgn{-1}p$, so if you square the right side above then you get $\pi^{p-1}\Gamma\_p(1/2)^2 = -p(-\sgn{-1}p) = \sgn{-1}p p$, which is the formula for the square of the mod $p$ quadratic Gauss sum that I mentioned at the start of this answer (when $q = p$).


