"Consecutive" square residues in odd-order finite fields Let $\mathbb F = \mathrm{GF}(p^r)$ be finite field for $p$ an odd prime, and define
$$
\begin{align}
Q &= \bigl\{ u^2 \,\big|\, u \in \mathbb F^\times \bigr\}  \\
N &= \mathbb F^\times \smallsetminus Q
\end{align}
$$
As $\mathbb F^\times$ is a cyclic group of even order, we have $\lvert Q \rvert = \lvert N \rvert = \tfrac{1}{2}(p^r - 1)$. A classic result of number theory for the case $r = 1$ is that
$$
\# \bigl\{ q \in Q \;\big|\; q+1 \in Q \bigr\} =
\begin{cases}
(p-5)/4, & \text{if $p \equiv 1 \pmod{4}$}; \\[1ex]
(p-3)/4, & \text{if $p \equiv 3 \pmod{4}$}.
\end{cases}
$$
Is there a corresponding result for $r > 1$ as well, and how is it shown?
 A: We can in fact adapt a standard proof of the result for $r = 1$, such as the one presented in this answer to the related question for the integers modulo $p$. Below, let $\chi: \def\F{\mathbb F}\F \to \{-1,0,+1\}$ be the quadratic character of $\F^\times$ extended to the whole field, so that
$$ \chi(a) = \begin{cases} +1 & \text{for $a \in Q$}, \\ -1 & \text{for $a \in N$}, \\ 0 & \text{for $a = 0$}. \end{cases} $$
Let $C = \# \bigl\{ q \in Q \,\big|\, q+1 \in Q \bigr\}$: then we have
$$ C = \sum_{a \in \F}\Bigl(\frac{1+\chi(a)}2 \Bigr)\Bigl(\frac{1+\chi(a{+}1)}{2}\Bigr) - \frac{1}{2} - \Bigl(\frac{1 +\chi(-1)}{4}\Bigr), $$
where the first subtracted term is to correct for the $a = 0$ contribution, and the second subtracted term is to correct for the $a = -1$ contribution. Since  $\lvert N \rvert = \lvert Q \rvert$, we have
$$ \sum_{a \in \F} \chi(a) = \sum_{a \in \F} \chi(a+1) = 0; $$
then because $\chi$ is a multiplicative homomorphism, and as $a^{-1} \in Q \iff a \in Q$, we have
$$\begin{align}
C
&= \frac{\lvert \F \rvert}{4} + \frac{1}{4}\sum_{a \in \F} \chi(a) \chi(a+1) -\frac{1}{2} - \Bigl(\frac{1 + \chi(-1)}{4}\Bigr)
\\&= \frac{1}{4}\biggl[\lvert \F \rvert + \sum_{a \in \F^\times} \chi\bigl(a^{-1}(a+1)\bigr) - 2 - \Bigl(1 + \chi(-1)\Bigr)\biggr]
\\&= \frac{1}{4}\biggl[ p^r  - 3 - \chi(-1) + \sum_{a \in \F^\times} \chi\bigl(1+a^{-1}\bigr) \biggr]
\\&= \frac{1}{4}\biggl[ p^r  - 3 - \chi(-1) + \!\sum_{\! b \in \F \smallsetminus \{1\}\!\!} \chi(b) \biggr]
\\&= \frac{1}{4}\Bigl[ p^r - 4 - \chi(-1) \Bigr],
\end{align} $$
which is precisely the same as for the case as $r=1$ (in which $\chi(-1) = +1$ for $p \equiv 1 \pmod{4}$ and $\chi(-1) = -1$ otherwise) except with an included exponent. In particular: as
$$ \chi(-1) = \begin{cases} +1 &\text{if $\F\:\!{:}\:\!\mathbb Z_p$ is an even degree extension or $p \equiv 1 \pmod{4}$}; \\ -1 &\text{otherwise}, \end{cases} $$
we have $\chi(-1) = +1$ if $p^r \equiv 1 \pmod{4}$, and $\chi(-1) = -1$ otherwise. Thus we have
$$
C =
\begin{cases}
(p^r-5)/4, & \text{if $p^r \equiv 1 \pmod{4}$}; \\[1ex]
(p^r-3)/4, & \text{if $p^r \equiv 3 \pmod{4}$}.
\end{cases}
$$
