Funnily enough this happens for all $p$ in the sequence A000057, this
sequence consists of primes $p$ for which the Fibonacci entry point is
$p+1$ (i.e. the first positive $n$ for which $p$ divides $F(n)$ is $p+1$,
where $(F(n))_{n \geq 0}$ is the Fibonacci sequence). Note that for all
$p$ in this list the polynomial $x^2 - x - 1$ is irreducible in $\mathbb{F}
_p[x]$ as is explained in section 4 of these notes.
The Conjecture
Take a prime number $p$ in A000057.
The first few of these primes are $$2,\ 3,\ 7,\ 23,\ 43,\ 67,\ 83,\ 103,\ 127,
\ 163,\dots.$$ Now let $F_{a,b}(n)$ be a sequence in $\mathbb{F}_p$ defined
by $$F_{a,b}(0) = a,\ F_{a,b}(1) = b \text{ and } F_{a,b}(n) + F_{a,b}(n+1) =
F_{a,b}(n+2).$$ We let $r_{a,b}(n)$ be the unique element of $\{0, \dots, p-1\}$
such that $r_{a,b}(n) \equiv F_{a,b}(n) \pmod{p}$. If we, for example, write down the
first couple of terms of the sequence $r_{0,1}(n)$ for $p = 7$ we
get $$ 0,\ 1,\ 1,\ 2,\ 3,\ 5,\ 1,\ 6,\ 0,\ 6,\ 6,\ 5,\ 4,\ 2,\ 6,\ 1,\dots $$ if we calculate
the sum of these $16$ values we get $49$. We can try this a couple of more times
for different values of $p$ and come to the conjecture that if $a,b$ are not both zero
we find that $$\sum_{n=0}^{2p+1} r_{a,b}(n) = p^2.$$
The Proof
First we observe that $F_{a,b}(n) + F_{a,b}(n + p + 1)$ seems to be zero for all
starting values $a,b$ and all $n$. To prove this we use the theory of characteristic
polynomials for recurrence relations to obtain a general formula for the value of
$F_{a,b}(n)$. The characteristic polynomial for this relation is given by $f(x)
= x^2 - x - 1$. Since we have chosen $p$ such that $f(x)$ is irreducible in
$\mathbb{F}_p[x]$ we have to go to the extension field $$\mathbb{F}_{p^2}:=
\mathbb{F}_p[x]/(f(x))$$ to obtain the roots $z_1,z_2$ such that $f(x) = (x-z_1)
(x-z_2)$. By comparing coefficients we see that $z_1 \cdot z_2 = -1$ and from the
theory of finite fields we know that the Frobenius map $x \mapsto x^p$ sends
$z_1$ to $z_2$ and vice versa. Now the general term $F_{a,b}(n)$ is given
by $$F_{a,b}(n) = c_1 \cdot z_1^n + c_2 \cdot z_2^n,$$ where $c_1,c_2$ are
constants in $\mathbb{F}_{p^2}$ that depend on the starting values $a,b$. We
can now calculate that
\begin{align*}
F_{a,b}(n+p+1) &= c_1\cdot z_1^{n+p+1} + c_2 \cdot {z_2}^{n+p+1} \\
&= c_1 \cdot z_1^n \cdot z_1 \cdot z_1^p + c_2 \cdot {z_2}^n \cdot {z_2} \cdot {z_2}^p \\
&= c_1 \cdot z_1^n \cdot (z_1 \cdot {z_2}) + c_2 \cdot {z_2}^n \cdot ({z_2} \cdot z_1) \\
&= - c_1 \cdot z_1^n - c_2 \cdot {{z_2}}^{n}\\
&= -F_{a,b}(n).
\end{align*}
This tells us that indeed $F_{a,b}(n) + F_{a,b}(n + p + 1) = 0$. Let us now consider
the sequences of the form $F_{0,a}(n)$, such that the first value is zero. We see that
$c_1 = -c_2$ such that the general formula is of the form $$F_{0,a}(n) = C(z_1^n -
z_2^n),$$ where $C \in \mathbb{F}_{p^2}$ depends on $a$. We want to know when
$F_{0,a}(n) = 0$ for the first time with $n > 0$ for non-zero $a$. Looking at the direct
formula we see that $F_{0,a}(n) = 0$ if and only if $z_1^n = z_2^n$. From Galois theory
we know that this happens if and only if $z_1^n \in \mathbb{F}_{p}$. Now using the fact
that $z_1^2 = z_1 +1$ we can prove by induction that $$z_1^n = \overline{F
(n)} \cdot z_1 + \overline{F(n-1)},$$ where $(F(n))_{n \geq 0}$ is now the regular
Fibonacci sequence. So we see that this is an element of $\mathbb{F}_{p}$ if and only if
$F(n)$ is divisible by $p$. So we are looking for the first positive $n$ such that
$F(n)$ is divisible by $p$. We have chosen $p$ precisely such that this value is $p+1$.
This means that the period of $F_{0,a}$ is at least $p+1$, but since $F_{0,a}(1) = a$
and $F_{0,a}(p+2) = -a$ and these are not equal, we see that the period of $F_{0,a}$
is $2(p+1)$. So we see that the pairs
$$
(F_{0,a}(0),F_{0,a}(1)),\ (F_{0,a}(1),F_{0,a}(2)),\ \dots, \ (F_{0,a}(2p+1),F_{0,a}(2p+2))
$$
are all different. We also see that the sequences $$F_{0,1},\ F_{0,2}, \dots,
F_{0,\frac{p-1}{2}}$$ are all different since a sequence of the form $F_{0,a}$ contains
only the two tuples $(0,a)$ and $(0,-a)$ with $0$ as the first entry of the pair. So if we
now count pairs we have found $$\frac{p-1}{2}\cdot 2(p+1) = p^2-1,$$ together with the
trivial sequence $F_{0,0}$ we see that we have counted all $p^2$ pairs and thus all
sequences $F_{a,b}$ are actually of the form $F_{0,c}$ for some $c$ possibly shifted.
Since the period of $F_{a,b}$ is $2(p+1)$ the sum is constant under the shifting of the
sequence such that $$\sum_{n=0}^{2p+1} r_{a,b}(n) = \sum_{n=0}^{2p+1} r_{0,c}(n),$$
for some $c$. We have seen that if $c \neq 0$ indeed $0$ only appears two times in the
sum. Since $F_{a,b}(n+p+1)=-F_{a,b}(n)$ we see that $r_{a,b}(n+p+1)+ r_{a,b}(n) = p$
if $F_{a,b}(n) \neq 0$ and thus we see that the sum is equal to
$$\sum_{n=0}^{2p+1} r_{0,c}(n) = p^2.$$
