Fibonacci Game Invariant Sum The Problem
Suppose I start with two natural numbers, $F_1=x \pmod 7$ and $F_2=y \pmod 7$ so that not both $x$ and $y$ are $0$ (and of course, $x,y<7$). From there, I form the sequence $F_{n+1}=F_n+F_{n-1}\pmod 7$. 
The interesting fact I noticed is that $\sum _1 ^{16} F_n=49$ always!
For example, if I started with, say $x=1$ and $y=4$, the first $16$ terms of my sequence would be $1,4,5,2,0,2,2,4,6,3,2,5,0,5,5,3$, and indeed, the sum is $49$.
Why is this? 
What I've Tried
We note that if we expand the first $16$ terms generally, as functions of $x$ and $y$, that $F_a+F_{8+a}\equiv 0 \pmod 7$ for all $1\le a \le 8$ regardless of what $x$ and $y$ are. Since $F_a+F_{8+a}$ can only take on values of $0$ and $7$, it is enough to show that of the first $8$ numbers $F_1$ through $F_8$, exactly one has to be $0$ (the rest being nonzero would show that the total sum is always 7*7+0*1=49). Of course, we could brute force, but is there a nice way to see what's going on?
 A: Because there only a limited number of possible pair values $\bmod 7$, and because given a certain pair you can work both forwards and backwards uniquely under Fibonacci rules, it is inevitable that you have cycles of values that return to their starting pair. I think there might be an element of small-number effect that the $48$ possible value pairs (since we exclude $0,0$) fall neatly into just $3$ cycles of length $16$, all including zeroes:
$$\begin{array}{c}
0 & 1 & 1 & 2 & 3 & 5 & 1 & 6 & 0 & 6 & 6 & 5 & 4 & 2 & 6 & 1 & (0)  \\
0 & 2 & 2 & 4 & 6 & 3 & 2 & 5 & 0 & 5 & 5 & 3 & 1 & 4 & 5 & 2 & (0)  \\
0 & 3 & 3 & 6 & 2 & 1 & 3 & 4 & 0 & 4 & 4 & 1 & 5 & 6 & 4 & 3 & (0)  \\
\end{array}$$
So every starting pair can be found in these cycles, which exhibit your symmetric effect of having $a_i\equiv -a_{i+8}$ (which also perpetuates through the Fibonacci recurrence).
So I don't know if that counts as a "nice way" - I certainly think there is more structure to be understood - but this is an interesting area that I was also looking at across bases (although I hadn't yet looked at variant starting values).
A: One way to prove your simplification (nice job btw!) [Not sure if this is nice enough, but it should be a little more satisfying than brute force]: Consider the group $G:=<A^2>\subset GL(2,\mathbb{F}_7)$ where $A=\begin{pmatrix}0&1\\1&1\end{pmatrix}$. Note that $A^2 \begin{pmatrix}F_n\\F_{n+1}\end{pmatrix}  = \begin{pmatrix} F_{n+2}\\ F_{n+3}\end{pmatrix}$ (mod $7$) for all $n\in \mathbb{N}$. Computing powers of $A$ mod $7$: $A^2= \begin{pmatrix}1&1\\1&2\end{pmatrix}$, $A^4 = \begin{pmatrix} 2&3\\3&5\end{pmatrix}$, $A^8 = -I$. Thus $A$ has order $16$ and $|G|=8$.
Let $G$ operate on $\mathbb{F}_7^2$ by multiplication mod $7$. It is easy to see that every element $(0,0)\neq (a,b)\in\mathbb{F}_7^2$ is only fixed by the identity element of $G$. So its orbit has $8$ elements, and since $|\mathbb{F}_7^2| = 49 = 8\cdot 6 + 1$, there are $6$ of these in total. Let's consider the case that some orbit contains an element of type $(0,a)$ or $(a,0)$. The Fibonacci sequences starting at these points mod $7$ look like this:
$0,a,a,2a,3a,5a,a,-a,0,-a,..$
$a,0,a,a,2a,3a,5a,a,-a,0,..$
where the dots indicate that the same sequence as before is repeated with just $a$ exchanged for $-a$. So the orbits (excluding the trivial one) of pairs in $\mathbb{F}_7^2$ ,which contain a zero in one of their pairs, contain exactly 2 zeroes (in two of their pairs). Since there are (excluding the pair $(0,0)$) exactly $6\cdot 2 = 12$ of these, each of the $6$ orbits contains exactly two of them, with the 'distance' (in the Fibonacci sequence) of one zero to the next being $8$ (following from the expansion of the sequence above). So any $8$ consecutive numbers in a Fibonacci sequence contain exactly one that is divisible by $7$.
