$y=1$. In either case, $y=z=1$. Then we may solve for our initial variables to show that $a=b$ and $c=a+b$. Restated, $a=k$, $b=k$, and $c=2k$.

The number $k$ in the cycle is the greatest common divisor of the sequence.

Because of the Fibonacci additive property, if any number divides two or more elements of the sequence (excluding the first two, which may be divisible by 3), it must divide all numbers in the sequence. Thus, $k$ must divide every element. The least of these elements, then, can only be $k$ itself, making it the greatest common divisor.

Will it be the case that all 3-free Fibonacci sequences end in cycles of length 3? We will build suspense by delaying this discussion, meanwhile we have a lemma about the length of any potential cycle:

\[thm:parity\] Any cycle in a 3-free Fibonacci sequence is of length $3n$ for some integer $n$.

Begin with any 3-free Fibonacci sequence, and divide out the highest power of 2 in the GCD of all its elements. The resulting sequence is a 3-free Fibonacci sequence with at least one odd element. It is clear that dividing or multiplying any number by 3 does not change its parity. Thus, 