TITLE: Does every prime divide some Fibonacci number?
QUESTION [67 upvotes]: I am tring to show that $\forall a \in \Bbb P\; \exists n\in\Bbb N : a|F_n$, where $F$ is the fibonacci sequence defined as  $\{F_n\}:F_0 = 0, F_1 = 1, F_n = F_{n-1} + F_{n-2}$ $(n=2,3,...)$. How can I do this?
Originally, I was trying to show that $\forall a\in\Bbb N\;\exists n\in\Bbb N:a|F_n$. I soon found out that if the $k$-th Fibonacci can be divided by $m$, then the $nk$-th Fibonacci can also be divided by $m$, and this can be reduced to my original problem in this post.

REPLY [6 votes]: Any linearly recursive integer sequence has the property that every large enough prime divides some term, as long as

some term of the sequence is equal to $0$. 

That follows from the solution of linear recursive sequences.  Number the sequence so that $S_0 = 0$, and consider primes $p$ not dividing any of the denominators or characteristic roots that appear in the algebraic solution of the recurrence (this is all but a finite number of primes).  If $k$ is chosen so that $\alpha^k=1 \mod p$ for all roots $\alpha$ of the characteristic polynomial, then $p|S_k$ if the roots are distinct, and $p|S_{kp}$ whether or not the roots are distinct. 
If the sequence has the additional property that

the recursion can be run backward (its extreme coefficients are $\pm 1$) 

then every prime divides some term of the sequence.
The second statement is what was proved in the other answers: when reduced mod $p$, a recursion that can be run in both directions is periodic and thus repeats the value of $0$.
The converse, that there must exist a term equal to $0$ in order to have a term equal to $0$ mod $p$ for all large $p$, is plausible but seems difficult to prove.