Primes and Fibonacci?

We seek to prove the following lemma, from which the answer almost immediately follows:

Lemma: The Fibonacci sequence is periodic $mod \ n$ for every positive integer $n$ .

Proof: Since $F_a$ $mod \ n$ can have at most $n$ different values, it follows that there are finite, namely at most $n^2$ possibilities $mod \ n$ for the number pair $(F_a, F_{a+1})$ . Therefore there exists a pair of integers $(k,l)$ with $l>k$ such that $F_k \equiv F_l$ and $F_{k+1} \equiv F_{l+1}$ $mod \ n$ . Notice that from the definition of the Fibonacci sequence the values of $F_{k+2}$ and $F_{l+2}$ depend only upon the previous two terms $mod \ n$ , so we must have $F_{k+2} \equiv F_{l+2}$ . Inductively we arrive at the conclusion that $F_{k+i} \equiv F_{l+i}$ for every nonnegative integer $i$ . Thinking backwards, the fact $F_{a-1}=F_{a+1}-F_a$ yields the value of $F_{a-1}$ to depend only upon the next two terms, hence $F_{k-1} \equiv F_{l-1}$ . Again, inductive reasoning implies that $F_{k-i} \equiv F_{l-i}$ $mod \ n$ for every nonnegative integer $i \leq k$ . This and the previous conclusion clearly yield that the sequence $F_a$ is periodic $mod \ n$ with a period of at most $l-k$ .

Let $P(x) = \displaystyle \sum_{i=0}^n a_ix^i$ . Observe that from the problem's condition $P(F_0) = P(0) = a_0$ is a positive prime. For the sake of convenience set $a_0 = p$ . Using the result of the lemma and the fact that $F_0 \equiv 0 \pmod{p}$ we can deduce that $F_k \equiv 0 \pmod{p}$ for infinitely many nonnegative integer $k$ . Since all of the terms of $P(F_k) = \displaystyle \sum_{i=0}^n a_i{F_k}^i$ are divisible by $p$ and the value of $P(F_k)$ must be a positive prime, the only possibility is that $P(F_k)=p$ for infinitely many $k$ . However, this would imply that the polynomial $P(x)-p$ has infinitely many roots, which is a clear contradiction. Henceforth there does not exist any polynomial with the desired property.