Which are those integer-coefficient polynomial functions that assign a positive prime number to every Fibonacci number? (The Fibonacci numbers are elements of a recursive sequence: $F_{0}=0, F_{1}=1$; for $n>1$ then $F_{n}=F_{n-1}+F_{n-2}$.)
Solution. Let $f$ be a polynomial function that satisfies the condition. Then $f\left(F_{0}\right)$ must be a positive prime number, let this be $p$. According to this, the constant term of the polynomial is $p$, since $f\left(F_{0}\right)=f(0)=p$. Now let's examine the remainder of the division of Fibonacci numbers by $p$. There are infinitely many consecutive pairs of Fibonacci numbers, but the remainders of the numbers in these pairs when divided by $p$ can only have at most $p^{2}$ possibilities, so there exist $i$ and $j$ such that for $i \neq j$, the remainders of $F_i$ and $F_j$ are the same, and the remainders of $F_{i+1}$ and $F_{j+1}$ are also the same. This means that the Fibonacci sequence modulo $p$ is periodic. 

Let $n_0$ be the period of this sequence. For $n > n_0$, the Fibonacci numbers modulo $p$ will repeat. Therefore, for $n > n_0$, $f(F_n) \equiv f(F_{n \mod n_0}) \pmod{p}$. Since $f(F_{n_0}) = p$, for $n > n_0$, $f(F_n) \equiv 0 \pmod{p}$. 

Now, let $n_1$ be the smallest integer such that $F_{n_1} > p$. For $n > n_1$, $f(x) > p$. Thus, substituting a Fibonacci number greater than $n_1$ and divisible by $p$ (which exists according to the above proof) into the polynomial, the value will be divisible by $p$, since in this case every term is divisible by $p$. Therefore, the polynomial $f$ assigns a prime number divisible by $p$ and greater than $p$ to this Fibonacci number, which is a contradiction. Thus, the polynomial can only be constant, i.e., the sought polynomials are: $f(x)=p$, where $p$ is a positive prime number.

Remark. Similarly to what is described in the first part of the solution, it can be proven that the well-known statement is true that for any positive integer $m$, the Fibonacci numbers modulo $m$ form a periodic sequence. In fact, this statement is true not only for the Fibonacci sequence but for all recursive sequences. The proof can be obtained by a straightforward modification of the solution described above.