Smallest possible value on Fibonacci Function Suppose $f$ is a polynomial with integer coefficients, such that for all non-negative integers $n$ the $n$-th Fibonacci number $u_n$ divides $f(u_{n+1})$. Find the smallest possible positive value of $f(4)$.

This is a really nice problem!  The answer is $255$.  First note that (by taking integer linear combinations of polynomials with the given property) the problem is equivalent to "find the GCD of all possible nonzero values of $f(4)$".  Now observe that $g(x)=x^4-1$ is one such polynomial (left as an exercise! see hint below), and $g(4)=255$.  So the answer is some divisor of $255 = 3 \times 5 \times 17$.  
To conclude, we need to show that $f(4)$ must always be divisible by $3$, by $5$, and by $17$. 
But $21 \mid f(34)$, so $f(34)$ is divisible by $3$; and $f(34) \equiv f(4) \pmod{3}$ (because $34-4$ is divisible by $3$), so $f(4)$ is also divisible by $3$.
Similarly $34 \mid f(55)$, so $f(55)$ is divisible by $17$; and $f(55) \equiv f(4) \pmod{17}$ (because $55-4$ is divisible by $17$), so $f(4)$ is also divisible by $17$.
Finally $55 \mid f(89)$, so $f(89)$ is divisible by $5$; and $f(89) \equiv f(4) \pmod{5}$ (because $89-4$ is divisible by $5$), so $f(4)$ is also divisible by $5$.  Done.
