Determine all polynomials $f$ with integer coefficients such that $f(p)$ is a divisor of $2^p-2$ for every odd prime $p$.

[I]Proposed by Italy[/i]
1. **Lemma:** Suppose we have some odd prime \( p \). If \( q \) is an odd prime such that \( q \mid f(p) \), then \( q \in \{3, p\} \).

   **Proof:** Suppose that \( q \ne p \). Now, we have 
   \[
   f(p) \equiv 0 \pmod{q},
   \]
   so for all primes \( p' \equiv p \pmod{q} \) (of which there are infinitely many by Dirichlet's theorem), we have 
   \[
   f(p') \equiv 0 \pmod{q}.
   \]
   Thus, \( q \mid 2^{p'} - 2 \) for all \( p' \equiv p \pmod{q} \). Let \( r \) be the order of \( 2 \) modulo \( q \). Then, \( r \mid p' - 1 \) for all primes \( p' \equiv 1 \pmod{q} \).

   Now, by Dirichlet's theorem, there are infinitely many primes \( p' \) such that \( p' \equiv p \pmod{q} \) and \( p' \equiv -1 \pmod{r} \). To comply with the above, we must have 
   \[
   1 \equiv -1 \pmod{r},
   \]
   so \( r = 2 \). Thus, 
   \[
   2^2 \equiv 1 \pmod{q},
   \]
   so \( q = 3 \), as desired. \(\blacksquare\)

2. Write \( f(x) = x^m g(x) \) where \( g(0) \ne 0 \). Then \( g \) also satisfies the problem condition, but \( p \nmid g(p) \), so all the prime factors of \( g(p) \) are \( 2 \) or \( 3 \). Indeed, \( g(p) \) cannot be divisible by \( 4 \), and \( g(p) \) is constant modulo \( 2 \), so we have either \( g(p) = 3^e \) for all primes \( p \), or \( g(p) = 2 \cdot 3^e \) for all primes \( p \). Both of these are not possible due to size reasons, so we must have \( g \) constant.

3. But we have \( 3^2 \nmid 2^3 - 2 \), so \( m \in \{0, 1\} \). The only constants that divide \( 2^p - 2 \) for all primes \( p \) are \( \pm 1, \pm 2, \pm 3, \pm 6 \). Thus, the solutions for \( f \) are in the set 
   \[
   \{\pm 1, \pm 2, \pm 3, \pm 6, \pm x, \pm 2x, \pm 3x, \pm 6x\}.
   \]

4. It is easy to verify that the only ones that work are 
   \[
   \boxed{\{\pm 1, \pm 2, \pm 3, \pm 6, \pm x, \pm 2x\}}.
   \]