Let $P$ be a polynomial with real coefficients. Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that there exists a real number $t$ such that
\[f(x+t) - f(x) = P(x)\]
for all $x \in \mathbb{R}$.
1. **Initial Setup:**
   Given the functional equation:
   \[
   f(x+t) - f(x) = P(x)
   \]
   for all \( x \in \mathbb{R} \), where \( P(x) \) is a polynomial with real coefficients and \( t \) is a fixed real number.

2. **Base Case:**
   Define \( f(x) \) on the interval \([0, t)\) arbitrarily. Let \( f(x) \) be any function on this interval.

3. **Inductive Step:**
   Use the given functional equation to extend \( f(x) \) to the entire real line. For \( x \in [0, t) \), define:
   \[
   f(x + t) = f(x) + P(x)
   \]
   This extends \( f(x) \) to the interval \([t, 2t)\).

4. **Further Extension:**
   Continue this process iteratively. For \( x \in [kt, (k+1)t) \) where \( k \) is an integer, define:
   \[
   f(x + t) = f(x) + P(x)
   \]
   This extends \( f(x) \) to the interval \([(k+1)t, (k+2)t)\).

5. **Negative Direction:**
   Similarly, for \( x \in [0, t) \), define:
   \[
   f(x - t) = f(x) - P(x - t)
   \]
   This extends \( f(x) \) to the interval \([-t, 0)\).

6. **General Form:**
   By induction, for any integer \( k \), we have:
   \[
   f(x + kt) = f(x) + \sum_{i=0}^{k-1} P(x + it)
   \]
   and for negative \( k \):
   \[
   f(x + kt) = f(x) + \sum_{i=k}^{-1} P(x + it)
   \]

7. **Conclusion:**
   The function \( f(x) \) can be constructed piecewise using the given functional equation. The general solution is:
   \[
   f(x) = g(x) + \sum_{i=0}^{\left\lfloor \frac{x}{t} \right\rfloor - 1} P(x - it)
   \]
   where \( g(x) \) is any function defined on the interval \([0, t)\).

The final answer is \( \boxed{ f(x) = g(x) + \sum_{i=0}^{\left\lfloor \frac{x}{t} \right\rfloor - 1} P(x - it) } \) where \( g(x) \) is any function defined on the interval \([0, t)\).