Let $ P(x) \in \mathbb{Z}[x]$ be a polynomial of degree $ \text{deg} P \equal{} n > 1$. Determine the largest number of consecutive integers to be found in $ P(\mathbb{Z})$.

[i]B. Berceanu[/i]
1. Let \( P(x) \in \mathbb{Z}[x] \) be a polynomial of degree \( n > 1 \). We need to determine the largest number of consecutive integers that can be found in \( P(\mathbb{Z}) \).

2. Consider the polynomial \( Q(x) = P(x+1) - P(x) \). The degree of \( Q(x) \) is \( n-1 \) because the leading term of \( P(x+1) \) and \( P(x) \) will cancel out, leaving a polynomial of degree \( n-1 \).

3. Since \( Q(x) \) is a polynomial of degree \( n-1 \), it can have at most \( n-1 \) roots. This means that there are at most \( n-1 \) values of \( x \) for which \( Q(x) = 0 \).

4. If \( Q(x) = 1 \) for \( n \) consecutive integers, then \( P(x+1) = P(x) + 1 \) for these \( n \) consecutive integers. This implies that \( P(x) \) increases by 1 for each consecutive integer \( x \).

5. Therefore, if \( P(x) \) increases by 1 for \( n \) consecutive integers, then \( P(x) \) will take on \( n \) consecutive integer values.

6. To see why this is the maximum, consider that if \( P(x) \) were to take on \( n+1 \) consecutive integer values, then \( Q(x) \) would have to be 1 for \( n+1 \) consecutive integers, which is not possible because \( Q(x) \) is a polynomial of degree \( n-1 \) and can have at most \( n-1 \) roots.

7. Therefore, the largest number of consecutive integers that can be found in \( P(\mathbb{Z}) \) is \( n \).

\(\blacksquare\)

The final answer is \( \boxed{ n } \).