How many integers $n$ are there such that $n^4 + 2n^3 + 2n^2 + 2n + 1$ is a prime number?
1. We start with the polynomial \( P(n) = n^4 + 2n^3 + 2n^2 + 2n + 1 \).
2. We need to determine for which integer values of \( n \), \( P(n) \) is a prime number.
3. First, we factorize the polynomial:
   \[
   P(n) = n^4 + 2n^3 + 2n^2 + 2n + 1
   \]
   Notice that:
   \[
   P(n) = (n^2 + 1)(n^2 + 2n + 1)
   \]
   We can rewrite \( n^2 + 2n + 1 \) as \( (n+1)^2 \), so:
   \[
   P(n) = (n^2 + 1)(n+1)^2
   \]
4. For \( P(n) \) to be a prime number, one of the factors must be 1 (since a prime number has exactly two distinct positive divisors: 1 and itself).
5. Consider the factor \( (n+1)^2 \):
   - If \( (n+1)^2 = 1 \), then \( n+1 = \pm 1 \), giving \( n = 0 \) or \( n = -2 \).
6. Check these values:
   - For \( n = 0 \):
     \[
     P(0) = 0^4 + 2 \cdot 0^3 + 2 \cdot 0^2 + 2 \cdot 0 + 1 = 1
     \]
     1 is not a prime number.
   - For \( n = -2 \):
     \[
     P(-2) = (-2)^4 + 2 \cdot (-2)^3 + 2 \cdot (-2)^2 + 2 \cdot (-2) + 1 = 16 - 16 + 8 - 4 + 1 = 5
     \]
     5 is a prime number.
7. Now consider the factor \( n^2 + 1 \):
   - If \( n^2 + 1 = 1 \), then \( n^2 = 0 \), giving \( n = 0 \). We already checked \( n = 0 \) and found that \( P(0) = 1 \), which is not prime.
8. For other values of \( n \):
   - If \( n \geq 1 \), then \( (n+1)^2 \geq 4 \), making \( P(n) \) composite.
   - If \( n \leq -3 \), then \( (n+1)^2 \geq 4 \), making \( P(n) \) composite.

Conclusion:
The only integer \( n \) such that \( P(n) \) is a prime number is \( n = -2 \).

The final answer is \( \boxed{ n = -2 } \).