We divide entire $ Z$ into $ n$ subsets such that difference of any two elements in a subset will not be a prime number. $ n$ is at least ?

$\textbf{(A)}\ 6  \qquad\textbf{(B)}\ 5  \qquad\textbf{(C)}\ 4  \qquad\textbf{(D)}\ 3  \qquad\textbf{(E)}\ \text{None}$
1. **Assume division into two subsets \( A \) and \( B \):**
   - Let \( a \in A \). Then \( a + 2 \) must be in \( B \) because the difference \( 2 \) is a prime number.
   - Similarly, \( a + 5 \) must be in neither \( A \) nor \( B \) because the difference \( 5 \) is a prime number.
   - This leads to a contradiction because every integer must belong to one of the subsets \( A \) or \( B \).

2. **Assume division into three subsets \( A, B, \) and \( C \):**
   - Let \( a \in A \). Then \( a + 2 \) must be in \( B \) and \( a + 5 \) must be in \( C \) because the differences \( 2 \) and \( 5 \) are prime numbers.
   - However, \( a + 7 \) must be in neither \( A, B, \) nor \( C \) because the difference \( 7 \) is a prime number.
   - This again leads to a contradiction because every integer must belong to one of the subsets \( A, B, \) or \( C \).

3. **Assume division into four subsets \( A, B, C, \) and \( D \):**
   - We can use equivalence classes modulo \( 4 \). Specifically, we can define the subsets as follows:
     \[
     A = \{ x \in \mathbb{Z} \mid x \equiv 0 \pmod{4} \}
     \]
     \[
     B = \{ x \in \mathbb{Z} \mid x \equiv 1 \pmod{4} \}
     \]
     \[
     C = \{ x \in \mathbb{Z} \mid x \equiv 2 \pmod{4} \}
     \]
     \[
     D = \{ x \in \mathbb{Z} \mid x \equiv 3 \pmod{4} \}
     \]
   - In this case, the difference between any two elements in the same subset is a multiple of \( 4 \), which is not a prime number (except for the trivial case of \( 4 \), which is not considered a prime number).

Therefore, it is possible to divide the set of integers \( \mathbb{Z} \) into four subsets such that the difference between any two elements in the same subset is not a prime number.

The final answer is \( \boxed{4} \).