Five, $f$ is a mapping from the set of natural numbers $N$ to set $A$. If for $x$, $y \in \mathbf{N}$, $x-y$ is a prime number, then $f(x) \neq f(y)$. How many elements does $A$ have at least?
Five $A$ has at least 4 elements.
Let $f(4 k+i)=i(i=1,2,3,4 . k=0,1,2, \cdots)$, then when $f(x)=f(y)$, $x-y$ is a multiple of 4, and not a prime number. That is, when $x-y$ is a prime number, $f(x) \neq f(y)$. Therefore, the minimum value of $|A| \leqslant 4$.
On the other hand, the 4 numbers $1,3,6,8$, any two of which have a difference that is a prime number. So $f(1), f(3), f(6)$, $f(8)$ are all different, $|A| \geqslant 4$.