1. For a finite set $A$, there exists a function $f: \mathbf{N}_{+} \rightarrow A$, with the following property: if $i, j \in \mathbf{N}_{+}$ and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements does the set $A$ have at minimum?
1. Since the absolute value of the difference between any two of $1,3,6,8$ is a prime number, by the problem's condition, $f(1), f(3), f(6), f(8)$ are distinct elements in $A$, thus $|A| \geqslant 4$. On the other hand, if we let $A=\left\{0,1,2,3 \mid\right.$, and if $x \in \mathbf{N}_{+}, x=4 k+r$, then $f(x)=r$, where $k \in \mathbf{N}, r=0,1,2,3$. For any $x, y \in \mathbf{N}_{+}$, if $|x-y|$ is a prime number, assume $f(x)=f(y)$, then $4 \mid |x-y|$, which contradicts the fact that $|x-y|$ is a prime number. Therefore, $f: \mathbf{N}_{+} \rightarrow A$ satisfies the problem's conditions and $|A|=4$, hence $A$ has at least 4 elements.