5. Call two vertices of a simple polygon "visible" to each other if and only if they are adjacent or the line segment connecting them lies entirely inside the polygon (except for the endpoints which lie on the boundary). If there exists a simple polygon with $n$ vertices, where each vertex is visible to exactly four other vertices, find all possible values of the positive integer $n$.

Note: A simple polygon is one that has no holes and does not intersect itself.
5. First, prove that there does not exist an $n>6$ that satisfies the problem.

Let $A_{1}, A_{2}, \cdots, A_{n}$ be all the vertices.
Lemma 1: Suppose $A_{i}$ is visible to $A_{i-1}, A_{j}, A_{k}, A_{i+1}$ (arranged clockwise, note that $A_{i-1}, A_{i+1}$ are adjacent to $A_{i}$). Then $A_{i-1}$ is visible to $A_{j}$, $A_{j}$ is visible to $A_{k}$, and $A_{k}$ is visible to $A_{i+1}$.

Proof of Lemma 1: Consider the triangulation of the polygon that includes the edges $A_{i} A_{j}$ and $A_{i} A_{k}$.
Lemma 2: Using the notation from Lemma 1, $A_{j} A_{k}$ is an edge.
Proof of Lemma 2: Assume $A_{j} A_{k}$ is a diagonal inside the polygon.

By Lemma 1, $A_{j}$ can see $A_{j-1}$. But $A_{j} A_{i}$ and $A_{j} A_{k}$ are internal diagonals, so $A_{j} A_{i-1}$ is an edge.

Thus, on the boundary of the polygon, there is only one vertex between $A_{i}$ and $A_{j}$.

Similarly, there is only one vertex between $A_{j}$ and $A_{k}$, and between $A_{k}$ and $A_{i}$ on the boundary of the polygon. This contradicts $n>6$.
Therefore, $A_{j} A_{k}$ is an edge, and $k=j-1$.
Lemmas 1 and 2 are proved.
Let $i$ be such that $A_{i-1}$ and $A_{i+1}$ are mutually visible (such an $i$ exists, for example, take the outermost triangle in the triangulation of the polygon).

By Lemma 2, $A_{i-1}$ can see $A_{i+2}$, $A_{i+1}$ can see $A_{i-2}$, and $A_{i-2}$ can see $A_{i+2}$. Thus, we have found four vertices such that both $A_{i-1}$ and $A_{i+1}$ can see these four vertices.

If $A_{i}$ can see some vertex, then $A_{i-1}$ and $A_{i+1}$ must also see it (by Lemma 1). Therefore, $A_{i}$ can see $A_{i-2}$ and $A_{i+2}$, which implies that $A_{i-2} A_{i+2}$ is an edge (by Lemma 2). Any convex polygon can serve as an example.

The remaining case is $n=6$. In this case, by Lemma 2, there exist vertices $A_{i}, A_{j}, A_{k}$ such that $A_{i} A_{j}$, $A_{j} A_{k}$, and $A_{k} A_{i}$ are internal diagonals of the polygon.

Without loss of generality, assume they are $A_{2}, A_{4}, A_{6}$ in the hexagon. Then, $A_{3}$ cannot be seen by $A_{6}$, which implies that either $\angle A_{2}$ or $\angle A_{4}$ is greater than $180^{\circ}$, and thus, $A_{3}$ cannot be seen by either $A_{1}$ or $A_{5}$. In fact, $A_{3}$ can be seen by the other four vertices, which is a contradiction.
Therefore, there does not exist a polygon satisfying the problem for $n=6$.
Thus, only $n=5$ is possible.