Prove that if a continuous map $F$ of an interval into itself has a periodic orbit of period three then it also has periodic orbits of least period $n$ for all positive integers $n$.

Explain briefly why there must be at least two periodic orbits of least period $5 .$

[You may assume without proof:

(i) If $U$ and $V$ are non-empty closed bounded intervals such that $V \subseteq F(U)$ then there is a closed bounded interval $K \subseteq U$ such that $F(K)=V$.

(ii) The Intermediate Value Theorem.]