# What can you say about the periods of a function with uniformly bounded periodic orbits?

Assume that all prime periods of periodic orbits of a continuous map $f:[0:1]\to [0:1]$ are uniformly bounded (i.e. there exists N such that the prime period of every periodic orbit of f is smaller than N). What can you say about periods of periodic orbits of f? For example, can f have a periodic orbit of period 2007? Of period 2048?

This is just a problem I found online at http://www.its.caltech.edu/%7Easgor/Ma4/ in relation to what I am currently studying, so hopefully I am not missing anything to do it. I'm not sure how to approach this though. The only thing I can think of is that since $f([0,1])\subset [0,1]$ and is continuous, due to the intermediate value theorem, $f$ has a fixed point. But I am not sure on how to work this in, or if it is relevant. I figure the interval $[0,1]$ is somehow relevant, but am not positive.

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I think this is an exercise in Sharkovskii's theorem. Suppose that $f$ has a periodic point $x$ with prime period $n$ which is not a power of $2$, so $n = 2^mk$, where $k\neq 1$ is odd. Then $y = f^{2^m}(x)$ has prime period $k$. By Sharkovskii's theorem, this implies that for every prime number $p>k$, there is a point $y$ with period $p$. This is, of course, necessarily the prime period of $y$. In particular, we see that $f$ cannot have uniformly bounded prime periods, since there are infinitely many prime numbers $>k$.
It follows that the only way that $f$ can have uniformly bounded prime periods is if all its periodic points have periods that are powers of $2$. So $f$ cannot have an orbit of period $2007$, but it can have a point of period $2048 = 2^{11}$.