Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 32A / b / iii / Solution

Take $n=2^k$. Every proper divisor of $n$ divides $n/2=2^{k-1}$, so the points fixed by $F^n$ that do not have exact period $n$ are precisely those fixed by $F^{n/2}$. Part (ii) therefore givesBecause different starting points are counted separately, no division by the period is made. This is the exact power-of-two periods of the doubling map.