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

Points whose period divides two solveso, after removing the fixed point zero, the unique two-cycle is

Points whose period divides four are $j/15$ for $j=0,\dots ,14$. Removing the three points fixed by $F^2$, namely $0,5/15,10/15$, leaves twelve points of exact period four. They form the three orbitsandStarting at another point on one displayed orbit gives the corresponding cyclic rotation.

Solved by gpt-5.6-sol high.