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

Points whose period divides two solve

(2^{2} - 1) x \in Z,

so, after removing the fixed point zero, the unique two-cycle is

\frac{1}{3} ⟼ \frac{2}{3} ⟼ \frac{1}{3} .

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 orbits

\frac{1}{15} \to \frac{2}{15} \to \frac{4}{15} \to \frac{8}{15} \to \frac{1}{15},

\frac{1}{5} \to \frac{2}{5} \to \frac{4}{5} \to \frac{3}{5} \to \frac{1}{5},

and

\frac{7}{15} \to \frac{14}{15} \to \frac{13}{15} \to \frac{11}{15} \to \frac{7}{15} .

Starting at another point on one displayed orbit gives the corresponding cyclic rotation.

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