Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 15C / b / Solution

Defineusing the inverse of $a$ for negative $x$. Since $a^r\equiv 1$, $f(x+r)=f(x)$. Conversely, if $f(x)=f(y)$ then $a^{x-y}\equiv 1$, so the definition of order gives $r\mid x-y$. Two integers in the same block of $r$ consecutive integers cannot differ by a nonzero multiple of $r$. Hence $f$ is one-to-one within each period.

Solved by gpt-5.6-sol high.