Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 11I / b / i / Solution

We prove that (i) implies (ii). Since $b$ is primitive modulo $p$, writeBeing primitive modulo $p^2$ is equivalent to $p\nmid c$. The binomial theorem then gives inductivelyIt follows that the order modulo $p^m$ is exactlyso $b$ is primitive modulo every $p^m$. This is lifting a primitive root to odd prime powers. The implication (ii)$\Rightarrow$(i) is immediate by taking $m=2$.