Let $p$ and $q$ be two distinct odd prime numbers. Determine the multiplicative order of $1+p q$ modulo $p^{2} q^{3}$.
We develop $(1+p q)^{k}$ using the binomial theorem. All terms of degree at least 3 are divisible by $p^{2} q^{3}$, so

$$
(1+p q)^{k} \equiv 1+k p q+\frac{k(k-1)}{2} p^{2} q^{2} \quad\left(\bmod p^{2} q^{3}\right)
$$

If $(1+p q)^{k} \equiv 1\left(\bmod p^{2} q^{3}\right)$, then $p^{2} q^{2}$ divides $k p q+\frac{k(k-1)}{2} p^{2} q^{2}$, so it divides $k p q$, so $p q$ divides $k$. The term $\frac{k(k-1)}{2} p^{2} q^{2}$ is therefore divisible by $p^{2} q^{3}$, so $p^{2} q^{3}$ divides $k p q$ and $p q^{2}$ divides $k$. We verify conversely that if $p q^{2}$ divides $k$ then $(1+p q)^{k} \equiv 1\left(\bmod p^{2} q^{3}\right)$, so the order sought is $p q^{2}$.