 Solution

By the Chinese remainder theorem, $b^{n-1}\equiv 1(\mathrm{mod}n)$ iff it holds modulo every $p_j$. Since $n-1=p_j{r}_j-1\equiv r_j-1(\mathrm{mod}{p}_j-1)$ and the order of $b$ modulo $p_j$ divides $p_j-1$, this is equivalent to ${\mathrm{ord}}_{p_j}(b)\mid r_j-1$ for every $j$.