Let $I$ be the ideal of $\mathbb{Z}[X]$ generated by $(X^2+2)(X+1)$ and $5$. Then, how many elements $f \in \mathbb{Z}[X]/I$ satisfy $f^{18}=1$?
I know that I can write $f(X)=aX^2+bX+c$ ($a,b,c=0\sim 4$) in $\mathbb{Z}[X]/I $, but it is too hard to calculate $f^{18}$.
The ring $R=\Bbb Z[X]/I$ is isomorphic to $\Bbb F_5\times \Bbb F_{25}$.
In $\Bbb F_5$, $t^{18}=1$ has two solutions, as $\gcd(5-1,18)=2$.
In $\Bbb F_{25}$, $t^{18}=1$ has six solutions, as $\gcd(25-1,18)=6$.
Therefore $R$ has twelve solutions to $t^{18}=1$.