How many idempotent elements are in Z_m [closed]

Question

How many idempotent elements are in Z_m or How many roots have this polynomial in Z_m

f(x)=x^2 +x procedure of proof is important for me.

Answer 1

Idempotent elements are roots of $g(x)=x^2-x$; my answer will apply equally well to $f(x)=x^2+x$.

The important step is the Chinese Remainder Theorem: one way of stating it is that if $m=p_1^{r_1}\times\cdots\times p_k^{r_k}$ is factored into powers of distinct primes, then the ring $Z_m$ is equal to the direct product of rings $Z_{p_1^{r_1}} \times \cdots \times Z_{p_k^{r_k}}$. So it suffices to count the number of roots in each ring $Z_{p_i^{r_i}}$ and then multiply those numbers together to obtain the number of roots in $Z_m$.

I think it will be easy to convince yourself that the polynomial $g(x)=x^2-x$ has exactly two roots in any ring of the form $Z_{p^r}$.

(By the way, the isomorphism between $Z_m$ and $Z_{p_1^{r_1}} \times \cdots \times Z_{p_k^{r_k}}$ is completely explicit, so this even gives a way to construct the idempotent elements, not just count them.)