Find all roots of $\,x^2\!\equiv x\pmod{\!900}$, i.e all idempotents in $\Bbb Z_{900}$

As you noticed, $900$ factors into $2^23^25^2$, and the relevance of this is that the Chinese remainder theorem gives us an isomorphism $\Bbb Z_{900}\cong\Bbb Z_{2^2}\times\Bbb Z_{3^2}\times\Bbb Z_{5^2}$. You can easily show that an idempotent of this ring must be of the form $(e_1,e_2,e_3)$ where each of the $e_i$ are idempotent in $\Bbb Z_{p^2}$. This reduces the problem to one in $\Bbb Z_{p^n}$, and then by implementing the Chinese remainder theorem you can lift the idempotents back to elements of $\Bbb Z_{900}$.

Following this strategy takes us past several insightful lemmas that can come in handy for other problems.

Understand $\Bbb Z_{p^n}$

By ideal correspondence, and our knowledge of ideals of $\Bbb Z$, the only proper ideals of $\Bbb Z_{p^n}$ are of the form $(p^k+p^n\Bbb Z)$ where $1\leq k\leq n$, and $(p+p^n\Bbb Z)$ is the unique maximal ideal of the whole ring. That makes it a prime ideal as well! Furthermore, you can see that $x^n=0+p^n\Bbb Z$ for every $x\in (p+p^n\Bbb Z)$, that is, every element there is nilpotent.

Useful facts about idempotents

If $e$ is an idempotent

1. then $1-e$ is idempotent too.
2. then $e(1-e)=0$.
3. and also nilpotent, then it is zero.

Together we get

For now, let me drop the $+p^n\Bbb Z$ in the coset notation to declutter the math. I mean that I'm writing $(p)$ rather than $(p+p^n\Bbb Z)$ in $\Bbb Z_{p^n}$.

Let $e$ be idempotent in $\Bbb Z_{p^n}$. Then since $e(1-e)=0\in (p)$ and $(p)$ is prime, one of $e$ or $1-e$ is in $(p)$.

If $e\in (p)$, then it's both an idempotent and a nilpotent, hence zero.

If instead $1-e\in (p)$, then by the same argument, $1-e=0$. But this is the same as saying $e=1$.

So in conclusion, you can see that the $e_i$ must be either the zero elements or the identity elements of their respective rings. In total that gives you $2^3$ choices of idempotents, each of which you can pull back to $\Bbb Z_{900}$ via your favorite implementation of the Chinese remainder theorem.

Hint $\ \,p\,$ prime, $\,p^n\mid e(1\!-\!e)\,\Rightarrow\, p^n\mid e\,$ or $\,p^n\mid 1\!-\!e,\ $ by $\ 1\!-\!e,\, e\,$ coprime, by $\ (1\!-\!e)+ e = 1$

So $\!\bmod p^n$ the only idempotents (roots of $\,e^2\equiv e)\,$ are the trivial idempotents $\,e \equiv 0,1.\,$

So, by CRT [cf, below] $\,e\,$ is idempotent mod $\,2^2 3^2 5^2\iff e \equiv 0,1\,$ mod $\,2^2,3^2,5^2,\,$ e.g.

$\ e \equiv (\color{#c00}1,0,0)\Rightarrow {\rm mod}\ 2^2\!:\ e\equiv \color{#c00}1 \equiv 3^2 5^2 k \equiv \color{#c00}k,\,$ so $\, e = 15^2k = 15^2(1\! +\! 4j)\equiv 225\pmod{\!900}$

So $\,(0,1,1) = 1-e \equiv 1-225 = 901-225 \equiv 676.\,$ Same for other $\,(e_1,e_2,e_2),\ e_i \in \{0,1\}$

Remark $\ $ Idempotents $\!\bmod n\,$ correspond to splittings of $n$ into $\color{#c00}{\rm co}\color{#0a0}{\rm prime}$ factors, e.g. above the idempotent $\,e = 225 = 3^2 5^2\,$ corresponds to the factorization $\,n = \color{#c00}{2^2}\!\times \color{#0a0}{3^2 5^2}\,$ where $\,\color{#c00}{e\equiv 1\pmod{\!2^2}},\,$ $\color{#0a0}{e\equiv 0\pmod{\!3^2 5^2}}.\,$ In fact some integer factorization algorithms work by searching for nontrivial idempotents mod $\,n,\,$ which immediately yield a factorization of $\,n\,$ (generally we can quickly factor $\,n\,$ given any polynomial which has more roots mod $\,n\,$ than its degree, so a nontrivial idempotent (or square-root) $\Rightarrow$ quadratic has $> 2$ roots, which splits $n$).

Generally we can find modular roots of polynomials using CRT as explained below, where above is the special case $f(x) = x^2 -x = x(x-1)$.

By CRT, each combination of a root $\,r_i\,$ mod $\,m\,$ and a root $\,s_j\,$ mod $\,n\,$ corresponds to a unique root $\,t_{ij}\,$ mod $\,mn\,$ i.e.

$$\begin{eqnarray} f(x)\equiv 0\pmod{mn}&\overset{\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod m\\f(x)\equiv 0\pmod n\end{array} \\ &\iff& \begin{array}{}x\equiv r_1,\ldots,r_k\pmod m\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod n\end{array}\\ &\iff& \left\{ \begin{array}{}x\equiv r_i\pmod m\\x\equiv s_j\pmod n\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\ &\overset{\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\ \end{eqnarray}\qquad\qquad$$

Some integer factorization algorithms work by searching for nontrivial idempotents mod $\,n,\,$ which immediately yield a factorization of $\,n\,$ (generally one can quickly factor $\,n\,$ given any polynomial which has more roots mod $\,n\,$ than its degree, so any nontrivial idempotent or nontrivial square-root will split $\,n,\,$ since it yields a quadratic with $3$ roots).