Roots over Z/pZ

We can multiply residu-classes, not "allways" divide them:
We can only divide classes when we're working in the group $(\mathbb{Z}/n\mathbb{Z})^*$ .

In your example you used n = 21, and 7|21. Observe that $3\cdot 7 \equiv 9\cdot 7 \equiv 0$ mod 21. But it's not true that $7\equiv 0$ mod 21.

You're getting into trouble when you're using elements that divide n.
The conclusion is: $x^a \equiv x$ mod $n \Leftrightarrow x^{a-1}\equiv 1$ mod n iff gcd(x,n) = 1.

We can draw the following conclusion:

$x^a\equiv x$ mod $n \Leftrightarrow x(x^{a-1}-1)\equiv 0$ mod n.

Hence we have the following cases:

Let $x\in \mathbb{Z}/n\mathbb{Z}$
1. x is a root if: $x\equiv 0$ mod n.
2. if $\phi(n)|a-1$ then all $x\in (\mathbb{Z}/n\mathbb{Z})^*$ are roots, ie. all $x\in \mathbb{Z}/n\mathbb{Z}$ with gcd(x,n) = 1. And x with gcd(x,n) $\neq 1$ can be roots.
3. $a-1|\phi(n)$ . Then only elements from the group $(\mathbb{Z}/n\mathbb{Z})^*$ that have order $a-1$ are roots. And x with gcd(x,n) $\neq 1$ can be roots.
4. If neither $a-1|\phi(n)$ or $\phi(n)|a-1$ then only x, with gcd(x,n) $\neq 1$ can be roots.

I hope you know the Euler totient function a little. For more about the euler totient function: Euler's totient function - Wikipedia, the free encyclopedia

Finding roots for all $x\in \mathbb{Z}/n\mathbb{Z}$ with gcd(x,n) $\neq 1$ is indeed a hard problem.