Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$ Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$

Im not sure should I use primitive root or quadratic residue.
For primitive root, $U_{33} = \{1,2,4,5,7,8,10,13,14,16,17,19,20,23,25,26,28,29,31,32\}$
Thus $\phi_{(3\cdot11)}=20=2^2\times{5}$
But it takes way too much time to test if each of them is primitive root or not...
But Im not quite familiar with the method of Quadratic Residue.
We have $x^8\equiv5\pmod3$ and $x^8\equiv 5 \pmod {11}$
so.... follows the rule of $x^2\equiv q \pmod n$,
we have $(x^4)^2 \equiv 5 \pmod 3$ and $(x^4)^2 \equiv 5 \pmod {11}$
and then how do I continue with it??
 A: I would think to work $\pmod 3$ and $\pmod {11}$, then combine the answers with the Chinese Remainder Theorem.  The easy one is $\pmod 3$, where there are only two to check.  We find $1^8 \equiv 1 \equiv 2^8 \not \equiv 5 \pmod 3$ so there are no solutions.
A: $$x^8 \equiv 5 \pmod{33} \implies x^8 \equiv 5 \pmod 3 \implies (x^4)^2 \equiv 2 \pmod 3 \implies \text{No solution}$$
In general, when you want to solve for $$x^m \equiv a \pmod n \,\,\, (\spadesuit)$$ and if $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, the idea is to first solve for $$x^m \equiv a \pmod {p_l^{a_l}} \,\,\, (\clubsuit)$$ You have a solution for your original problem $(\spadesuit)$ iff you have a solution for each $l \in \{1,2,\ldots,k\}$ in $(\clubsuit)$. Once you find solution for each $l$ in $(\clubsuit)$, put them together using Chinese Remainder theorem.
A: Hint $\rm\ mod\ 3\!:\ x\not\equiv 0\:\Rightarrow\:x\equiv \pm 1\:\Rightarrow\:x^2\equiv 1\:\Rightarrow\:x^{2n}\!\equiv 1\not\equiv 5$
A: Hint: Consider the equation modulo $3$. What are the solutions to $x^8 \equiv 5 \mod 3$?
Alternatively, you could reduce the above to $(x^4)^2 \equiv -1 \mod 3$, which means that for a solution, you need that $-1$  is a quadratic residue modulo $3$ and $x^4$ is its 'square root'. Is $-1$ a quadratic residue?
