How many elements $x$ in the field $\mathbb{ Z}_{11}$ satisfy the equation $x^{12} - x^{10} = 2$?

Solution 1:

Hint: every nonzero element $x \in \mathbb{Z}/11\mathbb{Z}$ satisfies $x^{10} = 1$.

Solution 2:

Hint: You just have to solve the congruence

$$x^{12}-x^{10}\equiv2\pmod{11}$$

Then apply the Fermat's little theorem.

Solution 3:

Obviously $x \equiv 0 \pmod{11}$ is not a solution, therefore $x$ is non-zero in $\mathbb{Z}_{11}$ and we can factorize this way:

$x^{12}-x^{10} \equiv 2 \pmod{11} \implies x^{10} ( x^{2}-1) \equiv2 \pmod{11}$ but $x^{10}\equiv 1 \pmod{11}$ for every non-zero $x \in \mathbb{Z}_{11}$ by Fermat's little theorem, or more generally using Euler's totient function $\varphi(n)$. That implies:

$x^2 - 1 \equiv 2 \pmod{11} \implies x^2\equiv 3 \pmod{11} \implies x \equiv 5,6 \pmod{11}$