Find all solutions of $x^{11} \equiv 1 \pmod {23}$. Justify your work
If I attempt to apply power of $11$ to all values from $1-23$, I get too large a value to then be able to see if it can be reduced $\pmod{23}$.
Is there a simpler way? any help would be appreciated
When doing modular arithmetic you never need to deal with really big numbers because you reduce along the way. So for $2^{11} \pmod{23}$ you compute $$ \begin{align} 2^2 & = 4\\ 2^3 & = 8\\ 2^4 & = 16\\ 2^5 & = 32 \equiv 9\\ 2^6 & \equiv 2 \times 9 = 18\\ 2^7 & \equiv 2 \times 18 = 36 \equiv 13 \end{align} $$ and so on.
There are other shortcuts that help with this particular problem, but the general principle (you never need big numbers) is worth remembering all the time.
By Euler's criterion, the solution is exactly the nonzero squares mod $23$.
So just compute $1^2, 2^2, \dots, 11^2 \bmod 23$.
Exponentiation by squaring is well-suited to such calculations (where needed) in modular arithmetic. As an example of how this can be used in this case:
\begin{align} 13^2 &= 169 \equiv 8 \bmod 23\\ 13^4 &\equiv 8^2 \equiv 18 \bmod 23\\ 13^8 &\equiv 18^2 \equiv (-5)^2 \equiv 25\equiv 2 \bmod 23\\ \text {and then}\qquad\\ 13^{11} = 13^8\cdot13^2\cdot13^1 &\equiv 2\cdot8\cdot 13 \equiv 208 \equiv 1\bmod 23\\ \end{align}
In this case though, given that $a^{22}\equiv 1 \bmod 23$ for any $a$ coprime to $23$ by Fermat's little theorem, we just need to find all $b$ such that $b\equiv a^2\bmod 23$ for some $a$. In place of the above calculation, we find $6^2=36\equiv 13 \bmod 23$ and thus necessarily $13^{11}\equiv 6^{22}\equiv 1 \bmod 23$.
Since $23$ is prime, $\Bbb Z_{23}^*\cong \Bbb Z_{22}$ is cyclic.
Find a generator. $5$ is one, since $5^{11}\equiv-1$ (and, $5^2\equiv 2$).
Now the other $9$ generators are the $5^k, $ where $(k,22)=1.$
So $5,5^3,5^5,5^7,5^9,5^{13},5^{15},5^{17},5^{19},5^{21}.$ That's $5,10,20,17,11,21,19,15,7,14.$
The rest are the solutions you're looking for, since by little Fermat we always have $x^{22}\equiv 1\implies (x^{11}+1)(x^{11}-1)\equiv 0.$
