Find base of exponentiation Given the two primes $23$ and $11$, find all integers $\alpha$ such that $\alpha^{11} \equiv 1 \mod 23$.
How to compute this? What to use?

As $\phi(23)=22$
and if ord$_{23}\alpha=a, a|22\implies a=1,2,11,22$
Now $2^2\equiv4,2^5\equiv9\implies2^{11}=2(9)^2\equiv1\implies$ord$_{23}2=11$
We know, ord$_ma=d,$ ord$_m(a^k)=\frac{d}{(d,k)}$ (Proof @Page#95)
So, ord$_{23}(2^k)=\dfrac{11}{(11,k)}$
$\implies$ord$_{23}(2^k)=\dfrac{11}{(11,k)}=11$ for $1\le k\le10$
and there will be exactly $\phi(11)=10\ \alpha$s  such that ord$_{23}\alpha=11$
