Can someone explain me the right steps for the solution of this congruence using the method of the lift?
$X^5 \equiv 1 \pmod {25}$
I know that I can write this as $f(x) = X^5-1 \equiv 0 \pmod {25}$.
Obviously $25=5^2$
Then I start solving $X^5-1 \equiv 0 \pmod {5}$.
Here I find that the only solution for this is $X_0\equiv 1 \pmod {5}$.
So I proceed with the lift $x \rightarrow x^2 $ finding $M=0$ and $f'(X_0)=5\equiv 0 \pmod {5}$. In this case there are $5$ solutions writable as $x=1+5t$ and those are the set $S=[ 1,6,11,16,21]$.
Am I right or wrong?