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We assume $p$ is an odd prime.Show that if $x$ is a primitive root of p, and $x^{p-1}$ is not congruent to 1 mod$p^2$, then x is a primitive root of $p^2$
What is the logic in this step?So we have $x^n=(pk+1)^{l}\equiv 1\pmod{p^2}$.
So we have $lpk+1\equiv 1\pmod{p^2}$.
Hello Poirot,What is the logic in this step?
Also, note the result is vacuously true when p=2.