Poirot1
- 243
- 0
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 $p^2$
Last edited:
Poirot said: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 $p^2$
caffeinemachine said: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,Poirot said:What is the logic in this step?
Also, note the result is vacuously true when p=2.