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Let $k\ge2$ and p an odd prime number. If $$g^{p-1}\equiv1[p]$$ and
$$g^{p-1} \not\equiv 1 [p^2]$$ then $$g^{p^k-1} \equiv 1[p^k]$$ I want to use this lemma: $$g^n \equiv 1[p^k] \Rightarrow \phi(p^k)|n$$ Where $\phi$ is the euler's phi function. I tried to work with this lemma, but i don't get results.

Branco Flores Rocha
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  • Here https://math.stackexchange.com/questions/227199/order-of-numbers-modulo-p2 we have proved, if modulo order of g is $$d\pmod p,$$ then the modulo order of $g$ will be $$d$$ or $$pd\pmod{p^2}$$ finally use https://math.stackexchange.com/questions/31679/if-g-is-a-primitive-root-of-p2-where-p-is-an-odd-prime-why-is-g-a-prim – lab bhattacharjee Mar 18 '20 at 01:22
  • If k=2 and p=5 and g=2 it is FALSE. Did you copy something incorrectly? – DanielWainfleet Mar 19 '20 at 01:07
  • @DanielWainfleet Perhaps it is $k \geq 3$ but unfortunately it was when I copied it. – Branco Flores Rocha Mar 21 '20 at 01:00
  • The 3rd line is incorrect. I intend to give a complete A later today. – DanielWainfleet Mar 22 '20 at 22:23

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