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Prove that there exists a primitive root $g$ modulo $p$ ($p$ an odd prime) such that $g^{p-1}\not\equiv 1 \pmod {p^2}$

So far, I have been able to prove that if $g$ is a primitive root modulo $p$ ($p$ an odd prime) and $g^{p-1}\equiv 1 \pmod {p^2}$, then $(g+p)^{p-1}\not\equiv 1 \pmod {p^2}$. I don't know how to continue? Any help I would appreciate.

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  • $\begingroup$ If you've proved this, then you need to simply note that if $g$ is ap rimitive rood modulo p, then so is $g+p$. $\endgroup$
    – D S
    Commented Feb 3 at 16:43
  • $\begingroup$ What you've proven suffices....where do you see a gap? $\endgroup$
    – lulu
    Commented Feb 3 at 16:44
  • $\begingroup$ @DS I know, but I don't see why it connects to the problem? $\endgroup$
    – user1277306
    Commented Feb 3 at 16:45
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    $\begingroup$ More clearly, if the primitive root $g$ satisfies the required condition, then we are done. Otherwise, the primitive root $g+p$ satisfies the given conditions. Again, we are done. $\endgroup$
    – D S
    Commented Feb 3 at 16:49
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    $\begingroup$ Note $g\equiv g+p\pmod p$ so $(g+p)^k \equiv g^k\pmod p$. So $g+p$ is a primitive root modulo $p$ if and only if $g$ is a primitive root modulo $p$. If you have proven what you say you have (and if you've proven any primitive roots exist at all) then I'd say you are done. $g + p$ is a primitive root modulo $p$ so that $(g+p)^{p-1}\not \equiv 1\pmod{p^2}$. $\endgroup$
    – fleablood
    Commented Feb 3 at 17:07

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