1
$\begingroup$

Let $a$ be a primitive root mod $p$. Prove that $a$ or $a+p$ is a primitive root mod $p^2$.

This is what I did so far:
$\phi(p^2)=p^2-p=p(p-1)$.

$(a+p)^{p-1}=a^{p-1} +(p-1)a^{p-2}p \pmod{p^2}$.

I'm stuck here...

$\endgroup$
1
$\begingroup$

You are almost there.

Hint: A primitive root $g \pmod{p}$ is also a primitive root $\pmod{p^2}$ if and only if $g^{p-1} \not \equiv 1 \pmod{p^2}$. Can $a^{p-1}$ and $(a+p)^{p-1}$ both be congruent to $1 \pmod{p^2}$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.