# If $p$ is an odd prime with primitive root $1<r<p$, is $r$ also a primitive root modulo $p^2$?

I use excel computed till $$p=23$$, it's true. But is this always true? if not, could you pls give a counter example?

• That's not very far. – Angina Seng Sep 20 at 10:20
• It may happen that $r^{p-1}\equiv 1\pmod{p^2}$, I suppose – Hagen von Eitzen Sep 20 at 10:30
• @AnginaSeng thanks for the hint, yeah I just need to go one step further :p – athos Sep 20 at 10:34
• Try Table[SubsetQ[PrimitiveRootList[Prime[n]^2], PrimitiveRootList[Prime[n]]], {n, 1,100}] on Mathematica. – Chrystomath Sep 20 at 10:43
• – lhf Sep 20 at 10:52

just to dot it that $$r=14$$ is a primitive root for $$p=29$$, but not for $$p^2=841$$. Instead $$r+p = 43$$ is a primitive root for $$841$$.