Prove that it is not a primitive root module $p^2$ I don't even know how to start to prove the following...
Let $p$ be an odd prime. Prove that if $a$ is a primitive root modulo $p$ then exactly one of the following integers
$a,a+p,a+2p,...,a+(p-1)p$ is not a primitive root module $p^2$.
Thanks a lot :)

As $a+rp\equiv a\pmod p$ and as $a$ is a primitive root $\pmod p,$
$a+rp,1\le r\le p-1$ are also primitive root $\pmod p,$
By Question about primitive roots of p and $p^2$, ord$_{p^2}(a+rp)=p(p-1)$ or $p-1$ for $0\le r\le p-1$
Now we have $a^{p-1}=1+kp$ where $k$ is some integer
$(a+rp)^{p-1}\equiv a^{p-1}+(p-1)a^{p-2}rp\pmod{p^2}\equiv1+kp-a^{p-2}rp$
Now this will be $\equiv1\pmod{p^2}\iff k\equiv a^{p-2}r\pmod p\iff r\equiv ka$ as $a^{p-1}\equiv1\pmod p$
Clearly, there one such $r,0\le r\le p-1$ such that ord$_{p^2}(a+rp)=p-1$
