If $r$ (mod $p$) is a primitive root, then $r+p$ is also a primitive root mod p

Question

I've struggled with this problem for a while...I tried to figure it out using a reduced residue system's theorem but I honestly have no idea how to work with the given hypothesis.

Answer 1

$r$ (mod $p$) is a primitive root.

This means that there is a $k$ with $r^k = a$ mod $p$ for all $a$ coprime to $p$.

$(r+p)^k = r^k$ mod $p$

which means that l exists such as $l = k$ with $(r+p)^l = a$ mod $p$ for all $a$ coprime to $p$.

Therefore $(r+p)$ is a promitive root as well.

Answer 2

Hint: By the binomial theorem, all the terms of $(r+p)^k$ except the first have a factor of $p$. Thus each of them is zero $\bmod p$.

Or even more elegantly, by the congruence power rule $(r+p)^k=r^k$. See @BillDubuque's comment below.