0

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.

Bill Dubuque
  • 272,048
SpongeBob1873
  • 3

2 Answers2

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.

Noureddine Ouertani
  • 380
  • Thanks a lot!! I'm not sure why $r^{k}=a$ mod $p$ for all $a$ coprime to $p$. – SpongeBob1873 Oct 31 '20 at 23:53
  • Isn't that the definition of a primitive root? https://en.wikipedia.org/wiki/Primitive_root_modulo_n – Noureddine Ouertani Oct 31 '20 at 23:54
  • Ohh thanks! The book I'm currently using states it a bit different, but now I understand that it is (and must be) basically the same... I guess I haven't seen it that way – SpongeBob1873 Nov 01 '20 at 00:02
  • @Sponge Key Idea: all of the (modular) definitions of "$r$ is a primitive root $!\bmod{!p}$" are statements concerning powers $, r^n\pmod{! p}.,$ Replacing $,r,$ by $,r_1 \equiv r\pmod{!p},$ doesn't alter the truth of such statements since $,r_1\equiv r\Rightarrow\ r_1^n\equiv r^n\pmod{!p},$ by the Congruence Power Rule. In particular this is true for your $, r_1 = r + p\ \ $ – Bill Dubuque Nov 01 '20 at 01:01
  • Thanks, @BillDubuque that's very useful! – SpongeBob1873 Nov 01 '20 at 03:11
1

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.