If $a\equiv 1\pmod {p^2},$ then $a\equiv 1\pmod p,$ for $\,a = r^{{\rm ord}(r)}$

Question

I've seen this in a lot of proofs for number theory-based problems, but I can't see where they are getting it from.

If $r^{\operatorname{ord}(r)} = 1\pmod{p}$ then $r^{\operatorname{ord}(r)}= 1\pmod{p^2}$.

thanks

EDIT:

Should have been, if $r^{\operatorname{ord}(r)} = 1\pmod{p^2}$, then $r^{\operatorname{ord}(r)} = 1\pmod{p}$

$2^2\equiv1\pmod3$ but $2^2\equiv4\pmod9$. What is your definition of $ord(r)$? Often $r^{ord( r)}\equiv1$ follows immediately from the definition — J. W. Tanner, Mar 01 '20 at 15:15
Well, $2$ is a primitive root $\pmod 3$ so the counterexample you were given applies. If you want a different counterexample, then note that $2$ is a primitive root $\pmod 5$ and that $2^4\equiv 1 \pmod 5$ but $2^4\not \equiv 1 \pmod {25}$ — lulu, Mar 01 '20 at 15:31
are you asking whether a primitive root mod $p$ is a primitive root mod $p^2$? cf. this — J. W. Tanner, Mar 01 '20 at 15:35
I have slightly edited the question to give some more context/ change the question. — Yloereps, Mar 01 '20 at 15:38
Wow yes, it really was that simple! I've been working on some problems for hours and just seemed to have a mental lapse. Thank you! — Yloereps, Mar 01 '20 at 15:45
I took the liberty of editing your title to match your corrected question — J. W. Tanner, Mar 01 '20 at 15:51

score 1 · Accepted Answer · answered Mar 01 '20 at 15:48

1

$r^d\equiv1\bmod p^2$ means $p^2|r^d-1$;

since $p|p^2$, this means $p|r^d-1$, so $r^d\equiv1\bmod p$.

answered Mar 01 '20 at 15:48

J. W. Tanner

1 Answers1