Proof about the length of the period $a\pmod m$

Asked Jul 10 '16 at 05:15

Active Jul 10 '16 at 11:48

Viewed 27 times

Theorem 4-3 if $a^u \equiv 1\pmod m$, then $ord_m\text{ } a|u$.

Proof: Put $ord_m\text{ }a=t$, and let $u=qt+r$, $0 \leq r\ <t$. Then

$$a^u =a^{qt+r}=((a^t)^q)*a^r \equiv a^r \equiv 1(mod\text{ } m)$$

and if r were different from zero, there would be a contradiction with the definition of t.

How does that show $ord_m a|u$?

edited Jul 10 '16 at 11:48

3SAT

asked Jul 10 '16 at 05:15

TheMathNoob

If $r=0,$ then $u=qt$ is divisible by $t=ord_m a.$ – awllower Jul 10 '16 at 05:36
why there would be a contradiction with the definition of t? – TheMathNoob Jul 10 '16 at 05:41
Because $t$ is the smallest positive integer $n$ such that $a^n\equiv 1\pmod m.$ But one is led to $a^r\equiv1\pmod m$ with $r\lt t.$ – awllower Jul 10 '16 at 05:42
how do we deduce $a^r \equiv 1(mod\text{ } m)$? – TheMathNoob Jul 10 '16 at 05:45
You deduced it in the third line, didn't you? – awllower Jul 10 '16 at 06:49
See this answer and this answer for more conceptual viewpoints. – Bill Dubuque Jul 10 '16 at 15:35
I can't understand why $(a^t)^q*a^r \equiv a^r\equiv 1(mod\text{ } m)$ – TheMathNoob Jul 10 '16 at 23:03
I got it, because $(a,m)=1$, so $(a^{tq},m)=1$ which implies that $a^{tq}$ has a multiplicative inverse right? – TheMathNoob Jul 10 '16 at 23:47

0 Answers0