2
$\begingroup$

In this bolg It says $x=ord_{n}b$ and $ord_n = $ the least positive integer x such that $b^x\equiv $ 1 (mod n) and below it says $b^x\equiv $ 1 (mod n) if and only if $ord_{n}b$ | x and then it gives proof for that. Aren't these values already equal? so of course they divide each other or I'm missing something?

$\endgroup$
1
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Apr 15, 2022 at 22:08

1 Answer 1

1
$\begingroup$

There could be lots of exponents that cause $b^x \equiv 1 \pmod{n}$. The order is the least positive of these.

For instance,

$$2^3 \equiv 2^6 \equiv 2^{15} \equiv 1 \pmod{7}$$

and note that $3$ divides each of $6$ and $15.$

$\endgroup$
1
  • $\begingroup$ Oh I completely missed this point. Thanks!! $\endgroup$ Apr 15, 2022 at 21:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .