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