# About The Order of an Integer

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?

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Apr 15, 2022 at 22:08

There could be lots of exponents that cause $$b^x \equiv 1 \pmod{n}$$. The order is the least positive of these.
$$2^3 \equiv 2^6 \equiv 2^{15} \equiv 1 \pmod{7}$$
and note that $$3$$ divides each of $$6$$ and $$15.$$