Prove that if numbers a and b are coprime, then there is such natural number n, that a^n =(modulus equal sign)= 1 (mod b)

Question

zyberg · Answer

As I understand, I need to try looking at a/b and then a^n/b.
What gets me really confused is how would I predict reminders of a^n/b? Or is it not needed?

welshfella · Answer

a^n  = 1 (mod b)     (modular equals)

as a consequence of this i know we can say the following 

a^n - 1 = bk    for some integer k


but what next i dont know.

loser66 · Answer

$a^n\equiv 1$ mod b , is what we have to prove. @welshfella

welshfella · Answer

yeah of course

I just wondered if that transformation might help in the proof

mathmate · Answer

@Zyberg
FYI  $\equiv$ is written in LaTeX as  \equiv .

sparrow2 · Answer

i remember that somewhere i learnt this:|dw:1465636649837:dw|