Question

HELP! primitive roots
http://prntscr.com/4lft8f

Answer 1

\(\large 2^n\equiv 1 \pmod {2^n-1}\)

Answer 2

so the order of \(2\) is \(n\) in \(\mod 2^n-1\)

Answer 3

hehe yeah

Answer 4

So \(n | \phi(2^n-1) \)

im starting to like these proofs xD

Answer 5

and also

2^pi(2^n-1) = 1 mod 2^n-1

thus 2|pi(2^n-1)

Answer 6

i meant not 2 lol

Answer 7

proofs are fun

Answer 8

somehow i don't see why the order of 2 is n
http://math.stackexchange.com/questions/926094/find-the-order-of-2-in-mod-2n-1

Answer 9

since

2^n-1=0 mod 2^n-1

Answer 10

yes, but how does that prove n is the smallest such integer ?

Answer 11

2^k-1 always < 2^n-1 when k<n :o

Answer 12

so
2^k-1 = 2^k-1 mod 2^k-1
since 2^k-1 <2^n-1 :O

do u see it now ?

Answer 13

typo...
2^k-1 = 2^k-1 mod 2^n-1

Answer 14

yes !

Answer 15

great ^^