number theory orders proof

dts5044

Junior Member
Joined
Mar 6, 2008
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76
stuck on a proof..any help would be great!

show, for any integer k>=0:

en(a^k)= en(a) / gcd( en(a), k)

where en(a) is the order of a modulo n, also (a,n) = 1


so far I have: let 'b' be the order of a^k mod n
let 'h' be the order of a mod n

so (a^b*k) congr. a^h congr. 1 mod n

so h | bk

also, a^(h*k) congr. a^(b*k) congr. 1 mod n

rearranging the exponents and setting equal modulo b gives: b|h

so I have h | bk and b|h...is there somewhere to go from here?
 
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