prove that if n1 and a0 are integers and d(a,n), then the additive order of a modulo n is n/d?
Yes. Though, If you have the skinny ones, You can use it as a curling iron, or make waves with it.
if d (a,n) then a md for some m with (m,n)1 and n d * (n/d) if the additive order of a is k then ka k md 0 (mod n) so km 0 (mod (n/d)) (m,n) 1 so (m,n/d) 1 so (k,n/d) n/d since k is the least such k, k n/d
