I am having trouble with this proof: show that if d|n then phi(d)|phi(n). I know that if d|n, then ad=n and that phi(ad)=(a,d)*phi(a)&phi(d)/phi((a,d)), but I can't seem to get anywhere with this info. Thanks for your help.
Right. Good point... So if d|n and d=p1^a1p2^a2...pk^ak, n=p1^b1p2^b2...pk^bk, then phi(d)=p1p2...pk*(p^k-1) and phi(n)=p1p2...pk*(p^k-1), so p1p2...pk|phi(n). Would that work?