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.
Your work isn't correct, but you're getting closer. Look at your expressions for phi(d) and phi(n). They're the same, which can't be right. Also, both expressions contain a "p", which wasn't used before.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?