1. I think you may have meant to say : in which case you can work it out like so - a method by Gauss that I learned from my book - and I'll go through it step by step since I am new to this myself:
Suppose and we have a class of numbers where holds any number less than that has its highest divisor as . Then we have a set of classes:
A number m only exists in if . Further, so a number m only exists in if and are relatively prime. By definition the number of integers less than and relatively prime to is so the number of elements in is .
From here it is a simple matter.
We just proved that and if you work it out manually with a small number like you will find that and with that we have proved
For the record I relied quite heavily on my book for much of the above so I cannot take credit for this work.
2. I know there is a much more complete way of proving this, but at the moment I can only think of a proof by contradiction:
Lets suppose works for even numbers as well. If , then we can use the equation where p is a prime and e is its power. If we work it out with (a prime) then while . We have reached a contradiction and we at least know for the even number 2 this does not work. The only problem is I do not know how to extend this to all even numbers. This is a very weak proof and I would like to see someone else write one better.
3. I can't figure this one out either.