Proof:phi(m) is always even if m>2

[cwatki14]

So I am aiming to prove that phi(m) is always even if m>2.
What I have thus far
If n is an integer such that (n,m)=1 where 1<n<m then (m-n,m)=1. Note: If (m-n,m) are not coprime this would imply that m-n divides m. This is a contradiction.
Now, consider the case where m is even and n=m/2. Clearly m/2 divides m and the gcd(m/2,m)=m/2. m/2=1 iff m=2.

I am just a little lost how I can use this information to imply that phi(m) is always even...

 

[ystael]

Note: If (m-n,m) are not coprime this would imply that m-n divides m. This is a contradiction.

I don't understand this part by itself. If m - n is not coprime to n, it is not coprime to n, nothing more. However, it is true that if (n, m) = 1 then (m - n, m) = 1; you just need a better proof.

As for the argument as a whole: you have the flesh, but you are missing the bones.

\phi(m) is the size of a set. What set? The set RP(m) of numbers 1 \leq n \leq m which are relatively prime to m.

How do you show that the size of a set is even? One way is to divide it into pairs.