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

1. Mar 11, 2010

### 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...

2. Mar 12, 2010

### ystael

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.