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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...
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...