Why is Euler's Totient function always even? I want to prove why $\phi(n)$ is even for $n>3$.
So far I am attempting to split this into 2 cases.
Case 1: $n$ is a power of $2$. Hence $n=2^k$. So $\phi(n)=2^k-2^{k-1}$. Clearly that will always be even.
Case 2: $n$ is not a power of $2$. This is where I am unsure where to go. I figure I will end up using the fact that $\phi(n)$ is multiplicative, and I think I'll get a $(p-1)$ somewhere in the resulting product which will make the whole thing positive, as $p$ is prime implies $(p-1)$ is even.
 A: You can do it via the formula as you do, but you can also simply use the definition that $\phi(n)$ is the number of numbers $k$, with $1 \le k \le n$, such that $\gcd(n, k) = 1$. 
Clearly, if $\gcd(k, n) = 1$, then $\gcd(n - k, n) = 1$ as well, so (for $n > 2$) all the numbers relatively prime to $n$ can be matched up into pairs $\{k, n-k\}$. So $\phi(n)$ is even.
A: Hint $\ $ The map $\,x\mapsto -x\pmod n\,$ has no fixed points so pairs-up the residues coprime to $n.\,$
Remark $\ $ Such use of reflections (or involutions) to pair-up terms frequently proves handy, e.g. see prior posts here on Wilson's theorem (in groups), esp. this one to start.
A: A proof using group theory: Let $Z_n$ denote the cyclic group of order $n$. There exists a nontrivial order 2 element of $Aut(Z_n)$ for all $n>2$, namely the (additive) inversion map. Since $|Aut(Z_n)|=\phi(n)$, this implies that $\phi(n)$ is even for all $n>2$.
A: $\varphi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_k})$ where $p_i$'s are prime factors of $n$.
Finally in numerator part every term of $(1-\frac{1}{p_i})$ is even, and all the pis in denominator will be cancelled by $n$ in numerator. So it is even.
A: Suppose $n>3$. If $n$ has an odd prime factor, say $p$; then $n=p^km,(m,p)=1$ and $\varphi (n)=\varphi(p^k)\varphi(m)=(p-1)p^{k-1}\varphi(m)$, with $p-1$ even. If $n$ has no odd prime factors, then $n=2^k$ with $k>1$ so $\varphi(2^k)=2^{k-1}$ is even. 
A: This answer will use some slightly more advanced machinery to get a short answer.
If $n\geq 3$ (you don't need to assume $n > 3$) then $-1\neq 1$ in $\mathbb{Z}/n\mathbb{Z}$, but $(-1)^2 = 1$, so $-1$ is an element of order $2$ in $(\mathbb{Z}/n\mathbb{Z})^{\times}$, which means that $|(\mathbb{Z}/n\mathbb{Z})^{\times}| = \varphi(n)$ is even by Lagrange.
A: I saw a proof related to group theory in the answers. I would like to comment on it since I have a related idea but I cannot due to my low reputation. His proof is a little bit advance and I could not understand it well as I am new to the field.
However, I came with my own proof:
From group Theory, let $Z_n$ be the group modulo n under addition. Then, every co-prime-to-n element in this group is paired with its inverse which is a co-prime as well[i.e. they have the same order = n]. Thus, the number of co-primes divisible by 2 as required.
I am not sure if this proof works well, but seems to have a good idea.
A: One very intuitive proof is to notice that
$$\left(\frac n2 +k \right)+\left(\frac n2 -k \right)=n$$
So, if $d$ divdes $n$, then $d$ divides both or none of $\left(\frac n2 +k \right)$ and $\left(\frac n2 -k \right)$. So, either there is at least one $d$ which divides all of $n$, $\left(\frac n2 +k \right)$ and $\left(\frac n2 -k \right)$, or there is no such $d$.
This means either both or none of $\left(\frac n2 +k \right)$ and $\left(\frac n2 -k \right)$ are coprime to $n$.
Note that we don't even need $\frac n2$ to be an integer. All we need are to have $\left(\frac n2 +k \right)$ and $\left(\frac n2 -k \right)$ to be integers.
