Why is Euler's Totient function always even?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.

Solution 5:

$\varphi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_k})$ 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.