Euler Phi Function is Even for Argument greater than 2

Theorem
Let $n \in \Z: n \ge 1$.

Let $\phi \left({n}\right)$ be the Euler $\phi$ function of $n$.

Then $\phi \left({n}\right)$ is even iff $n > 2$.

Proof
We have $\phi \left({1}\right) = 1$ from the definition, and $\phi \left({2}\right) = 1$ from Euler Phi Function of a Prime.

Now let $n \ge 3$. There are two possibilities:

Odd Prime Divisor
$n$ has (at least one) odd prime divisor $p$, say.

From the corollary to Euler Phi Function of an Integer: Corollary, it follows that:
 * $p - 1$ divides $\phi \left({n}\right)$

But as $p$ is odd, $p - 1$ is even and hence:
 * $2 \mathop \backslash \left({p - 1}\right) \mathop \backslash \phi \left({n}\right)$

and so $\phi \left({n}\right)$ is even.

No Odd Prime Divisor
Now suppose $n$ has no odd prime divisors.

Then its only prime divisor must be $2$ and so $n = 2^k$ where $k > 1$.

Then from Euler Phi Function of a Prime:
 * $\phi \left({n}\right) = 2^k \left({1 - \frac 1 2}\right) = 2^{k-1}$

where $k-1 > 0$.

Hence $\phi \left({n}\right)$ is even.