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 $n > 2$.

Proof
We have from the definition of Euler $\phi$ function:
 * $\phi \left({1}\right) = 1$

and from Euler Phi Function of Prime Power: Corollary:
 * $\phi \left({2}\right) = 1$

Now let $n \ge 3$.

There are two possibilities:

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

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

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

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

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

Then its only prime factor must be $2$.

Thus:
 * $n = 2^k$

where $k > 1$.

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

where $k-1 > 0$.

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