Euler Phi Function is Even for Argument greater than 2

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

Let $\map \phi n$ be the Euler $\phi$ function of $n$.

Then $\map \phi n$ is even $n > 2$.

Proof
We have from the definition of Euler $\phi$ function:
 * $\map \phi 1 = 1$

and from Euler Phi Function of Prime Power: Corollary:
 * $\map \phi 2 = 1$

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 $\map \phi n$

But as $p$ is odd, $p - 1$ is even and hence:
 * $2 \divides \paren {p - 1} \divides \map \phi n$

and so $\map \phi n$ 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:
 * $\map \phi n = 2^k \paren {1 - \frac 1 2} = 2^{k - 1}$

where $k-1 > 0$.

Hence $\map \phi n$ is even.