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:


 * $n$ has (at least one) odd prime divisor $p$, say.

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

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


 * 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 $k-1 > 0$ and hence is even.