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.