Euler Phi Function Even for Argument Greater than 2

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[edit] 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.


[edit] 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:


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.


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.

\blacksquare

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