- #1
crypto_rsa
- 4
- 0
I know it's a dumb question but I can't figure out why the totient of n is always even (I've read in a book that it "follows immediately from the definition of the totient function", so it should not require any theorem to prove). It is clear to me that it holds true for
n = pk, where p is a prime, because
phi(pk) = pk - 1(p - 1) and (p - 1) is even
But why is it true in the general case? I think I could use multiplicativity of phi() to prove it but I am confused by the "follows from definition" note.
n = pk, where p is a prime, because
phi(pk) = pk - 1(p - 1) and (p - 1) is even
But why is it true in the general case? I think I could use multiplicativity of phi() to prove it but I am confused by the "follows from definition" note.