Re: [math-fun] Is this true?

Perhaps you left out some hypothesis? Let n = 3 then gcd(3,phi(3)) = 1 but mod 3: 0^0 = 1, 1^1 = 1, 2^2 = 4 = 1 so x^x = 2 mod 3 has no solution. On Sun, Mar 13, 2016 at 10:54 AM, David Wilson <davidwwilson@comcast.net> wrote:

x^x == k (mod n) is solvable for all k iff gcd(n, phi(n)) = 1.