13 Mar
2016
13 Mar
'16
11:31 a.m.
On Sun, Mar 13, 2016 at 7:54 AM, David Wilson <davidwwilson@comcast.net> wrote:
x^x == k (mod n) is solvable for all k iff gcd(n, phi(n)) = 1.
Yes; in particular, since n and phi(n) are coprime, there's a number m less than n*phi(n) such that m == k (mod n) and m == 1 (mod phi(n)). -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com