Re: [math-fun] Is this true?

13 Mar 2016


      What about 5^5 == 2 (mod 3)?
...
-----Original Message-----
From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On
Behalf Of W. Edwin Clark
Sent: Sunday, March 13, 2016 12:49 PM
To: math-fun
Subject: 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.
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