N NOX Andrew Dec 2009 226 90 May 15, 2010 #1 I read two numbers \(\displaystyle a\) and \(\displaystyle b\) are coprime if \(\displaystyle b^{a - 1} = 1\) (mod a). Is it me or does this test not hold when a = 6 and b = 5, which are coprime? \(\displaystyle 5^{6 - 1} = 1\) (mod 6) \(\displaystyle 5^5 = 1\) (mod 6) \(\displaystyle 3125 = 1\) (mod 6) 3125 mod 6 is 5, which does not equal 1, so the test fails. Is it me or the test?

I read two numbers \(\displaystyle a\) and \(\displaystyle b\) are coprime if \(\displaystyle b^{a - 1} = 1\) (mod a). Is it me or does this test not hold when a = 6 and b = 5, which are coprime? \(\displaystyle 5^{6 - 1} = 1\) (mod 6) \(\displaystyle 5^5 = 1\) (mod 6) \(\displaystyle 3125 = 1\) (mod 6) 3125 mod 6 is 5, which does not equal 1, so the test fails. Is it me or the test?

chiph588@ MHF Hall of Honor Sep 2008 1,163 429 Champaign, Illinois May 16, 2010 #2 Carmichael number - Wikipedia, the free encyclopedia