20110331, 15:22  #23 
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Jul 2009
Dumbassville
2^{6}×131 Posts 
Okay thanks I now think I know how to correct it to get a general principle. it depends on a1 I believe for the first a that works .
the reason I saw this is 1 = (21) I tested it on 3^5 and it appears to work. so they will have gaps of a_{1}1 to work. 
20110331, 16:01  #24 
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Jul 2009
Dumbassville
2^{6}×131 Posts 
I actually know a bit of how to prove this just not in a way most would care for.

20110331, 16:05  #25  
Aug 2006
3·5^{2}·79 Posts 
Quote:
Working with integers mod m, I think that if x^a = x and x is nonzero (that is, not divisible by m) then you can conclude that x^(a1) = 1 (mod m). Can someone confirm? I'm a little sick and my mind is hazy today. 

20110331, 16:10  #26  
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Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20110331 at 16:20 

20110331, 17:24  #27  
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Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
x^(a1)%z = ((x^a)%z)/x assuming ((x^a)%z) = x you would be correct as x/x = 1 for non 0 values of x. Last fiddled with by science_man_88 on 20110331 at 17:27 

20110331, 18:26  #28  
Nov 2003
7460_{10} Posts 
Quote:
x is not divisible by m (5 is not divisible by 10). You need (x,m) = 1. 

20110331, 18:40  #29 
Aug 2006
3×5^{2}×79 Posts 
Right, of course. I can't see or think straight today...

20110331, 20:42  #30 
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Jul 2009
Dumbassville
2^{6}·131 Posts 

20110518, 15:42  #31  
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Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20110518 at 15:43 

20110518, 16:59  #32 
Jun 2003
37·127 Posts 

20110518, 17:25  #33 
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Jul 2009
Dumbassville
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