20110518, 17:28  #34  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
http://en.wikipedia.org/wiki/Multiplicative_order and solving for k=p for mod q and base a or what ever. 

20110518, 21:41  #35  
Aug 2005
Seattle, WA
1,567 Posts 
Quote:
So for example, the order of 2 mod 7 is 3, since 2^3 == 1 (mod 7) and there is no smaller positive exponent for which that would be true. On the other hand, the order of 3 mod 7 is 6 (try looking at the powers of 3 mod 7 to verify this for yourself). An important fact about multiplicative orders is that if b^k == 1 (mod n), then the order of b mod n must divide k. See if you can prove this for yourself. For example, 4^6 == 1 (mod 21), so the order of 4 mod 21 must divide 6. And indeed, you can check that it's 3. 

20110518, 21:44  #36  
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Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20110518 at 21:46 

20110518, 23:42  #37 
Aug 2006
13445_{8} Posts 
sm: Try
znorder(Mod(2,101)) and the like. 
20110519, 00:05  #38 
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Jul 2009
Dumbassville
20300_{8} Posts 

20110722, 19:40  #39 
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Jul 2009
Dumbassville
2^{6}·131 Posts 
Lucas Lehmer test
okay I'm back , this time about something I see that might drop a step in the LL test:
is there a way to test to find y: IF then if so we can take the LL test down one step , because if then 
20110722, 21:41  #40  
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Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
http://mathforum.org/library/drmath/view/66757.html if applied generally instead of just to 2 it might be able to get it back to the first would that speed things up ? though I'm kinda confused about what I found. Last fiddled with by science_man_88 on 20110722 at 21:42 

20110723, 01:57  #41 
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Jul 2009
Dumbassville
2^{6}·131 Posts 
just realized something using PARI:
Code:
(22:43)>for(x=1,7,print((x^2)%7"::"x",")) 1::1, 4::2, 2::3, 2::4, 4::5, 1::6, 0::7, 
20110724, 01:09  #42  
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Jul 2009
Dumbassville
20300_{8} Posts 
Quote:


20110726, 02:02  #43 
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Jul 2009
Dumbassville
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