20120916, 09:57  #1 
Dec 2011
24_{16} Posts 
Prime conjecture
Let m be a positive odd integer.
If 2^{m} = 2 (mod. m(m1)) then m is a prime. True or False? This is true for m = 3, 7, 19, 43, 127, 163, 379, 487, 883, 1459, 2647, 3079, 3943, 5419, 9199, 11827, 14407, 16759, 18523, 24967, 26407, 37339, 39367. These are all primes and there are no other values lessthan 40 003. Is it possible to prove the conjecture? 
20120916, 11:56  #2 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·3^{2}·5^{2}·7 Posts 
I have no idea if it is provable or not, but there are only two more solutions below 10^9:
42463, 71443 
20120916, 11:59  #3  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
6 = 2 mod (93=6) which is false 14 = 2 mod (497=42) which is false 38 = 2 mod (36119=342) which is false the main problem is that m(m1) = m^2m which is larger than 2m once m>3 so all m>3 fail this test. or at least that's my interpretation of it. I looked up your phrasing and found it on mathhelpforum.com and apparently it's 2^m not 2m as it reads here. Last fiddled with by science_man_88 on 20120916 at 12:06 

20120916, 13:31  #4  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×3^{2}×5^{2}×7 Posts 
Quote:
There are indeed a lot more solutions for that up to 10^9. Here are the first few (all prime): 42463, 71443, 77659, 95923, 99079, 113779, 117307, 143263, 174763, 175447, 184843, 265483, 304039, 308827, 318403, 351919, 370387, 388963, 524287, 543607, 554527, 581743, 585847, 674083, 688087, 698923, ... Last fiddled with by retina on 20120916 at 13:36 

20120916, 13:40  #5  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20120916, 14:28  #6  
Dec 2011
2^{2}×3^{2} Posts 
Prime conjecture
Quote:


20120916, 14:56  #7 
Einyen
Dec 2003
Denmark
110001111001_{2} Posts 
nevermind, was not correct.
Last fiddled with by ATH on 20120916 at 15:06 
20120916, 15:05  #8 
Dec 2011
2^{2}·3^{2} Posts 
Mersenne prime sequence
If it is possible to prove the following conjecture then I can prove the existance of an infinite sequence of Mersenne primes and I can represent them as an algebraic expression. However, they are not representable as numbers.
Let m be a positive odd integer. If 2^{m} = 2 (mod. m(m1)) then m is a prime. True or False? This is true for m = 3, 7, 19, 43, 127, 163, 379, 487, 883, 1459, 2647, 3079, 3943, 5419, 9199, 11827, 14407, 16759, 18523, 24967, 26407, 37339, 39367, and these are all prime. I can find no composite integers for which the statement holds. Is it possible to prove the conjecture? Last fiddled with by Stan on 20120916 at 15:09 Reason: Add icon 
20120916, 15:28  #9 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
see also:
http://mersenneforum.org/showthread.php?p=311833 http://mersenneforum.org/showthread.php?p=311851 and probably more. oh yeah I forgot I originally found it correctly on: http://mathhelpforum.com/numbertheo...onjecture.html Last fiddled with by science_man_88 on 20120916 at 15:31 
20120916, 16:15  #10  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
M_{m} = 1 mod (m(m1)) we know: M_{m} = 1 mod m, if m is prime so M_{m} would need the correct residue mod (m1) for this to work. 

20120916, 20:09  #11 
∂^{2}ω=0
Sep 2002
República de California
2^{3}·1,459 Posts 
Dude, did you really need to start 3 separate threads on this? I merged all of 'em here.

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