20201023, 09:12  #23  
Romulan Interpreter
Jun 2011
Thailand
2^{3}·19·61 Posts 
Quote:
How do you "produce a composite number from a prime number"? Do you multiply it with 7? Or subtract 53? Can you prove that, by doing so, you don't end with another number which is still prime? And you want to do that by "using 2^p1". How? You take 2^p1 and hit your prime number in the head with it until it becomes composite? And what the hack is "contains prime factors" having to do with the way you "produced" it? All composite numbers contain prime factors, regardless of how you "produced" them. All prime numbers "contain prime factors" too (themselves). Man, how old are you? You may be like 12 years old or so, learning the basics right now, and not having the proper way to express yourself in English yet, and in that case, you may be a genius and become great in the future. At that age I didn't have any idea about all this stuff. If that's the case, my apologies. But if you are older than 18, stop the funking trolling and go learn the lingo, if you want to talk math. P.S. I never pretended to be a decent person. Last fiddled with by LaurV on 20201023 at 09:40 

20201023, 09:40  #24 
"Viliam FurÃk"
Jul 2018
Martin, Slovakia
3×127 Posts 
I would love a way to give a "like" or some kind of positive evaluation of my opinion to the post, without having to write another one, saying I like it. Hmmm, Xyzzy ? Last fiddled with by Viliam Furik on 20201023 at 09:41 
20201023, 11:56  #25  
Feb 2017
Nowhere
2^{5}·3^{3}·5 Posts 
Quote:
Let p > 2 be prime. The smallest conceivable composite factor of 2^{p}  1 is (2p+1)^{2}. So if q = 2*k*p + 1 divides 2^{p}  1, and k < 2*p + 2, then q is prime. Alas, it is more than likely that if k < 2*p + 2 and q = 2*k*p + 1 is prime, that q does not divide 2^{p}  1. Especially if q is congruent to 3 or 5 (mod 8) 

20201023, 12:00  #26 
Romulan Interpreter
Jun 2011
Thailand
2^{3}·19·61 Posts 

20201023, 12:28  #27 
Feb 2017
Nowhere
2^{5}·3^{3}·5 Posts 

20201025, 05:35  #28  
Dec 2017
240_{10} Posts 
Quote:


20201025, 19:53  #29 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·3,109 Posts 
A red card given

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