20200820, 20:47  #1 
Jun 2019
2·17 Posts 
Mersenne factorization by (nxy+x+y)
Let Mersenne number 2^{n} 1 if 2^{n} 1 composite 2^{n} 1 = n^{2}xy + (x+y)n + 1 so 2^{n} /n = (n^{2}xy + (x+y)n) /n = nxy+x+y Finding the x and y we can factor the number into a product (nx)+1 and (ny)+1 example 2^{11}1 = 2047 (20471) /2= 186 186 = nxy+x+y = 11* 8*2 + 8+2 X= 8 Y=2 and 2047 = (88+1)*(22+1) Difficulty and complexity (nxy+x+y) like a Diophantine equation Are there any solutions? sory for my english 
20200820, 22:20  #2 
Jul 2018
Martin, Slovakia
7·29 Posts 
Obvious mistakes
I would like to point out a few mistakes.
1. (2^n) is never divisible by (n), (2^n2) is divisible by (n), when (n) is prime (btw, it is because of Little Fermat theorem) 2. (20471) /2= 186; you probably meant (20471)/11 = 186. Apart from these typos, I guess I will leave the topic for other guys. 
20200820, 22:27  #3  
Jun 2019
2×17 Posts 
Quote:
Last fiddled with by baih on 20200820 at 22:29 

20200821, 00:38  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
23A3_{16} Posts 
Please demonstrate the power of this method on a tiny number 2^12771.
it is composite. Show us. 
20200821, 01:09  #5  
Jun 2019
100010_{2} Posts 
Quote:
The difficulty is the same as the difficulty of (Trial division) But it may help in some cases If someone found a solution to the equation c=nxy+x+y Last fiddled with by baih on 20200821 at 01:17 

20200821, 03:21  #6 
Mar 2019
1100100_{2} Posts 

20200821, 03:42  #7  
"Rashid Naimi"
Oct 2015
Remote to Here/There
11110000011_{2} Posts 
Quote:
The problem is you need bruteforce (trying different integers for a solution) and the combinations are astronomically large. You might have some fun with WolframAlpha: https://www.wolframalpha.com/input/?...er+the+integer https://www.wolframalpha.com/input/?...er+the+integer Good luck, try expanding the concept. You might get something interesting or at worst expand your thinkingpower in the process. 

20200825, 07:55  #8 
Romulan Interpreter
Jun 2011
Thailand
37·239 Posts 

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