Mersenne factorization by (nxy+x+y)
[CENTER]Let Mersenne number 2[SUP]n[/SUP] 1
if 2[SUP]n[/SUP] 1 composite 2[SUP]n[/SUP] 1 = n[SUP]2[/SUP]xy + (x+y)n + 1 so 2[SUP]n[/SUP] /n = (n[SUP]2[/SUP]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[SUP]11[/SUP]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?[/CENTER] sory for my english 
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. 
[QUOTE=Viliam Furik;554433]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.[/QUOTE] thanks i mean (2^n)2 
Please demonstrate the power of this method on a tiny number 2^12771.
it is composite. Show us. 
[QUOTE=Batalov;554448]Please demonstrate the power of this method on a tiny number 2^12771.
it is composite. Show us.[/QUOTE] non 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 
[QUOTE=baih;554454]If someone found a solution to the equation c=nxy+x+y[/QUOTE]
There's infinitely many solutions: c=x=y=0, x=y=1 and c=n+2, and so on. What is the purpose of this is equation and what constraints are you placing on the variables? 
[QUOTE=baih;554421][CENTER]Let Mersenne number 2[SUP]n[/SUP] 1
if 2[SUP]n[/SUP] 1 composite 2[SUP]n[/SUP] 1 = n[SUP]2[/SUP]xy + (x+y)n + 1 so 2[SUP]n[/SUP] /n = (n[SUP]2[/SUP]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[SUP]11[/SUP]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?[/CENTER] sory for my english[/QUOTE] That's a good find. I think I have a similar post here somewhere. 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: [url]https://www.wolframalpha.com/input/?i=186+%3D+11xy%2Bx%2By+solve+over+the+integer[/url] [url]https://www.wolframalpha.com/input/?i=%28%282%5E12771%291%29+%2F2%3D+1277xy%2Bx%2By+solve+over+the+integer[/url] Good luck, try expanding the concept. You might get something interesting or at worst expand your thinkingpower in the process. 
[QUOTE=a1call;554470]I think I have a similar post here somewhere.[/QUOTE]Except your post was leftaligned, therefore easier to read, haha. This is just some rubbish thrown in the middle of the screen, impossible to read.

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