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Old 2020-08-20, 20:47   #1
baih
 
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Jun 2019

2·17 Posts
Default Mersenne factorization by (nxy+x+y)

Let Mersenne number 2n -1

if 2n -1 composite



2n -1 = n2xy + (x+y)n + 1


so 2n /n

= (n2xy + (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


211-1 = 2047
(2047-1) /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
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Old 2020-08-20, 22:20   #2
Viliam Furik
 
Jul 2018
Martin, Slovakia

7×29 Posts
Default Obvious mistakes

I would like to point out a few mistakes.

1. (2^n) is never divisible by (n), (2^n-2) is divisible by (n), when (n) is prime (btw, it is because of Little Fermat theorem)
2. (2047-1) /2= 186; you probably meant (2047-1)/11 = 186.

Apart from these typos, I guess I will leave the topic for other guys.
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Old 2020-08-20, 22:27   #3
baih
 
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Jun 2019

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Default

Quote:
Originally Posted by Viliam Furik View Post
I would like to point out a few mistakes.

1. (2^n) is never divisible by (n), (2^n-2) is divisible by (n), when (n) is prime (btw, it is because of Little Fermat theorem)
2. (2047-1) /2= 186; you probably meant (2047-1)/11 = 186.

Apart from these typos, I guess I will leave the topic for other guys.
thanks i mean (2^n)-2

Last fiddled with by baih on 2020-08-20 at 22:29
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Old 2020-08-21, 00:38   #4
Batalov
 
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"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

3×3,041 Posts
Default

Please demonstrate the power of this method on a tiny number 2^1277-1.
it is composite.
Show us.
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Old 2020-08-21, 01:09   #5
baih
 
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Jun 2019

2216 Posts
Default

Quote:
Originally Posted by Batalov View Post
Please demonstrate the power of this method on a tiny number 2^1277-1.
it is composite.
Show us.
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

Last fiddled with by baih on 2020-08-21 at 01:17
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Old 2020-08-21, 03:21   #6
mathwiz
 
Mar 2019

6416 Posts
Default

Quote:
Originally Posted by baih View Post
If someone found a solution to the equation c=nxy+x+y
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?
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Old 2020-08-21, 03:42   #7
a1call
 
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"Rashid Naimi"
Oct 2015
Remote to Here/There

3×641 Posts
Default

Quote:
Originally Posted by baih View Post
Let Mersenne number 2n -1

if 2n -1 composite



2n -1 = n2xy + (x+y)n + 1


so 2n /n

= (n2xy + (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


211-1 = 2047
(2047-1) /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
That's a good find. I think I have a similar post here somewhere.

The problem is you need brute-force (trying different integers for a solution) and the combinations are astronomically large.

You might have some fun with Wolfram-Alpha:

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 thinking-power in the process.
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Old 2020-08-25, 07:55   #8
LaurV
Romulan Interpreter
 
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Jun 2011
Thailand

22·3·11·67 Posts
Default

Quote:
Originally Posted by a1call View Post
I think I have a similar post here somewhere.
Except your post was left-aligned, therefore easier to read, haha. This is just some rubbish thrown in the middle of the screen, impossible to read.
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