mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Miscellaneous Math

Reply
 
Thread Tools
Old 2017-07-30, 00:29   #1
carpetpool
 
carpetpool's Avatar
 
"Sam"
Nov 2016

1000110002 Posts
Post A second proof for the Lucas-Lehmer Test

The Lucas-Lehmer test is a test for Mersenne Numbers 2^n-1.

2^n-1 is prime if and only if 2^n-1 divides S(4, n-2).

Here S(4, n) = S(4, n-1)^2-2

starting with S(4, 0) = 4

now suppose we replace S(4, 0) = 4 with S(x, 0) = x.

We get the following polynomial sequence

S(x, 0) = x
S(x, 1) = x^2-2
S(x, 2) = x^4-4*x^2+2
S(x, 3) = x^8-8*x^6+20*x^4-16*x^2+2
S(x, 4) = x^16-16*x^14+104*x^12-352*x^10+660*x^8-672*x^6+336*x^4-64*x^2-2
...

Now we see that the discriminant D of S(x, n) = 2^r where r = (n+1)*2^n-1

Since the exponent r is odd, each prime factor q dividing S(x, n) has the form k*2^(n+1)+-1.

Now assume 2^n-1 divides S(4, n-2).

This shows that each prime factor of 2^n-1 must have the form k*2^(n-1)+-1. Since k*2^(n-1)+-1 > sqrt(2^n-1) is always true for k > 0, there is no prime factor less than or equal to sqrt(2^n-1) with that form. By trial division test if c is composite, there exists a prime p < sqrt(c) that divides c. Therefore, 2^n-1 must be prime. Please feel free to comment, suggest, improve or ask on this. Thanks!!!

Last fiddled with by carpetpool on 2017-07-30 at 00:31
carpetpool is offline   Reply With Quote
Old 2017-07-30, 00:44   #2
Batalov
 
Batalov's Avatar
 
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3

2·3·5·7·43 Posts
Default

Quote:
Originally Posted by carpetpool View Post
(A)... if and only if ...(B)
If you are trying to provide a proof, then where is the second half?
Batalov is offline   Reply With Quote
Old 2017-07-30, 09:21   #3
R. Gerbicz
 
R. Gerbicz's Avatar
 
"Robert Gerbicz"
Oct 2005
Hungary

17·79 Posts
Default

Quote:
Originally Posted by carpetpool View Post
Now we see that the discriminant D of S(x, n) = 2^r where r = (n+1)*2^n-1
Prove it.
Quote:
Originally Posted by carpetpool View Post
Since the exponent r is odd, each prime factor q dividing S(x, n) has the form k*2^(n+1)+-1.
This is very false, it fails even for n=3: 2 divides S(3), but you can't write 2 in the k*2^4+-1 form.
R. Gerbicz is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Modifying the Lucas Lehmer Primality Test into a fast test of nothing Trilo Miscellaneous Math 25 2018-03-11 23:20
Lucas-Lehmer test proof etc. science_man_88 Miscellaneous Math 48 2010-07-14 23:33
proof the lucas lehmer test kurtulmehtap Math 13 2009-10-02 12:12
Lucas-Lehmer Test storm5510 Math 22 2009-09-24 22:32
Lucas-Lehmer Test proof alpertron mersennewiki 16 2006-03-18 07:23

All times are UTC. The time now is 21:42.

Thu Jun 4 21:42:40 UTC 2020 up 71 days, 19:15, 0 users, load averages: 1.29, 1.52, 1.49

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.