mersenneforum.org A second proof for the Lucas-Lehmer Test
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 2017-07-30, 00:29 #1 carpetpool     "Sam" Nov 2016 22·83 Posts 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
2017-07-30, 00:44   #2
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

3×29×113 Posts

Quote:
 Originally Posted by carpetpool (A)... if and only if ...(B)
If you are trying to provide a proof, then where is the second half?

2017-07-30, 09:21   #3
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

156810 Posts

Quote:
 Originally Posted by carpetpool 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 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.

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