Re: Mersenne: M(2) - composite?!

2001-07-31 Thread Peter-Lawrence . Montgomery

> From: Nathan Russell <[EMAIL PROTECTED]>

> Hi all,
> 
> I'm a bit puzzled.  The other day, I donated blood and kept my mind
> busy by doing LL tests on a few exponents mentally.  I kept getting
> the result that the LL test for M(2) ends up resulting in a repeating
> value of -1, and certainly cannot ever become zero.  Am I missing
> something really obvious?  I confirmed it on paper later to make sure
> I didn't make a mistake in the mental math.  

 
One proof of the LL test notes that if we define

a[0] = 4
a[j+1] = a[j]^2 - 2(j >= 0)
then 
a[j] = (2 + sqrt(3))^(2^j) + (2 - sqrt(3))^(2^j) (j >= 0)

This is a simple induction argument.  Next, if

   alpha = (1 + sqrt(3))/sqrt(2)
   beta = (1 - sqrt(3))/sqrt(2)
then
   alpha^2 = 2 + sqrt(3)
   beta^2  = 2 - sqrt(3)
   alpha*beta = -1
so
   a[j-1] = alpha^(2^j) + beta^(2^j)(j >= 1)

If N = 2^p - 1 is prime, and p >= 3 is odd, then N == 7 (mod 24).
Therefore 2 is a quadratic residue (indeed, 2^((p+1)/2) is a square
root of 2 modulo N) but 3 is a quadratic nonresidue.  
alpha and beta are algebraic conjugates in GF(N^2).
So alpha^N = beta and beta^N = alpha (mod N).  
Since alpha*beta = -1, this implies

   alpha^(N+1) == beta^(N+1) == -1 (mod N)

and a[p-1] == alpha^(N+1) + beta^(N+1) == -2 (mod N) .
Therefore a[p-2] == 0 (mod N).

When p = 2 and N = 3, we can neither claim that 2 is a 
quadratic residue nor that 3 is a non-residue.
Both alpha and beta reduce to 1/sqrt(2) mod 3.
As in the case where p is odd, alpha and beta
are in the extension field GF(N^2) and not the base field GF(N).
Unlike that case, alpha = beta rather than alpha = beta^N and beta = alpha^N.
The proof falls apart.

   Peter Montgomery


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Re: Mersenne: M(2) - composite?!

2001-07-31 Thread Daran

-Original Message-
From: Nathan Russell <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
Date: 31 July 2001 19:51
Subject: Mersenne: M(2) - composite?!

>Hi all,
>
>I'm a bit puzzled.  The other day, I donated blood and kept my mind
>busy by doing LL tests on a few exponents mentally.  I kept getting
>the result that the LL test for M(2) ends up resulting in a repeating
>value of -1, and certainly cannot ever become zero.  Am I missing
>something really obvious?  I confirmed it on paper later to make sure
>I didn't make a mistake in the mental math.

>From <http://mersenne.org/math.htm> (brackets and exponentiation added for
the sake of clarity)

The Lucas-Lehmer primality test is remarkably simple. It states that 2**P-1
is prime if and only if S(p-2) is zero in this sequence: S(0) = 4, S(N) =
S(N-1)**2 - 2 mod 2**P-1.

If P=2 then S(P-2) is S(0) is 4 which is clearly not 0.  It's not even 0 mod
3

However <http://www.utm.edu/research/primes/mersenne.shtml#test> does state
the P must be odd.  So I guess the mersenne.org page is wrong.

>Nathan

Daran


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Re: Mersenne: M(2) - composite?!

2001-07-31 Thread CARLETON GARRISON

How did you get  -1?

Lucas-Lehmer Test: For p odd, the Mersenne number 2p-1 is prime if and only
if 2p-1 divides S(p-1) where S(n+1) = S(n)2-2, and S(1) = 4.

You missed the premise "for p odd".

And even so, S(p-1)/(2p-1) with p = 2; S(1)/3 = 4/3.

Carleton

- Original Message -
From: "Nathan Russell" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, July 31, 2001 2:19 PM
Subject: Mersenne: M(2) - composite?!


> Hi all,
>
> I'm a bit puzzled.  The other day, I donated blood and kept my mind
> busy by doing LL tests on a few exponents mentally.  I kept getting
> the result that the LL test for M(2) ends up resulting in a repeating
> value of -1, and certainly cannot ever become zero.  Am I missing
> something really obvious?  I confirmed it on paper later to make sure
> I didn't make a mistake in the mental math.
>
> Nathan
> _
> Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
> Mersenne Prime FAQ  -- http://www.tasam.com/~lrwiman/FAQ-mers
>

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Mersenne: M(2) - composite?!

2001-07-31 Thread Nathan Russell

Hi all,

I'm a bit puzzled.  The other day, I donated blood and kept my mind
busy by doing LL tests on a few exponents mentally.  I kept getting
the result that the LL test for M(2) ends up resulting in a repeating
value of -1, and certainly cannot ever become zero.  Am I missing
something really obvious?  I confirmed it on paper later to make sure
I didn't make a mistake in the mental math.  

Nathan
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