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CRGreathouse 2010-08-12 14:55

Assumptions: m, t, p, n are positive integers, p > 1.

[QUOTE=science_man_88;225054][TEX]24m= 6tp+(p-7)\right m=px+c [/TEX]
[TEX] 24m=6tp-(p+7)\right m=px+c ? [/TEX][/QUOTE]

So we have 24m + 7 being an integer with [TEX]24m+7=p(6t\pm1)[/TEX] as appropriate.

[QUOTE=science_man_88;225054][TEX]if(px+c==(4^n-1)/3,(4^\strike n -1)/3[/TEX][/QUOTE]

So if m is of the form (4^n-1)/3, you're saying that 2n+3 is not a Mersenne exponent. That is, if 24 * (4^n-1)/3 + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 8 * (2^(2n) - 1) + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.

science_man_88 2010-08-12 15:21

[QUOTE=CRGreathouse;225074]Assumptions: m, t, p, n are positive integers, p > 1.



So we have 24m + 7 being an integer with [TEX]24m+7=p(6t\pm1)[/TEX] as appropriate.



So if m is of the form (4^n-1)/3, you're saying that 2n+3 is not a Mersenne exponent. That is, if 24 * (4^n-1)/3 + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 8 * (2^(2n) - 1) + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.[/QUOTE]


something tells me you are getting at a trivial answer ?

3.14159 2010-08-12 15:32

[QUOTE=CRGreathouse]What script are you using to check? My isSPRP gives
[/QUOTE]

Nono, it's not PARI. It's a separate application. It's supposed to be an implementation of the Miller-Rabin primality test. I in fact used PARI to verify that it was indeed an error.

3.14159 2010-08-12 15:36

[QUOTE=CRGreathouse]So if m is of the form (4^n-1)/3, you're saying that 2n+3 is not a Mersenne exponent. That is, if 24 * (4^n-1)/3 + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 8 * (2^(2n) - 1) + 7 is composite, 2n+3 is not a Mersenne exponent. That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.
[/QUOTE]

Those are all trivially correct. There is no grand discovery here.

CRGreathouse 2010-08-12 15:41

[QUOTE=3.14159;225082]Those are all trivially correct. There is no grand discovery here.[/QUOTE]

The difficult discovery that I made was determining sm88's method. Once that was accomplished, it was not hard to determine that the method came down to "Mersenne numbers are prime iff they are prime".

CRGreathouse 2010-08-12 15:42

[QUOTE=3.14159;225081]Nono, it's not PARI. It's a separate application. It's supposed to be an implementation of the Miller-Rabin primality test. I in fact used PARI to verify that it was indeed an error.[/QUOTE]

Really? Interesting.

kar_bon 2010-08-12 15:45

[QUOTE=CRGreathouse;225074]That is, if 2^(2n+3) - 1 is a composite, 2n+3 is not a Mersenne exponent.[/QUOTE]

But this is the other way round: M(n)=2^n-1 can be prime, so n must be a prime!

3.14159 2010-08-12 15:54

[QUOTE=CRGreathouse]The difficult discovery that I made was determining sm88's method. Once that was accomplished, it was not hard to determine that the method came down to "Mersenne numbers are prime iff they are prime".
[/QUOTE]

Wow. That was all that it was? A tautology/circular argument? :lol:!

[QUOTE=CRGreathouse]Really? Interesting.
[/QUOTE]

You can get it for yourself [URL="http://www.naturalnumbers.org/"]here[/URL].

I suspect it of being a kook site, but, seeing as their applets work correctly to some extent, I am undecided on that matter.

Try testing it out quickly (Note: Small integers only, please), and if you can catch 25326001 as an error, please note it!

3.14159 2010-08-12 15:56

[QUOTE=kar_bon]But this is the other way round: M(n)=2^n-1 can be prime, so n must be a prime!
[/QUOTE]

Karsten, I found that you were right about the differences in testing times based on the k-values used.

CRGreathouse 2010-08-12 16:01

[QUOTE=kar_bon;225085]But this is the other way round: M(n)=2^n-1 can be prime, so n must be a prime![/QUOTE]

The claim (once all the window-dressing was removed) was that if 2^n - 1 is prime, then 2^n - 1 is prime.

Or rather, it was a restricted case of this: if n > 3 is an odd number such that 2^n - 1 is prime, then 2^n - 1 is prime.

3.14159 2010-08-12 16:03

[QUOTE=CRGreathouse]The claim (once all the window-dressing was removed) was that if 2^n - 1 is prime, then 2^n - 1 is prime.
[/QUOTE]

Call the press! We're going to be filthy rich!

Amirite?

[SPOILER]Bullshit aside.. [/SPOILER]

CRG, have you managed to do some testing on 25326001 using that app? Does it say that it's a 7-SPRP?


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