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Old 2006-07-05, 22:12   #1
jasong
 
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"Jason Goatcher"
Mar 2005

66638 Posts
Default M_n, S_n, and the k in k*2^n+/1

While I don't understand why the fact that M_n divides S_n means M_n is prime, and I definitely don't understand Fast Fourier Transforms, I at least know how M_n and S_n are generated. Or at least I understand how they would be generated if we could store the rather extreme S_n number around where we're at now with the exponents.

Here's my question: In terms of solving whether or not k*2^n+/-1 is prime, as opposed to the simpler 2^n+/-1, if I were to solve the problem manually(yes, I know this is impossible in my lifetime if I were to use pen and paper), how does the k come into the equation?

I don't totally understand why modular arithmetic solves the problem. Well, I understand HOW it works, I just don't know WHY it works. If I need to know WHY modular arithmetic works to know how k enters the equation, then I apologize for wasting your time. Otherwise, I'd appreciate an explanation.
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Old 2006-07-11, 11:59   #2
alpertron
 
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Aug 2002
Buenos Aires, Argentina

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Default

For numbers of the form n=h*2k+1 where h is not too high (h < 2k) you can use Proth's theorem:

Start with a=3. Then perform the modular exponentiation a(n-1)/2 (mod n). If the result is n-1, the number is prime. If the result is not 1 or n-1, the number is composite, otherwise, increment the variable a and perform the modular exponentiation again.

You can read the proof on Chris Caldwell's site about primes

Last fiddled with by alpertron on 2006-07-11 at 12:04
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