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Re: [PrimeNumbers] Checking Large "Prime Numbers"?

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  • Bob Gilson
    Thanks to everyone for the great response - now for the joys of PARI/GP Regards Bob ... Woo-woo! Brownie-points for Phil! I was thinking of answering by
    Message 1 of 9 , May 9, 2006
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      Thanks to everyone for the great response - now for the joys of PARI/GP

      Regards

      Bob

      Phil Carmody <thefatphil@...> wrote:
      --- Alan Eliasen <eliasen@...> wrote:
      > Phil Carmody wrote:
      > > --- Bob Gilson <bobgillson@...> wrote:
      > > > A colleague of mine claimed the other day that
      > > > 5, followed by one billion 9's, and 6, followed by 999,999,999 zeroes,
      > with
      > > > a further last digit being 1, are in fact twin primes.
      > > > How does anyone go about refuting or confirming such allegations?
      > >
      > > With Pari/GP in a fraction of a second:
      > > ? test(p)=centerlift(6*Mod(10,p)^1000000000)^2
      > > ? forprime(p=2,100000,if(test(p)==1,print(p)))
      >
      > Impressive timings! This is the only response that actually seemed to
      > answer the original question, *how does one go about it* rather than just
      > enigmatically listing factors, which does not help the original poster, nor
      > answer the question posed.

      Woo-woo! Brownie-points for Phil!

      I was thinking of answering "by evaluating the expressions for the two numbers
      modulo 31", which could have been a middle-ground between my useful :-) and
      everyone else's useless :-P answers.

      > Could you explain the mathematics behind this one
      > (especially why you use the centerlift function and what it does) for those
      > not familiar with Pari/GP?

      It simply picks a distinguished member of the set of numbers == a (mod b) in
      the range (-b/2, b/2] rather than [0,b). So rather than +1 and p-1 you'll have
      -1 and +1. Hence the square to subsequently turn both of those into 1.

      > I would also be interested if the others who posted results would answer
      > the original question--how one goes about testing claims like this
      > (efficiently, I hope. I know how to do it several brute-force ways.)
      > Thanks!

      Essentially, the same way as the above, I'd bet.

      If you've been given a large number that is claimed to be prime, and it's not
      obviously a product of hand-crafted secret numbers, then the best way to
      counter the claim of primality is almost always to find a small factor. The
      best way to find a small factor is to evaluate the expression for it modulo the
      small primes in turn.

      Phil

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    • Thomas Hadley
      May I add a little more explanation to Phil s Pari script for us who are still novices? We re trying to find factors of n +/- 1 where n=6*10^1000000000.
      Message 2 of 9 , May 10, 2006
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        May I add a little more explanation to Phil's Pari script for us who are
        still novices?
        We're trying to find factors of n +/- 1 where n=6*10^1000000000. Phil's
        script calculates
        n (mod p) for primes p from 2 to 100000 and then squares it. If the
        result is 1, then n (mod p)
        is either -1 or +1. If it is -1, p is a factor of n+1 and if n(mod p) is
        +1, then p is a
        factor of n-1. Squaring combined these two tests into one.

        Here was Phil's script:

        test(p)=centerlift(6*Mod(10,p)^1000000000)^2
        forprime(p=2,100000,if(test(p)==1,print(p)))

        For Pari novices, like myself, this is a good example of how to use IntMod
        types, which
        is what you get with the Mod(x,p) function. When you do arithmetic
        functions on an IntMod, the
        result is always calculated modulo p, so it never gets too big. Now,
        10^1000000000 is too big
        for Pari to handle, but Mod(10,p)^100000000 does not get too big. Pari
        will do this
        exponentiation without overflowing anything. Same with the multiply by 6.

        An IntMod type is always in the range of 0 to p-1 (mod p), and lift()
        converts that type to
        an integer in that range. But centerlift( ) converts it to an integer in
        the
        range (-b/2, b/2], as Phil explained. p-1 is now -1, p-2 would be -2,
        etc.

        Phil could have made his test use lift() and then compare test(p) to 1 OR
        p-1 which would
        require a temp variable but eliminate needing to square. Only Phil would
        know which
        would be faster. Here's how that could be implemented.

        test(p)=lift(6*Mod(10,p)^1000000000)
        forprime(p=2,100000,temp=test(p);\
        if(temp==1,print(p," is a factor of n-1"));\
        if(temp==(p-1),print(p," is a factor of n+1")))

        I had to put \ at the end of some lines -- I still don't know the rules
        about when
        they are necessary.

        Hope this helps.

        Tom Hadley

        primenumbers@yahoogroups.com wrote on 05/09/2006 10:01:04 AM:

        > Thanks to everyone for the great response - now for the joys of PARI/GP
        >
        > Regards
        >
        > Bob
        >
        > Phil Carmody <thefatphil@...> wrote:
        > --- Alan Eliasen <eliasen@...> wrote:
        > > Phil Carmody wrote:
        > > > --- Bob Gilson <bobgillson@...> wrote:
        > > > > A colleague of mine claimed the other day that
        > > > > 5, followed by one billion 9's, and 6, followed by 999,999,999
        zeroes,
        > > with
        > > > > a further last digit being 1, are in fact twin primes.
        > > > > How does anyone go about refuting or confirming such
        allegations?
        > > >
        > > > With Pari/GP in a fraction of a second:
        > > > ? test(p)=centerlift(6*Mod(10,p)^1000000000)^2
        > > > ? forprime(p=2,100000,if(test(p)==1,print(p)))
        > >
        > > Impressive timings! This is the only response that actually seemed
        to
        > > answer the original question, *how does one go about it* rather than
        just
        > > enigmatically listing factors, which does not help the original
        poster, nor
        > > answer the question posed.
        >
        > Woo-woo! Brownie-points for Phil!
        >
        > I was thinking of answering "by evaluating the expressions for the two
        numbers
        > modulo 31", which could have been a middle-ground between my useful :-)
        and
        > everyone else's useless :-P answers.
        >
        > > Could you explain the mathematics behind this one
        > > (especially why you use the centerlift function and what it does) for
        those
        > > not familiar with Pari/GP?
        >
        > It simply picks a distinguished member of the set of numbers == a (mod
        b) in
        > the range (-b/2, b/2] rather than [0,b). So rather than +1 and p-1
        you'll have
        > -1 and +1. Hence the square to subsequently turn both of those into 1.
        >
        > > I would also be interested if the others who posted results would
        answer
        > > the original question--how one goes about testing claims like this
        > > (efficiently, I hope. I know how to do it several brute-force ways.)
        > > Thanks!
        >
        > Essentially, the same way as the above, I'd bet.
        >
        > If you've been given a large number that is claimed to be prime, and
        it's not
        > obviously a product of hand-crafted secret numbers, then the best way to
        > counter the claim of primality is almost always to find a small factor.
        The
        > best way to find a small factor is to evaluate the expression for
        itmodulo the
        > small primes in turn.
        >
        > Phil
        >


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