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18031Re: [PrimeNumbers] Checking Large "Prime Numbers"?

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  • Thomas Hadley
    May 10 10:21 AM
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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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