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22533Re: Algebraic factoring

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  • kraDen
    Feb 2, 2011
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      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
      >
      >
      >
      > --- In primenumbers@yahoogroups.com,
      > Jack Brennen <jfb@> wrote:
      >
      > > Actually, the original number 16^137-1 is equal to 2^548-1,
      > > correct? x^548-1 is easily seen to be divisible by x^4-1,
      > > which is divisible by x^2+1.
      > > So 2^548-1 is divisible by 2^2+1.
      Thankyou

      > Thanks Jack for explaining to my good friend Ken,
      > who had seemed to believe that I had cheated
      > by computing (shame!) the integer 2^2+1 and
      > then actually using trial division (crime!)
      I certainly didn't mean to imply this.
      I was sure you had done it algebraically.
      It just wasn't apparent to me how.
      As I said I found 4 factors but then got stuck.
      I desired learning how to derive the additional factors (for future use, originality is not my strong point but I willingly "stand on the shoulders of giants").
      cheers
      Ken
      > by whatever integer 2^2+1 turned out to be.
      > I never did any such heinous thing, honest, Guv.
      >
      > David
      >
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