22533Re: Algebraic factoring
- Feb 2, 2011--- In email@example.com, "djbroadhurst" <d.broadhurst@...> wrote:
> --- In firstname.lastname@example.org,
> 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.
> Thanks Jack for explaining to my good friend Ken,I certainly didn't mean to imply this.
> 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 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").
> by whatever integer 2^2+1 turned out to be.
> I never did any such heinous thing, honest, Guv.
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