Re: Quadratic residue method for finding primes

Bob Terwilliger wrote:
> Arturo Magidin wrote:

<deleted>
> >
> > Note that we know exactly which primes have 2 as a quadratic residue:
> > namely, those which are congruent to 1 or -1 modulo 8. Likewise, if we
> > replace 2 with any prime r, then we can figure out exactly which
> > primes have r as a quadratic residue by the use of Quadratic
> > Reciprocity; and they will all lie in certain arithmetic progressions
> > modulo r or modulo 4r (depending on whether r is congruent to 1 modulo
> > 4 or to 3 modulo 4).
> >
>
> Yes, isn't this a result of Dirichlet? I'm starting to sense analytic
> number theory on the horizon!

You all are running off into boringly old mathematics.

I've already taken my idea of focusing on n^2 - r to get a result
related to Goldbach's conjecture, which can be found on my blog. I've
decided not to post it on Usenet for now, as I'm less and less thrilled
with reply quality.

It's a remarkably simple result too, which is amazing to me.

In any event, r=17, and n^2 - 17, offers the question of long runs with
various residues where you can have a very high probability of finding
only primes.

Looks like focusing on r of the form

r = 2^k C + 1

where k is a natural number and C is a natural number, while r is not a
square, offers significant masking.

Note that of course r=17, is just using k=4, C=1.

The area seems wide open, and as I do research on the web it looks like
people just focused on other things--like the things the two
sci.math'ers who came over to reply here have been talking about, as
they try to head off into familiar, old, and to me, boring territory.

BIG money may be in the new area.

Or maybe not, but it's fresh territory--an almost unknown in modern
mathematics at this level of simplicity.

Oh, and I should emphasize I quickly found this intriguing result
related to Goldbach's conjecture, of all things.

It is clearly a wide open area that somehow got missed. Good luck!

James Harris
Received on Mon May 1 02:03:36 2006