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Re: Composites, and neat relation
Posted:
Sep 12, 2004 9:27 PM


jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409121043.5988edb8@posting.google.com>... > So someone pointed out that there's the trivial relation > > [x] + [x + 1/2] = [2x] where you're in reals, > > and I started thinking about > > [x] + [x + 1/k] = [2x] > > also in reals, with k>1, and it turns out you need x>1 as well, which > I was thinking about didn't put down before.
You can have x less than 1 as what's needed is
xk + 1 >= k
where that might be useful in using what might be a probabilitic prime test as way to factor as I'll explain below... > Using that with x = N/2kj, where N, k and j are naturals, and N>=2kj, > you get > > [(N+2j)/2kj] = [N/kj]  [N/2kj]
What's really neat here is that I've split kj.
> > and to test it, I'll use N=13, k=2, j=3, which gives > > [19/12] = [13/6]  [13/12] > > so that works. > > That should say something about what j can be in general given a > composite, but I'd think it'd just repeat what you can find by other > means. > > Still I'm mulling it over and don't mind tossing it out early. > > After all, if kj is prime, then only j = 1, or j=k, will work,
Actually, only j=1 will *always* work, if kj is prime.
> assuming that j is the one that equals 1.
It has to be.
> Is there some way then, given some composite C, to play with > > [(N+2j)/2C] = [N/C]  [N/2C] > > where N>2C, to determine if j can be other than 1 or C? > > If so, then it's a prime test. >
It should actually work. Basically you should be able to sample with various N's and calculate what j will work.
The more N's you samplepossibly randomlythe closer to the actual value of j you should get.
So, it would actually factor kj, and if kj is prime, your j would consistently decrease as it approached 1.
Neat. I wonder how well it works.
James Harris



