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Topic: Composites, and neat relation
Replies: 18   Last Post: Sep 14, 2004 3:38 AM

 Messages: [ Previous | Next ]
 JAMES HARRIS Posts: 9,787 Registered: 12/4/04
Re: Composites, and neat relation
Posted: Sep 12, 2004 9:27 PM

jstevh@msn.com (James Harris) wrote in message news:&lt;3c65f87.0409121043.5988edb8@posting.google.com&gt;...
&gt; So someone pointed out that there's the trivial relation
&gt;
&gt; [x] + [x + 1/2] = [2x] where you're in reals,
&gt;
&gt; and I started thinking about
&gt;
&gt; [x] + [x + 1/k] = [2x]
&gt;
&gt; also in reals, with k&gt;1, and it turns out you need x&gt;1 as well, which
&gt; I was thinking about didn't put down before.

You can have x less than 1 as what's needed is

xk + 1 &gt;= k

where that might be useful in using what might be a probabilitic prime
test as way to factor as I'll explain below...

&gt; Using that with x = N/2kj, where N, k and j are naturals, and N&gt;=2kj,
&gt; you get
&gt;
&gt; [(N+2j)/2kj] = [N/kj] - [N/2kj]

What's really neat here is that I've split kj.

&gt;
&gt; and to test it, I'll use N=13, k=2, j=3, which gives
&gt;
&gt; [19/12] = [13/6] - [13/12]
&gt;
&gt; so that works.
&gt;
&gt; That should say something about what j can be in general given a
&gt; composite, but I'd think it'd just repeat what you can find by other
&gt; means.
&gt;
&gt; Still I'm mulling it over and don't mind tossing it out early.
&gt;
&gt; 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.

&gt; assuming that j is the one that equals 1.

It has to be.

&gt; Is there some way then, given some composite C, to play with
&gt;
&gt; [(N+2j)/2C] = [N/C] - [N/2C]
&gt;
&gt; where N&gt;2C, to determine if j can be other than 1 or C?
&gt;
&gt; If so, then it's a prime test.
&gt;

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 sample--possibly randomly--the 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

Date Subject Author
9/12/04 JAMES HARRIS
9/12/04 The Last Danish Pastry
9/12/04 Jim Burns
9/12/04 Tim Smith
9/13/04 David C. Ullrich
9/12/04 Nate Smith
9/12/04 C. BOND
9/12/04 Dik T. Winter
9/12/04 Dik T. Winter
9/12/04 JAMES HARRIS
9/13/04 Paul Murray
9/13/04 JAMES HARRIS
9/13/04 C. BOND
9/13/04 Jim Burns
9/14/04 Nate Smith
9/13/04 Dik T. Winter
9/14/04 Paul Murray
9/13/04 Dik T. Winter
9/13/04 David C. Ullrich