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Composites, and neat relation
Posted:
Sep 12, 2004 2:43 PM


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.
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]
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, assuming that j is the one that equals 1.
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.
James Harris



