Re: JSH: Nearly done
From: Rick Decker (rdecker_at_hamilton.edu)
Date: 01/28/05
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Date: Thu, 27 Jan 2005 21:47:31 -0500
Tim Peters wrote:
> ...
>
> [Rick Decker]
>
>>>You can *always* find rational solutions for x. That's not a deep
>>>result.
>
>
> [JSH]
>
>>>>Maybe for really big numbers rational x's become difficult to find.
>>>>
>>>>But at this point, I don't see a mathematical reason why they should.
>>>>
>>>>I think it would be quite useful if one of you could settle the
>>>>question.
>
>
> [Rick Decker]
>
>>>Done. I'll give a proof if anyone's interested.
>
>
> Please do. I was hoping James could be tempted into a bit of civility by
> saying "yes, please do" himself, but days have passed. At least he didn't
> post to call you a liar.
For your (and the two or three others who might be interested) I just
posted a proof in this thread. As for the "liar" response, I've
gotten used to them (my favorite, which I have on my office door,
calls my behavior "comptemptible" and alleges that he made efforts
to get me in trouble at work by taking away my attempts at
"plausible deniability", whatever that means).
I originally thought the lack of response lately was due to James having
killfiled me, but I now think it's due to the fact that he doesn't have
a news client and so has to rely on Google groups. This makes it easy to
miss responses that are a day or so late in coming, which I think might
explain why he doesn't excoriate me as much as he could. Not that it
matters--I have tenure, the ultimate source of hammer-immunity.
>
> [Nora Baron]
>
>>Harris still seems to think that finding rational x might be a
>>problem.
>
>
> So leave him behind <wink>.
>
> ...
>
>
>>Given a rational x, Harris will then look at its numerator
>>and try to find primes which divide M. Of course I would
>>think you need some assurance that such primes are not also
>>in the denominators of the things that x is multiplied by -
>>I would guess that is very unlikely in a real live problem
>>because those factors will be extremely large.
>
>
> Rick is just looking for x such that 1 < gcd(numerator(x), M) < M (I don't
> know what James is looking for). When that holds, we've found a non-trivial
> factor, and all the equations leading up to it become irrelevant (so, e.g.,
> it doesn't matter if that factor also appears as a factor of some
> denominator along the way).
>
Quite right. In fact, since gcd is so efficient, it doesn't even
pay to reduce the fractions, since you might pick up the right
factor that way, given that it seems to me that all James is
doing amounts to nothing more than obfuscated trial divisors.
<snip>
Regards,
Rick
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