Tuesday, March 10, 2009


JSH: Competency problem

I've started looking at x and y more closely, including giving r(v) and t(v), where x = r(v)/t(v) on my math blog, and the more I look at them and consider the mathematical proof that underpins this research, my take on the postings I've seen is there is a massive competency problem from posters, as in lack of competency.

These equations CAN be kind of complicated to handle, to understand what to do, but for instance, there are two sets. The plus or minus is all in combination as in if you have + with one of them, then it's plus down the line, and the -/+ if it's negative it's negative for all of them.

THAT is not hard. Or you can simply derive the two sets from the underlying equations, no biggee.

Looking at r(v) + t(v) and r(v) - t(v), I'm not seeing anything that says these quadratics can't deliver,

Like here's an example of what I mean:

r(v) = [+/-((D-1)f_1 + (D-1)f^2*v^2) - [+/-2Dv +/- (D-1)(f_1 - f_2*v^2 - 2v) -/+(f_1 + f_2*v^2)]]

where f_1*f_2 = D-1.

All you need is abs(r(v)) < D, and you're guaranteed a non-trivial factorization if D is an odd composite.

Now the way the plus or minuses work you actually have just two equations:

r(v) = [((D-1)f_1 + (D-1)f^2*v^2) - [2Dv + (D-1)(f_1 - f_2*v^2 - 2v) - (f_1 + f_2*v^2)]]


r(v) = [-((D-1)f_1 + (D-1)f^2*v^2) - [-2Dv - (D-1)(f_1 - f_2*v^2 - 2v) + (f_1 + f_2*v^2)]]

and you need r(v) < D, so you can solve the quadratic for that, or if r(v) tends to be negative r(v) > - D.

That's it.

Now those are simpler so I can ask you yahoos if your position is that given r(v) above, is v always non-rational? Or is there never an integer v such that r(v) < D, or r(v) > -D, with those equations?

That's an easier question than the bigger one of whether or not I solved the factoring problem.

Now in my mind, many of you are just incompetent at mathematics, but you talk a lot on newsgroups.

So that's what you are: people who cannot really do math, but people who will despite that, talk a lot on newsgroups.

You're incompetent.

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