Re: Factorizaton idea, revisited

From: Abraham Buckingham (twizlewink_at_hotmail.com)
Date: 05/19/04 

Date: 18 May 2004 19:53:15 -0700

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0404271913.4fa86e5b@posting.google.com>...
> Consider
>
> (jk - Tk + T)(jk + Tk + T) = T^4
>
> where T = M + 1, or T = M - 1, where M is some integer to be factored.

Ok, M will be the number 8 then, and so T = 8 + 1 = 9

> Solving for k gives
>
> k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2)

solving for k gives me k = 72/j from the original equation.

> and the choice for T means that you can factor T^4 so that you have
>
> jk - Tk + T = f_1 and
>
> jk + Tk + T = f_2, where
>
> f_1 f_2 = T^4,

This is obvious from your first sequence, and in this case simplifes
to jk - 9k + 9 = f_1 and jk + 9k + 9 = f_2.

> and solve for j, and you'll get rational j's that are not integers for
> the interesting solutions.

How? Should I use my k = j/72 substitute it and then solve? This gives
me 72 - (9*72)/j + 9 = f_1 and 72 + (9*72)/j + 9 = f_2. Simlifying I
get f_1 = 81 - 648/j and f_2 = 81 + 648/j therefore f_1 f_2 = 81^2 -
(648/j)^2 = 9^4. Therefore 81^2 - 9^4 = (648/j)^2 = 0 therefore j =
+/- 648
 
> Using those j's you should be able to factor M rather easily, as then
> you have
>
> k = (-j(M+1) +/- (M+1)^2 sqrt(j^2 - M(M+2))/(j^2 - (M+1)^2)
>
> using T = M+1.

Ok if I should be able to factor M "easily" at this point I expect you
shouldn't have any trouble showing he how I should continue. What does
the relationship between j and k have to do with the factors of M? How
do I get from M = 8 to the 2*2*2 factoring we're aiming for? So far I
haven't seen an occurance of the number 2 anywhere in relation to M,
which I should have by now right? I mean I should be looking at a
final equation looking something like M = 2*c where c is some constant
and 2 is the first divsor of M this we find, but that hasn't happened
yet. Neither j or k works out to be a factor of M, so I'm confused.
 
> I want readers to consider the *insulting* posts that came in response
> to my previous posting, as I want you to think about those people.
>
>
> James Harris

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