Thursday, April 29, 2004
Factorizaton idea, revisited
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.
Solving for k gives
k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2)
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,
and solve for j, and you'll get rational j's that are not integers for the interesting solutions.
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.
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.
(jk - Tk + T)(jk + Tk + T) = T^4
where T = M + 1, or T = M - 1, where M is some integer to be factored.
Solving for k gives
k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2)
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,
and solve for j, and you'll get rational j's that are not integers for the interesting solutions.
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.
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.