visu
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CRGreathouse said:On the bottom of the first page, you write:
1/2(d6)x2 + (d4-(1/2d6))x + d1 = y2
Is this
\frac12d_6x^2+(d_4-\frac12d_6)x+d_1=y^2
or
\frac{1}{2d_6}x^2+\left(d_4-\frac{1}{2d_6}\right)x+d_1=y^2
or something else?
Also, where does this quadratic come from? Does x have some value (it isn't used before the equation) or is this simply a member of R[x]? *Why* do we want the left to be a square?
Kittel Knight said:Your "integer factorization algorithm" depends on the "generic two integer variable equation solver", which depends on prime factoring of integers,etc...
Six of one, half-dozen of the other...
visu said:The first of the two equations you wrote.Apologies for not being clear.
CRGreathouse said:Not a problem. I'll note that 1/2\cdot d_6=1 since d6 is always 2, so the equation
\frac12d_6x^2+\left(d_4-\frac12d_6\right)x+d_1=y^2
simplifies to
x^2+(d_4-1)x+d_1=y^2
which has roots
2y=1-d_4\pm\sqrt{d_4^2-2d_4+1-4d_1}