## Maybe progress in applying algebraic factoring to integer factoring

Expand Messages
• The following might or might not be progress in factoring theory. We seek the factors of z = r^4 + d3 r^3 + d2 r^2 + d1 r + d0, where d3,d2,d1,d0 are not
Message 1 of 1 , Apr 13, 2012
• 0 Attachment
The following might or might not be progress in factoring theory.

We seek the factors of z = r^4 + d3 r^3 + d2 r^2 + d1 r + d0,
where d3,d2,d1,d0 are not necessarily between 0 and (r-1), inclusively.

Suppose the factors are of the form

(r^2 + a1 r + a0) ( r^2 + b1 r + b0) = r^4 + d3 r^3 + d2 r^2 + d1 r + d0.

(r^2 + a1 r + a0) ( r^2 + b1 r + b0) = r^4 + (a1+b1) r^3 + (a1 b1 + a0
+b0) r^2 + (a0 b1 + b0 a1) r + a0 b0

It is not valid to equate coefficients of powers of r directly.

However, we can introduce parameters t1, t2, t3 such that

a1 + b1 = d3 - t3

a1 b1 + a0 + b0 = d2 - t2 + t3 r

a0 b1 + b0 a1 = d1 - t1 + t2 r

a0 b0 = d0 + t1 r

I've noticed that

(d3 - t3) + (d2 - t2 + t3 r) + (d1 - t1 + t2 r) + (d0 + t1 r) + 1

= (d3 + d2 + d1 + d0) + (r-1) (t1 + t2 + t3) + 1

= a1 + b1 + a1 b1 + a0 + b0 + a0 b1 + b0 a1 + a0 b0 + 1

= a1 b1 + a0 b1 + b0 a1 + a0 b0 + a1 + a0 + b1 + b0 + 1

= (a1 + a0)(b1 + b0) + (a1 + a0) + (b1 + b0) + 1

= (a1 + a0 + 1) (b1 + b0 + 1)

We pull from these equalities the conclusion

(a1 + a0 + 1) (b1 + b0 + 1) = (d3 + d2 + d1 + d0) + (r-1) (t1 + t2 + t3) +1

Perhaps we can figure out how to use this constraint on the coefficients

to solve the equation

(r^2 + a1 r + a0) ( r^2 + b1 r + b0) = r^4 + d3 r^3 + d2 r^2 + d1 r + d0.

for a1, a0, b1, b0, t1, t2, t3.

Kermit Rose
Your message has been successfully submitted and would be delivered to recipients shortly.