ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Tue, 01 Jan 2013 14:49:28 +0100Solving two-variate polynomial identities https://ask.sagemath.org/question/9674/solving-two-variate-polynomial-identities/Hi, some help is appreciated concerning the following search.
Suppose **P1**,**P2**,**P3** are two-variate polynomials with integer coefficients in A,B. I'm searching for all sets (**P1**,**P2**,**P3**) such that:
i) **P1**(A,B)+**P2**(A,B)=**P3**(A,B)
ii) Greatest common denominator **P1** and **P2** equals 1 (Thus gcd( **P1**,**P2** )=1)
iii) The product **P1** * **P2** * **P3** can be divided by AB(A+B) (Thus gcd( **P1**.**P2**.**P3**, AB(A+B) )=AB(A+B))
Some well-known identities are A^2 + B(2A+B) = (A+B)^2, (B-A)^2 + 4AB = (A+B)^2 and
(A+2B)A^3 + B(2A+3B)^3 = (A+B)(A+3B)^3. Most interesting are polynomials with *only* linear factors such as: 16(A+B)B^3 + A(3A+4B)^3+(A+2B)(3A+2B)^3 and 27(B-A)(A+B)^5 + (3A+2B)A^3(3A+5B)^2 = (2A+3B)B^3(5A+3B)^2.
I'm curious if via Sage one could develop a generating algorithm (maybe there is a connection to Graphs and/or Combinatorics ...). N.B.: A [related question](http://ask.sagemath.org/question/1452/polynomial-identity) was raised earlier.
Thanks in advance for any support!
RolandRolandbTue, 01 Jan 2013 14:49:28 +0100https://ask.sagemath.org/question/9674/