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Solving two-variate polynomial identities

asked 12 years ago

Rolandb gravatar image

updated 9 years ago

FrédéricC gravatar image

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 was raised earlier.

Thanks in advance for any support!

Roland

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answered 5 years ago

Max Alekseyev gravatar image

updated 5 years ago

The problem is quite vague as there exist lots of polynomials satisfying the conditions. Here is just one way to get some of them.

First, decide what factors of AB(A+B) divide what polynomials. For example, if we fix that AP1, BP2, and (A+B)P3, then we can generate them as follows:

  1. Take any polynomial Q in A and B, and set P3=(A+B)Q. Initially set both P1 and P2 equal to 0.
  2. For each term cAiBj in P3:
  3. (i) if i=0, add m to P2
  4. (ii) if j=0, add m to P1
  5. (iii) otherwise partition c=c1+c2 (which can be done in many ways), and add c1AiBj to P1 and add c2AiBj to P2
  6. If gcd(P1,P2)=1, report P1, P2, and P3.
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Asked: 12 years ago

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Last updated: Oct 04 '19