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Solving two-variate polynomial identities

asked 2013-01-01 07:49:28 -0600

Rolandb gravatar image

updated 2015-10-20 14:30:54 -0600

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 2019-10-04 08:05:25 -0600

Max Alekseyev gravatar image

updated 2019-10-04 08:10:32 -0600

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 $A\mid P1$, $B\mid P2$, and $(A+B)\mid P3$, 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 $c A^i B^j$ in $P3$:
  3. (i) if $i=0$, add $m$ to $P2$
  4. (ii) if $j=0$, add $m$ to $P1$
  5. (iii) otherwise partition $c = c_1 + c_2$ (which can be done in many ways), and add $c_1A^i B^j$ to $P1$ and add $c_2A^i B^j$ to $P2$
  6. If $\gcd(P1,P2)=1$, report $P1$, $P2$, and $P3$.
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Asked: 2013-01-01 07:49:28 -0600

Seen: 175 times

Last updated: Oct 04