Given a polynomial in two variables $P(x,y)$ with even exponents, can there be another polynomial $Q(x,y)$ such that $P(x,y)=(x^3+y^3-1)Q(x,y)$?
The original problem I am trying to solve asks:
- Is there a polynomial $P(x,y)=1+\sum_{i=1}^nz_ix^{2\alpha_i}y^{2\beta_i}$ where $z_i$ is an integer and $\alpha_i,\beta_i$ are natural numbers.
Such that:
- Whenever $(x,y)$ is a solution to $x^3+y^3-1=0$, it is also a solution to $P(x,y)=0$
I think that $P$ cannot exist because the locus of solutions to $P(x,y)=0$ contains the solutions to $x^3+y^3-1=0$ so $P=(x^3+y^3-1)Q$ for some $Q$. But any monomial $cx^ny^m$ in $Q$ will give rise to terms $cx^{n+3}y^m$ and $-cx^ny^m$ in $P$, which contradicts the premise that $P$ contains only even powers.
However, I am not certain if this holds logically, because $P$ can be manipulated with $x^3=1-y^3$ without changing the locus of solutions of interest.
The answer to your question is, in general, no. For instance, if $P(x,y)=x^2+y^2$, there is no polynomial $Q(x,y)$ such that $P(x,y)=(x^3+y^3-1)Q(x,y)$. However, the answer to the original problem is yes. An example of such a polynomial is \begin{align} P(x,y)&=(x^3+y^3-1)(x^3-y^3+1)(x^6-y^6-1-2y^3) \\ &=(x^6-(y^3-1)^2)(x^6-y^6-1-2y^3) \\ &=(x^6-y^6-1+2y^3)(x^6-y^6-1-2y^3) \\ &=(x^6-y^6-1)^2-4y^6 \\ &=1-2x^6-2y^6-2x^6y^6+x^{12}+y^{12}. \end{align}