Find $P,Q,R \in \mathbb{R}[x,y]$ satisfying an identity

Question

Find polynomials $P,Q,R \in \mathbb{R}[x,y]$ such that $(x+y+1)^2P(x,y) + x^2Q(x,y) + y^2R(x,y) = 1$.

All I have tried is bashing but doesn't work well. Is there a smart combination here?

Any help appreciated!

Answer 1

One way to think about it is that you want $(x+y+1)^2P(x,y)$ to have a constant term of $1$, and otherwise have only terms divisible by $x^2$ or $y^2$ (so that they can be cancelled by a corresponding term in either $x^2Q(x,y)$ or in $y^2R(x,y)$). Which is to say, no $x, y$ or $xy$ terms.

Let's take a closer look at what's stopping us. To do that, we need to expand $(x+y+1)^2$ to get $$ x^2+y^2+1+2x+2y+2xy $$ We can ignore the $x^2$ and $y^2$ terms for now, as they ultimately won't be in our way, but for now they are distracting. We need to eliminate the $x, y$ and $xy$ terms, but $P$ must at the same time have a constant term of $1$. How can we make that happen?

We know that $(a+b)(a-b)=a^2-b^2$, which looks like an excellent way to make terms of higher degree while not leaving any lower degree terms behind. Maybe we can use that somehow? We know $P$ must have constant term $1$, so that can be our $a$. Let the remaining problematic terms be our $b$, and see what happens: $$ (1+2x+2y+2xy)(1-(2x+2y+2xy))=1-(2x+2y+2xy)^2\\ =1-4x^2-4y^2-4x^2y^2-8xy-8x^2y-8xy^2 $$ That's quite a lot of terms, but note that we are down to a single problematic term: $-8xy$. Again, we ignore the non-problematic terms and calculate: $$ (1-8xy)(1+8xy)=1-64x^2y^2 $$ That's it. All together, our $P$ has become $$ P(x, y)=(1-2x-2y-2xy)(1+8xy) $$ As for finding $Q$ and $R$, you will have to calculate the full product $(x+y+1)^2P(x,y)$ and then just pick off terms one by one to include in either $Q$ or $R$ as you see fit.