Additional and Multiplication of Polynomials

Question

I was stuck in a problem which asks me to find a degree 2 polynomials $a(x), b(x), c(y), d(y)$ such that $a(x)c(y) + b(x)d(y) = 1 + xy + xy^2$ or claims that it is not exsited. And it asks to prove the answer.

My idea is to write these polynomials as: $a(x) = a_0 + a_1x + a_2x^2, b(x) =b_0 + b_1x + b_2x^2...., c(x)=c_0..., d(x)=d_0...$

Then, I expand the left-hand side of the equation and use the undetermined coefficient method to match the coefficients. The result is messy. I have so many equations like a2c2 + b2d2 = 1 to solve.

Next, my idea is to use the fact that: for a degree $2$ polynomials, the coefficient of its second-degree term should be nonzero. I try to find the conflicts between these equations so that I can conclude that the polynomials do not exist, but I failed.

I've searched for this question but found no answers. Thus, could anyone give me some hints? I am a new learner to mathematical proofs and is taking the application of Linear Algebra now. Hope you could provide a comparatively easy-understanding method for me. Thanks.

Answer 1

Let $z$ be root for $d$, and let $y=z$. Then $$a(x)c(z) + \underbrace{b(x)d(z)}_{=0} = 1 + xz + xz^2$$

If $c(z)=0$ then $1 + xz + xz^2=0$ for all $x$ which is impossible.

So since $c(z)\ne 0$ we have $$a(x) = {1\over c(z)}(1 + xz + xz^2)$$

which is valid for all $x$ so $a$ is linear and thus there are no such functions.

Answer 2

Assume there would exist a solution $$ a(x)b(y)+c(x)d(y) = 1+x(y+y^2) $$ with polynomials $a,b,c,d$ each of degree two. Then we may suppose without restriction that $a,c$ are monic. We derivate twice with respect to $x$, getting $2(b+d)=0$, so $$ (a(x)-c(x))b(y) = 1+x(y+y^2) $$ for all real $x,y$. The polynomial $a-c$ must have degree one, else there is no $x$ in the right hand side. But then setting $y=0$ (or $y=-1$) we get $(a-c)b(0)=1$. Contradiction.