Polynomials that form $1+xy+x^2 y^2$ Show that there is no polynomials $a(x), b(x) \in R[x]$ and $c(y), d(y) \in R[y]$ such that $1+xy +x^2 y^2 = a(x) c(y) + b(x) d(y) $
 A: Expanding on @MikeDaas's comment we have$$\begin{align}1&=a(x)c(0)+b(x)d(0),\\1+x+x^2&=a(x)c(1)+b(x)d(1),\\1-x+x^2&=a(x)c(-1)+b(x)d(-1)\\\implies x&=\frac{c(1)-c(-1)}{2}a(x)+\frac{d(1)-d(-1)}{2}b(x),\\x^2&=\frac{c(1)-2c(0)+c(-1)}{2}a(x)+\frac{d(1)-2d(0)+d(-1)}{2}b(x).\end{align}$$Since $1,\,x,\,x^2$ are all linear combinations of $a,\,b$, these two functions span a $3$-dimensional space, a contradiction.
A: In the original relation, plug in $y=0$, We get:
$$ 1 =  a(x)c(0) +b(x)d(0).$$
Again in the original relation, take the derivative wrt $y$ and then plug in $y=0$. We get:
$$ x = a(x)c'(0) +b(x)d'(0). $$
One more time, take the derivative wrt to $y$ twice in the original relation and we get:
$$ x^2 = 1/2\left(a(x)c''(y) +b(x)d''(y)\right). $$
We also see from the original relation, by plugging in $x=1$, that $c$ and $d$ have at most degree $2$, that is their second derivatives are constant.
Hence $1,x,x^2$ is spanned over $\mathbb{R}$ by two polynomials in $x$, $a$ and $b$, which is absurd (same argument as the one in answer above).
