Generalization of Rational Root Theorem [closed]

Can the rational root theorem in univariate variable polynomial equations be extended to more than one variable? For instance could we apply it to $$f(x,y)=1+x+xy-2y^2+4x^2$$ to find all possible rational roots? I am sure the obvious answer is no but is there any hint on how to find all possible rational roots say of this example?

Solution 1:

As said in the comments: the answer is NO in general.

For your particular case, you can do the following thing.

We have $$\begin{array}{lll}f(x,y)&=&-2(y-\frac{x}{4})^2+\frac{x^2}{2}+4x^2+x+1\cr &=&-2(y-\frac{x}{4})^2+\frac{9}{2}x^2+x+1\cr &=&-2(y-\frac{x}{4})^2+\frac{9}{2}(x+\frac{1}{9})^2+\frac{17}{18}.\end{array}$$

Soving $f(x,y)=0$ where $x,y\in\mathbb{Q}$ is the same as solving $-2u^2+\frac{9}{2}v^2=-\frac{17}{18}$,setting $u=y-\frac{x}{4}$ and $V=x+\frac{1}{9}$ (it is easy to derive $x,y$ from $u,v$. This is equivalent to $-36 u^2+81v^2=-17$, that is $(3u-9v)(3u+9v)=17,$ which is easy to solve. For $t\in\mathbb{Q}^\times$, this is equivalent to solve $3u-9v=t, 3u+9v=\frac{17}{t}$.

You then get $u,v$ as a rational fraction of $t$, then you get $x,y$.

[The idea is that you have an affine conic, so you can reduce it to a "diagonal form", and then reduce to the easier case to a rational conic $ax^2+by^2=c$. This situation is well understood from quadratic form specialist, and you have systematic methods to solve this even if it can be a bit tricky. Here we are in a particularly nice situation.]