# Roots of rational equation with multiple variables?

Let's say we have a rational polynomial in $k$ variables. We are only interested in rational solutions. If $k = 1$, name the variables ${x}$, if $k = 2$, name them ${x,y}$.

For $k = 1$, it can be done very fast. The Rational Root Theorem gives a set of candidates. But what for $k=2$? How can I split a polynomial into factors in this case?

Example: $x^2-y^2$ should be split to $(x-y)(x+y)$.

• There is no reason to post this question on StackOverflow; it is very much a number-theoretic question. Aug 12, 2012 at 14:53

Polynomials in more than one variable do not generally split into factors, even if you allow complex coefficients. For example, $x^2 + y^2 - 1$ doesn't split in this way.
Note that Fermat's Last Theorem can be phrased as the problem of finding rational points on the family of Fermat curves $x^n + y^n = 1$, so there's no reason to expect that this is an easy problem if you believe that Fermat's Last Theorem is difficult.
• Oh, one more question! Restricting my solutions to $\mathbb{Z}$ leads to the same issues, right? I think fermats theorem is usually stated for numbers in $\mathbb{Z}$... Aug 12, 2012 at 19:38
• @Johannes: they are actually worse because of Matiyasevich's theorem (en.wikipedia.org/wiki/Diophantine_set). Finding a rational solution to $x^n + y^n = 1$ is more or less the same as finding an integer solution to $X^n + Y^n = Z^n$ (multiply by a common denominator). Aug 12, 2012 at 19:47