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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 $(x-y)(x+y)$.

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There is no reason to post this question on StackOverflow; it is very much a number-theoretic question. – Qiaochu Yuan Aug 12 '12 at 14:53
up vote 4 down vote accepted

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.

For quadratic polynomials you can appeal to the Hasse-Minkowski theorem, the Chevalley-Warning theorem, and Hensel's lemma to determine when a solution exists; this argument is described in more detail in the beginning of Cassels' Lectures on Elliptic Curves.

Beyond the quadratic case, this problem is open. Already for cubic polynomials in two variables it is not known whether there exists an algorithm which provably solves this problem, although there appear to be algorithms which work reasonably well in practice. Bjorn Poonen's Computing Rational Points on Curves contains a good discussion of the issues involved. See also, for example, this MO question.

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.

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Thanks! Only to be sure: With "the problem" you mean: Given a Polynomial, find one root? Or do you mean: is there even a root? – Johannes Aug 12 '12 at 15:06
@Johannes: my understanding is that these are both open. – Qiaochu Yuan Aug 12 '12 at 15:07
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 $\mathbb{Z}$... – Johannes Aug 12 '12 at 19:38
@Johannes: they are actually worse because of Matiyasevich's theorem ( 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). – Qiaochu Yuan Aug 12 '12 at 19:47

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