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I can show that there infinitely many solutions to this equation. Is it possible that the set of rational solutions is dense?

edited Oct 13 at 17:33

René Pannekoek
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asked Aug 9 at 16:52

Marc Chamberland
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Dense in what? It can't be dense in $R^3$ because there won't be any solutions with distance less than 1 from (0,0,1000) – David Aug 9 at 17:05

Do you mean are the rational solutions dense in the real solutions? – sobe86 Aug 9 at 17:05

He probably means Zariski dense. – Felipe Voloch Aug 9 at 17:27

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@Marc: Dense in the surface, I think we got that, but in which topology? – Felipe Voloch Aug 9 at 20:08

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$x^3 + y^3 + z^3 - 3xyz$ is the determinant of a $3\times 3$ circulant matrix, and thus factors as $(x+y+z) (x + \omega y + \omega^2 z) (x+\omega^2 y + \omega z)$ where $\omega$ is a cube root of unity. If $x + \omega y + \omega^2 z = \alpha \in {\bf Q}(\omega)$ then $x + \omega^2 y + \omega z$ is the complex conjugate $\bar \alpha$, so $x+y+z = 1/(\alpha \bar\alpha)$. Conversely, given any nonzero $\alpha \in {\bf Q}(\omega)$ we solve these linear equations in $x,y,z\in\bf Q$ to obtain the general solution of $x^3 + y^3 + z^3 - 3xyz = 1$. – Noam D. Elkies Aug 9 at 21:37

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coma · Accepted Answer · 2012-08-10 01:29:23Z UTC

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I think this surface has a rational parameterization in terms of (a,b), given by:

$x = (1 + a + a^2)^2/(9 (3 + a (6 + (-1 + a)^2 a)) b^2) + ((-2 + (-2 + a) a) b)/(1 + a + a^2)$

$y = (1 + a + a^2)^2/(9 (3 + a (6 + (-1 + a)^2 a)) b^2) + (b + 2 a b)/(1 + a + a^2)$

$z = (1 + a + a^2)^2/(9 (3 + a (6 + (-1 + a)^2 a)) b^2) + (b - a^2 b)/(1 + a + a^2)$

For rational (a,b), this should give you a dense set of rational points...

Let me explain where this parameterization comes from, so it'll be clear that this indeed shows that rational solutions are dense.

Consider the (linear, rational) change of variables:

$x=p+r$,

$y=q+r$,

$z=r−p−q$.

The equation then simplifies to: $p^2+pq+q^2=1/(9r)$. Since the quadratic is psd, for real points we need $r>0$. Each slice (for fixed $r$) is just an ellipse. If $r$ is the square of a rational number (dense on $\mathbb{R}$), we can parameterize all the rational solutions for that slice. Taking the union over real slices (for all $r>0$), we're done.

edited Aug 10 at 1:29

answered Aug 9 at 21:27

coma
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How are we supposed to be parsing an expression of the form $u/v/w$? – Qiaochu Yuan Aug 9 at 21:33

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@QiaochuYuan: This is $$ x={\frac { \left( 1+a+{a}^{2} \right) ^{2}}{ 9 \left( 3+a \left( 6+ \left( -1+a \right) ^{2}a \right) \right) {b}^{2}}}+{\frac { \left( -2+ \left( -2+a \right) a \right) b}{1+a+{a}^{2}}}$$ $$ y ={\frac { \left( 1+a+{a}^{2} \right) ^{2}}{ 9 \left( 3+a \left( 6+ \left( -1+a \right) ^{2}a \right) \right) {b}^{2}}}+{\frac {b+2\,ab} {1+a+{a}^{2}}}$$ $$ z = {\frac { \left( 1+a+{a}^{2} \right) ^{2}}{9 \left( 3+a \left( 6+ \left( -1+a \right) ^{2}a \right) \right) {b}^{2}}}+{\frac {b-{a}^{2 }b}{1+a+{a}^{2}}} $$ – Robert Israel Aug 9 at 21:48

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It still doesn't prove density of the rational points in the reals. Any cubic surface over Q with a rational point admits a 6:1 rational map from P^2 to it (in particular, a map with Zariski dense image). However, there are cubic surfaces over Q with a rational point, but with the rational points non-dense in the reals. (See Swinnerton-Dyer, "Two special cubic surfaces".) – René Pannekoek 0 secs ago – René Pannekoek Aug 9 at 23:41

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Nice example, thanks! But I don't think it applies here. Let me explain where this parameterization comes from. Consider the (linear, rational) change of variables: $x = p+r, y=q+r, z = r-p-q$. The equation then simplifies to: $p^2 + p q+ q^2 = 1/r$. Since the quadratic is psd, for real points we need $r>0$. Each slice (for fixed r) is just an ellipse. If $r$ is the square of a rational number (dense on R), we can parameterize all the rational solutions for that slice. Taking the union over real slices, we're done. Does this make sense? (I'm running out of space here...) – coma Aug 10 at 1:21

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I see. It looks like you're considering the pencil of planes through the line $\ell:X+Y+Z=W=0$ on $S$, which cut out the family of conics you're describing. Nice! – René Pannekoek Aug 10 at 2:24

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René Pannekoek · Answer 2 · 2012-08-10 01:15:32Z UTC

The answer is yes, the rational points on your surface lie dense in the real topology.

Let's consider the projective surface $S$ over $\mathbb{Q}$ given by $X^3+Y^3+Z^3-3XYZ-W^3=0$. It contains your surface as an open subset, so to answer your question we might as well show that $S(\mathbb{Q})$ is dense in $S(\mathbb{R})$. Observe that $S$ has a singular rational point $P = (1:1:1:0)$. Since $P$ is singular, the intersection of $S$ with a plane $V$ that contains $P$ is a singular cubic $C_V$, with the rational point $P$ on it. It is a well-known and easy fact that on such curves, the rational points are dense (except if $C_V$ consists of three lines conjugate over $\mathbb{Q}$, but there are only finitely many $V$ for which this happens).

Well, in order to approximate any real point $R \in S(\mathbb{R})$ to within distance $\epsilon$, we just pick a plane $V_0$ defined over $\mathbb{Q}$, which we may choose such as to be within distance $\epsilon$ to $R$. But then also the rational points on $C_{V_0}$, which lie dense in its real locus, lie within $\epsilon$ of $R$.

Rational solutions to x^3 + y^3 + z^3 - 3xyz = 1

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