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



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


