# Solutions to a diophantine equation

I tried to find integer solutions to the following diophantine equation $$x^3 - 3y^3 + 5z^3 - 3xy^2 + 3x^2y + 9xz^2 + 7x^2z + 3yz^2 - 3y^2z + xyz = 0$$ but was unable to do so. I suspect that there are no (nontrivial) solutions, but am unable to prove it. How would one prove this?

• Is there any significance to this equation, or is it more the case of being just something suspected to have no solutions? – coffeemath Jul 11 '14 at 4:36
• You might want to add the tag elliptic curves. As my answer describes, the general theory of such cubic equations doesn't really belong to elementary number theory, but rather to the theory of elliptic curves. – user160609 Jul 11 '14 at 4:51
• Not any $x^2z$ term -- is that intended? – coffeemath Jul 11 '14 at 5:04
• No it is not intended. – Mayank Pandey Jul 11 '14 at 5:53
• To the OP: I saw you accepted my answer. I wonder --- were you able to carry out any of these steps. (I am just curious whether your curve actually has local points, and so lies in Sha, or not. Or maybe you found a global point after all?) – user160609 Jul 12 '14 at 2:36

I don't know about your equation in particular, but in general this can be delicate. If you dehomogenize the equation (say by dividing through by the $z$ variable), you get a plane cubic, and you are looking for rational solutions.

The first thing to do is to test if this cubic is smooth; this is straightforward. If it's not smooth, the whole thing becomes comparatively easy, since singular cubics are rational curves.

If the cubic is smooth, then the question is harder. To begin with, you look over $\mathbb R$ and over $\mathbb Q_p$ for each prime $p$ and see if you have solutions there; this is again easy, at least in principal. If there aren't solutions over one of these fields, then there are no rational solutions, and you're done.

If there are solutions over $\mathbb R$ and each $p$-adic field, then you have a cubic curve which is so-called locally trivial, so it is an element in the Shafarevic--Tate group of its Jacobian.

You can compute this Jacobian (see e.g. this paper); it will be another plane cubic, which has a rational point at infinity. Let's call it $E$ (for elliptic curve). And your plane cubic (lets call it $C$) is an element of order dividing $3$ in the Shafarevic--Tate group of $E$. What you want to know is whether or not it's the trivial element (i.e. whether or not it has a rational point).

Now since your curve is explicit and its coefficients are not too large, I think that it's likely that $E$ will appear in standard tables of elliptic curves, and you can see whether its Shafarevic--Tate group contains non-trivial $3$-torsion elements. If it doesn't, then actually your curve $C$ has a rational point. If it it's Sha does contain non-trivial $3$-torsion elements, then you have to determine whether your curve $C$ actually corresponds to one of these non-trivial $3$-torsion points.

This should be checkable, but at this point I'm not quite sure how you're supposed to actually check this. [Actually, I guess in principle what you do is use the action of $E$ on $C$ to construct a certain $1$-cocyle for an appropriate Galois group which describes (a preimage of) $C$ in the Selmer group. You also look at a table of elliptic curves to find all the rational points on $E$, so that you know how much of the Selmer group comes from Mordell--Weil. And then you determine whether your cocycle is one of those which does come from a rational point. I think this should all be effective in principle, but I've never carried it out, and I don't know how horrendous it is in practice. Presumably there are sensible short-cuts and reinterpretations that yield a more efficient approach than the one I've just described.]

Summary: First check smoothness, then check over $\mathbb R$ and all $\mathbb Q_p$. If $C$ is smooth and has points over all these local fields, the question can become quite delicate.