# Why there isn't any integer solutions in non-zero integers for $z^3 = 3(x^3 +y^3+2xyz)$?

Consider the following Diophantine equation

$$z^3 = 3(x^3 +y^3+2xyz)$$

Is there any elementary proof for the non solubility in positive integers for this Diophantine equation, where $x, y$ and $z$ are pairwise coprime positive integers?

I have proved the impossibility of solution of this Diophantine equation in positive integers, but that took quite a few pages and I consider this too long. I hope for a much more elementary and shorter proof for this puzzle.

Hint: It should be noted that is also true for the non-availability of the similar form (replacing coefficient 3 by 1) as the following:

$$z^3=x^3+y^3+2xyz,$$ but this case is very simple to prove, where $x, y\;\&\; z$ are nonzero integers.

• Could you just describe your proof in short in the question? – TheRandomGuy Apr 4 '16 at 9:47
• Evidently the only solutions are $(t,-t,0).$ Thus the only primitive solutions are $(1,-1,0)$ and $(-1,1,0).$ Where did you get this problem? There is a repeat of it, about 12 hours before this comment. Here: math.stackexchange.com/questions/1731616/… – Will Jagy Apr 7 '16 at 18:47
• I would like credible and/or official sources for the question. – Will Jagy Apr 7 '16 at 18:54
• The problem leads to finding rational points on the curve $3U^3 +3V^3 +6UV−1=0$, which is birationally equivalent to the elliptic curve $Y^2 +Y=X^3 −270X−1708$ with trivial torsion and rank zero, so it has no rational points. Thus, the original equation has no solutions, except those with z=0. – duje Apr 7 '16 at 19:37
• @duje thank you. It turns out that, if the coefficient $3$ is replaced by other integers, solutions are possible, I put a computer search at math.stackexchange.com/questions/1731616/… Mysterious that the ratios $9$ and $27$ are readily found. – Will Jagy Apr 7 '16 at 22:28

The problem leads to finding rational points on the curve $3U^3+3V^3+6UV-1 = 0$, which is birationally equivalent to the elliptic curve $Y^2+Y = X^3-270X-1708$ with trivial torsion and rank zero, so it has no rational points.