prove if $xyk \neq 0$, then: $x^3=3(k+xy)(k-xy-y^3)$ has no integral solutions.

Question

Let $\gcd(x,y)=1, k \in \mathbb{Z}$ and $x \equiv 0 \pmod 3$.

Show that if $xyk \neq 0$, then: $$x^3=3(k+xy)(k-xy-y^3)$$ has no integral solutions.

Any hints? I keep getting lost in my reasoning.

Answer 1

Writing the quadratic formula for $k,$ we find that $k$ is an integer if and only if $$ 12(x+y^2)^3 - 3 y^6 $$ is an integer square.

IF $y$ IS ZERO THIS MEANS ANY $x = 3 u^2$ WORKS.

When $y \neq 0,$ call that square $s^2,$ since $s=3t$ we get $ 12(x+y^2)^3 - 3 y^6 = 9 t^2, $ or $ 4(x+y^2)^3 - y^6 = 3 t^2. $ The only restrictions are that the variables $x,y,k$ not be zero. It becomes convenient to write $w = x + y^2$ and $4 w^3 - y^6 = 3 t^2,$ and we are asking whether we must have $w = y^2.$ The best order for this is $$ \color{red}{ y^6 - 4 w^3 + 3 t^2 = 0.} $$

From COHEN PAGES 396 397, definition of twisted projective middle of page 396, this result is Proposition 6.5.9(3)(d) near the top of page 397, we find that all solutions are twists of $$ (-12, 144, \pm 1728). $$ Taking his $\lambda = -1/12,$ all solutions are equivalent to $$ (1,1,\mp 1). $$

Which is to say, the only integer solutions to $$ \color{red}{ y^6 - 4 w^3 + 3 t^2 = 0} $$ have $w = y^2$ and $t = \pm y^3.$ Given $w = x + y^2,$ this means $x=0.$

EDIT: sad, just checked, page 397 is not shown, because that is where the good bits are...part (3)(d) says IF $ab^2 c^3 / 432 = m^6$ is a sixth power, we have the two solutions $(bc,-12bcm^2, \pm36b^2m^3),$ which are unique up to twisted projective equivalence.

The proof, quite short, is also on page 397.

THIS OTHER REALLY LONG URL is, with any luck, a pdf of a later paper that includes the result. Yep, Theorem 2.5.2, no new solutions occur when allowing Gaussian integers, just the ones Cohen indicated.