I would like to know if it is possible to prove that there are no integer solutions to:
$3n(4x^3-n^3)=y^2$, where $x$, $y$ and $n$ are all positive integers and $x>n$.
I have no idea how to start, so any comments are welcome.
Thank you and regards, Marcos.
Since $n\neq 0$, multiply by $144/n^4$ and set $X=12x/n$, $Y=12y/n^2$ to get the Elliptic curve $$ E:Y^2 = X^3 - 432 $$ So any solutions $(x,y,n)$ must also be a rational point $(X,Y) = (12x/n,12y/n^2)$ on $E$.
It is known that this curve has precisely 3 rational points. One way is to see this is by reformulating it as $$ (36+Y)^3 + (36-Y)^3 = 216Y^2+93312 = (6X)^3 $$ and then Fermat's Last Theorem forces at least one of $6X, 36+Y,36-Y$ to be zero. If $X=0$ there is no solution, hence $Y=\pm 36$, which in turn forces $X=12$. Hence the three points on $E$ are $$ (X,Y) = \mathcal O, (12,36),(12,-36) $$
Therefore solutions $(x,y,n)$ must satisfy $$ (12,\pm 36) = (\frac{12x}{n},\frac{12y}{n^2}) $$ Finally, equating $12 = 12x/n$ gives $x=n$, therefore there are no solutions since we want $x>n$.
A more direct/concise way is by computing $$ (3n^2 + y)^3 + (3n^2 - y)^3 = 54n^6 + 18n^2y^2 = 216n^3x^3 = (6nx)^3, $$ then by FLT at least one of $3n^2+y,3n^2-y,6nx$ is zero. Similarly $6nx\neq 0$, so we get $y=\pm 3n^2$. Both of them forces $x^3=n^3$, so $x=n$ means no solutions to original equation.