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.

  • 1
    $\begingroup$ Where did this even come from? Have you tried anything already? Like have you tried to see when this equation holds modulo some integer, for example $3$ or $4$? $\endgroup$ – SmileyCraft Jan 2 '19 at 2:25
  • 1
    $\begingroup$ I am not an expert on number theory, so the only thing I tried is a script that checked the first 10000 numbers for n, x and y... 3n(4x^3-n^3) is the discriminant of a quadratic equation whose solutions need to be integers, that is the origin of the expression. Therefore, proving that there are no integer solutions to this expression, it is possible to show that the quadratic equation does not have any solutions. $\endgroup$ – Marcos Jan 2 '19 at 2:30
  • $\begingroup$ Where does the problem come from? Random idea: expand the cube and then try to match the 3 expressions with $y^2$ in all possible ways. Say, $3n = y^2$ and rest of the expression is equal to 1. See if integer solutions are possible. Try all scenarios. You will always be able to determine values for all variables (3 variables, 3 equations) $\endgroup$ – Makina Jan 2 '19 at 2:42
  • 1
    $\begingroup$ Thank you for all your comments. I will try what you propose and see if I am able to make some progress. $\endgroup$ – Marcos Jan 2 '19 at 2:51
  • $\begingroup$ Marcos, the best thing would be for you to edit in the original problem, which you describe as some sort of quadratic. The problem you put above may be hard to settle, as any $n$ leads to a solution with $x=n, y=3n^2,$ and simply demanding $x>n$ does not clearly lead to a way to rule out solutions $\endgroup$ – Will Jagy Jan 2 '19 at 3:25

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.

  • $\begingroup$ That is great, thank you for this amazing answer! $\endgroup$ – Marcos Jan 3 '19 at 9:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.