Prove that there are no integer solutions to this equation












4












$begingroup$


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.






number-theory diophantine-equations







share|cite|improve this question











$endgroup$








  • 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 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 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 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 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 at 3:25
















4












$begingroup$


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.






number-theory diophantine-equations







share|cite|improve this question











$endgroup$








  • 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 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 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 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 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 at 3:25














4












4








4





$begingroup$


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.






number-theory diophantine-equations







share|cite|improve this question











$endgroup$




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.






number-theory diophantine-equations




number-theory diophantine-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 2:24









Henning Makholm

244k17312556




244k17312556










asked Jan 2 at 2:21









MarcosMarcos

303




303








  • 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 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 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 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 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 at 3:25














  • 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 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 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 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 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 at 3:25








1




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 at 2:25




$begingroup$
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 at 2:25




1




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 at 2:30






$begingroup$
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 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 at 2:42






$begingroup$
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 at 2:42






1




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 at 2:51




$begingroup$
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 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 at 3:25




$begingroup$
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 at 3:25










1 Answer
1






active

oldest

votes


















3












$begingroup$

Since $nneq 0$, 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 $6nxneq 0$, so we get $y=pm 3n^2$. Both of them forces $x^3=n^3$, so $x=n$ means no solutions to original equation.






share|cite|improve this answer











$endgroup$













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












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Since $nneq 0$, 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 $6nxneq 0$, so we get $y=pm 3n^2$. Both of them forces $x^3=n^3$, so $x=n$ means no solutions to original equation.






share|cite|improve this answer











$endgroup$













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
















3












$begingroup$

Since $nneq 0$, 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 $6nxneq 0$, so we get $y=pm 3n^2$. Both of them forces $x^3=n^3$, so $x=n$ means no solutions to original equation.






share|cite|improve this answer











$endgroup$













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














3












3








3





$begingroup$

Since $nneq 0$, 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 $6nxneq 0$, so we get $y=pm 3n^2$. Both of them forces $x^3=n^3$, so $x=n$ means no solutions to original equation.






share|cite|improve this answer











$endgroup$



Since $nneq 0$, 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 $6nxneq 0$, so we get $y=pm 3n^2$. Both of them forces $x^3=n^3$, so $x=n$ means no solutions to original equation.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 3:13

























answered Jan 3 at 3:04









Yong Hao NgYong Hao Ng

3,7091222




3,7091222












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


















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
















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




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


















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