Number theory vs cubing












0












$begingroup$


Solve this diophantide equation
$$4y^3-3=k^2$$



My way was not successful
First see that $k^2$ is odd, so I assumed that $k=2n+1$.



After simplifying I got:



$$y^3=n^2+n+1$$



And I am stuck here again.
Who has a great solution?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is an example of Mordell's equation for an elliptic curve. It has been solved ("great solution") here at this site already. Have a look at this question and related ones (like this one).
    $endgroup$
    – Dietrich Burde
    Dec 29 '18 at 15:56












  • $begingroup$
    By examination of simple instances of your simplification, $y=7, n=18$ is one solution, yielding $k=37$. There may be others.
    $endgroup$
    – Keith Backman
    Dec 29 '18 at 16:47
















0












$begingroup$


Solve this diophantide equation
$$4y^3-3=k^2$$



My way was not successful
First see that $k^2$ is odd, so I assumed that $k=2n+1$.



After simplifying I got:



$$y^3=n^2+n+1$$



And I am stuck here again.
Who has a great solution?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is an example of Mordell's equation for an elliptic curve. It has been solved ("great solution") here at this site already. Have a look at this question and related ones (like this one).
    $endgroup$
    – Dietrich Burde
    Dec 29 '18 at 15:56












  • $begingroup$
    By examination of simple instances of your simplification, $y=7, n=18$ is one solution, yielding $k=37$. There may be others.
    $endgroup$
    – Keith Backman
    Dec 29 '18 at 16:47














0












0








0





$begingroup$


Solve this diophantide equation
$$4y^3-3=k^2$$



My way was not successful
First see that $k^2$ is odd, so I assumed that $k=2n+1$.



After simplifying I got:



$$y^3=n^2+n+1$$



And I am stuck here again.
Who has a great solution?










share|cite|improve this question











$endgroup$




Solve this diophantide equation
$$4y^3-3=k^2$$



My way was not successful
First see that $k^2$ is odd, so I assumed that $k=2n+1$.



After simplifying I got:



$$y^3=n^2+n+1$$



And I am stuck here again.
Who has a great solution?







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 Dec 29 '18 at 16:12









amWhy

192k28225439




192k28225439










asked Dec 29 '18 at 15:50









HeartHeart

20516




20516








  • 1




    $begingroup$
    This is an example of Mordell's equation for an elliptic curve. It has been solved ("great solution") here at this site already. Have a look at this question and related ones (like this one).
    $endgroup$
    – Dietrich Burde
    Dec 29 '18 at 15:56












  • $begingroup$
    By examination of simple instances of your simplification, $y=7, n=18$ is one solution, yielding $k=37$. There may be others.
    $endgroup$
    – Keith Backman
    Dec 29 '18 at 16:47














  • 1




    $begingroup$
    This is an example of Mordell's equation for an elliptic curve. It has been solved ("great solution") here at this site already. Have a look at this question and related ones (like this one).
    $endgroup$
    – Dietrich Burde
    Dec 29 '18 at 15:56












  • $begingroup$
    By examination of simple instances of your simplification, $y=7, n=18$ is one solution, yielding $k=37$. There may be others.
    $endgroup$
    – Keith Backman
    Dec 29 '18 at 16:47








1




1




$begingroup$
This is an example of Mordell's equation for an elliptic curve. It has been solved ("great solution") here at this site already. Have a look at this question and related ones (like this one).
$endgroup$
– Dietrich Burde
Dec 29 '18 at 15:56






$begingroup$
This is an example of Mordell's equation for an elliptic curve. It has been solved ("great solution") here at this site already. Have a look at this question and related ones (like this one).
$endgroup$
– Dietrich Burde
Dec 29 '18 at 15:56














$begingroup$
By examination of simple instances of your simplification, $y=7, n=18$ is one solution, yielding $k=37$. There may be others.
$endgroup$
– Keith Backman
Dec 29 '18 at 16:47




$begingroup$
By examination of simple instances of your simplification, $y=7, n=18$ is one solution, yielding $k=37$. There may be others.
$endgroup$
– Keith Backman
Dec 29 '18 at 16:47










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