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?
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1This 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). – Dietrich Burde Dec 29 '18 at 15:56
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By examination of simple instances of your simplification, $y=7,\ n=18$ is one solution, yielding $k=37$. There may be others. – Keith Backman Dec 29 '18 at 16:47