I am trying to solve the equaiton $n^3+2019 n=k^2$, where $n$ and $k$ be two positive integral numbers. I tried with Mathematica and get two solution $k = 78, n = 3$ and $k = 17498, n = 673$. How can I find all solutions of the given equation?

  • $\begingroup$ if i were you, i would try to find values of $n$ such that ${{n}^{3}}+2019n$ is a complete square $\endgroup$
    – logo
    Jun 10, 2019 at 15:51
  • $\begingroup$ Yes. Thank you very much. $\endgroup$ Jun 10, 2019 at 15:53
  • $\begingroup$ You can also solve the cubic equation using this: en.m.wikipedia.org/wiki/Cubic_function $\endgroup$
    – Duns
    Jun 10, 2019 at 16:20
  • 4
    $\begingroup$ You are looking for integer points on an Elliptic curve. This is a difficult problem in general. $\endgroup$
    – Somos
    Jun 10, 2019 at 16:43

1 Answer 1


Clearly n and k have a common divisor like c. Also $2019=3\times 673$. Let $n=n_1 c$ and $k=k_1c$ such that $(k_1, n_1)=1$ then we have:

$$n_1^3 c^3 +3\times 673 n_1 c=k_1^2 c^2$$

That indicates $3\times 673 n_1 c$ must also be divisible by $c^2$ and this is possible only if $c=3$ or $c=673$, or $c=2019$.Let n itself be the common divisor then

$c=3$, $n=c=3$$k=78$


$c=2019$$n_1(2019 n_1^2 +1)=k_1^2$, but $(k_1, n_1)=1$ and $[n_1, (2010 n_1^2 +1)]=1$, it is not known this equation can have integer solutions. So it seems there is no more integer solutions to this question.



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