How can I find all solutions of this equation?

Question

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?

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

$c=3$, $n=c=3$ ⇒ $k=78$

$c=673$ ⇒ $n=673$ ⇒$k=17498$

$c=2019$ ⇒ $n_1(2019 n_1^2 +1)=k_1^2$, 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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