# How can I find all solutions of this equation?

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?

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

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.