Finding solutions to the diophantine equation $xy(x+y) = n$.

Question

I initially wanted to prove that there are no integer solutions for the equation $xy(x+y) = 4$, but I got intrigued by the general case as I noticed that there tends to be solutions when $n$ is in the form $2^a$. For example, $4 \times 4 \times(4+4)=2^7$. I have tried using the quadratic formula to solve for x in terms of y to reduce the original equation to a single-variable equation, but it gets far too messy:

$$ x = \frac{-y^2 \pm \sqrt{y^4 + 4ny}}{2y} \\ (-\frac{y^2}{2} \pm \frac1{2}\sqrt{y^4+4ny})(\frac{y}{2} \pm \frac1{2y}\sqrt{y^4+4ny})=n $$

I also see that $xy(x+y)$ is a multiplication of three almost independent (since addition makes factoring hard) terms, so we would have to somehow split numbers into 3 divisions.

Are there any concise special cases for n? And how would you approach disproving the existence of solutions given a specific n?

Answer 1

Above equation shown below

$xy(x+y) = n$ ------$(1)$

Equation $(1)$ has parametric solution & given below:

$x= k(k+1)(3k^2-4k-2)$

$y=5(-k)^3(k+1)$

$n=10k^5(k+1)^5(3k^2-4k-2)$

For $k=2$, we get:

$(x,y,n)=[(12),(-120),(155520)]$

Answer 2

This may help:
If $xy(x+y)=n$ then $(x+y)^3=3n+x^3+y^3$ which means $3n=(x+y)^3-x^3-y^3$ which shows that $3n$ is of a special kind, since we know that not every number is represented with $3$ cubes (with mixed sign).