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?

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    $\begingroup$ Degree is already $3$ ; at least homogenous polynomial :) ; could be hard to classify numbers that can be represented that way. $\endgroup$
    – Peter
    Jul 24, 2018 at 16:58
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    $\begingroup$ Simple observations: Odd $n$ can never work. $2p^a$ works iff $a\equiv 0\pmod 3$ or $p=3$, $a\equiv 1\pmod 3$. (In particular $2^a$ works iff $a\equiv 1\pmod 3$. $\endgroup$ Jul 24, 2018 at 17:15
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    $\begingroup$ On the last question: "And how would you approach disproving the existence of solutions given a specific n?" - if the $n$ is not too big (like $n=4$), we can just test all possible divisors of $n$ for $x$ and then solve for $y$ to see if it's an integer. $\endgroup$
    – lisyarus
    Jul 24, 2018 at 17:30
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    $\begingroup$ See OEIS sequence A088915. $\endgroup$ Jul 24, 2018 at 17:38

2 Answers 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).


Above equation shown below

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

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

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



For $k=2$, we get:


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    $\begingroup$ When I plug $k=2$ into your formula I get $n=1920$, which does not satisfy the original equation. Certainly there are other solutions, so this approach will not prove we cannot make a particular $n$ $\endgroup$ Jul 26, 2018 at 15:30
  • $\begingroup$ @Ross Millikan. Note that formula for OP equation has been corrected-Sam $\endgroup$
    – Sam
    Jul 28, 2018 at 13:52

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