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Let $x,y,z$ be 3 non-zero integers defined as followed:

$$(x+y)(x^2-xy+y^2)=z^3$$

Let assume that $(x+y)$ and $(x^2-xy+y^2)$ are coprime and set $x+y=r^3$ and $x^2-xy+y^3=s^3$

Can one write that $z=rs$ where $r,s$ are 2 integers? I am not seeing why not but I want to be sure.

Matt Samuel
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    FWIW, note that $(x+y)(x^2-xy+y^2)=x^3+y^3$, and so by Fermat's Last theorem there aren't any integer solutions. Of course, proving FLT is not elementary :) but the $n=3$ case isn't too hard. – PM 2Ring Nov 07 '15 at 03:28

2 Answers2

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Yes, there exist such integers $r$ and $s$. It is simplest to use the Fundamental Theorem of Arithmetic (Unique Factorization Theorem). The result is easy to prove for negative $z$ if we know the result holds for positive $z$. Also, the result is clear for $z=1$. So we may assume that $z\gt 1$.

Let $z=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, where the $p_i$ are distinct primes. Then $z^3=p_1^{3a_1}\cdots p_k^{3a_k}$.

Because by assumption $x+y$ and $x^2-xy+y^2$ are relatively prime, the primes in the factorization of $z^3$ must split into two sets, the ones that "belong to" $x+y$ and the ones that belong to $x^2-xy+y^2$. Because any prime has exponent divisible by $3$, each of $x+y$ and $x^2-xy+y^2$ is a perfect cube.

The rest is easy. From $z^3=r^3s^3$, it immediately follows that $z=rs$.

Remark: Note that $x$ and $y$ relatively prime does not imply $x+y$ and $x^2-xy+y^2$ are relatively prime. They could be both divisible by $3$.

André Nicolas
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Yes.

If $z^3 = r^3s^3$ we can take cube roots of both sides to get $z = rs$. It's valid to do this because $n$ is the only solution to $({n^3})^{1/3}$ for real $n$. If we go to complex numbers we have to be more careful because cube roots have 3 solutions.

PM 2Ring
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