I wanted to know, how can i prove that the product of two natural numbers with their sum cannot be the third power of a natural number.

Any help appreciated.



You're talking about the solvability of the Diophantine equation $xy(x+y)=z^3$.

If a prime $p$ divides $x$ and $y$, then it must divide $x+y$ and $z$ and we can cancel $p^3$ in the equation. So we can assume that $x$ and $y$ are coprime. This implies that $x$, $y$, $x+y$ are coprime. Therefore, if their product is a cube, then each factor must be a cube: $x=u^3$, $y=v^3$, $x+y=w^3$. But then $u^3+v^3=w^3$, which has no integer solutions, as proved by Euler.

  • $\begingroup$ @lal bhattacharjee u guessed a fairly tough one too, could u provide some help on the one u guessed or anyone else. $\endgroup$ – Shobhit Jun 20 '13 at 16:16
  • $\begingroup$ @Shobhit, what do you mean? $\endgroup$ – lhf Jun 20 '13 at 16:16
  • $\begingroup$ (x+1) *(y+1) = z^3 +1. solutions to this equation??? $\endgroup$ – Shobhit Jun 20 '13 at 16:22
  • 2
    $\begingroup$ @lhf You forgot the trivial solutions... $\endgroup$ – fretty Jun 20 '13 at 16:23
  • 2
    $\begingroup$ @Shobhit, ask a separate question. $\endgroup$ – lhf Jun 20 '13 at 16:23

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