# The product of two natural numbers with their sum cannot be the third power of a natural number.

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.

Thanks.

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So, you're talking about the solvability of the Diophantine equation $xy(x+y)=z^3$? –  lhf Jun 20 '13 at 16:06
@lhf has guessed it right. –  Ramanujan Jun 20 '13 at 16:10

## 1 Answer

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.

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