How to solve $y^3=x(x+1)$ where $x$ and $y$ are integers ? Can you help me ?

Thanks :)


closed as off-topic by gen-z ready to perish, mihaild, John Omielan, Shogun, Lee David Chung Lin Jul 16 at 1:50

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  • $\begingroup$ What do you mean “solve”? $\endgroup$ – gen-z ready to perish Jul 15 at 19:48
  • $\begingroup$ With a little effort you should be able to find one or two integer solutions, whose inclusion would improve the Question to asking if one can prove them to be the only solutions. $\endgroup$ – hardmath Jul 15 at 21:34

Hint: notice that $gcd(x,x+1)=1$ and thus $x=a^3$ and $x+1=b^3$



If two numbers $u$, $v$ are coprime and $uv$ is a $r$-th perfect power, then both $u$ and $v$ are $r$-th perfect powers.


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