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I can't seem to find the way to solve the following equation so help would be much appreciated..

$x^2+y^2=x^3+y^3$ over $\mathbb{Q}$

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    $\begingroup$ A promising looking beginning: If $x=p/q$ and $y=r/s$ in lowest terms, then your equation rearranges to $\frac{p^2(q-p)}{q^3}=\frac{r^2(r-s)}{s^3}$ in which each side is still in lowest terms. So you have $q=s$ and are left with an integer equation for the numerators ... $\endgroup$ – hmakholm left over Monica Feb 17 '14 at 13:07
  • $\begingroup$ I'm sorry but I don't get why this rearrangement gives q=s. Edit: ok I get it now $\endgroup$ – user129426 Feb 17 '14 at 13:13
  • $\begingroup$ $\underline{x^3+y^3}=(x+y)(\underline{x^2+y^2}-xy)$. $\endgroup$ – Lucian Feb 17 '14 at 13:20
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Take $\lambda=(t^2+s^2)/(t^3+s^3)$ for some $ t,s \in \Bbb Z$ such that $s \neq -t$ then $(\lambda t,\lambda s)$ is a solution.If $(x,y)$ is a non-trivial solution take $\lambda'=y/x$ ( Note that $\lambda' \neq 0$ )then $$x^3+y^3=x^2+y^2$$ $$\implies x(\lambda'^3+1)=(\lambda'^2+1)$$ $$\implies x=(\lambda'^2+1)/(\lambda'^3+1)$$ Put $\lambda'=s'/t'$ for $s',t' \in \Bbb Z$. We get the same form as above so those are the only solutions.

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