# Prove that if $x,y$ are two rational numbers such that $x^3+y^3=2xy$ then $1-xy$ is a perfect square of a rational number. [closed]

Prove that if $$x,y$$ are two rational numbers such that $$x^3+y^3=2xy$$ then $$1-xy$$ is a perfect square of a rational number.

Thank you for all solutions.

## closed as off-topic by user21820, Carl Mummert, Theoretical Economist, GNUSupporter 8964民主女神 地下教會, D_SOct 3 at 20:17

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• Maybe you can start by replacing $x$ by $a/b$ and $y$ by $p/q$ and see what equations can come out – Mark Sep 23 at 20:49
• Note that $x=1$ and $y=1$ is a solution! I suspect this is the only rational solution, in which case $1-xy=0^2$ does in fact hold. The hard part is proving that it is the only rational solution. – Zubin Mukerjee Sep 23 at 21:03
• I have an idea to substitute 1 by $\frac{x^3+y^3}{2xy}$ into the expression $1-xy$ to get a perfect square of a rational number but I cannot go on. – Blind Sep 23 at 21:04
• How about x=y=0? – Blind Sep 23 at 21:05

The result is trivial unless $$x,\,y$$ are non-zero, so that $$y=tx$$ with $$t$$ a non-zero rational number. Then $$x^3(1+t^3)=2tx^2$$ and $$x=\frac{2t}{1+t^3},\,1-xy=\frac{(1+t^3)^2-4t^3}{(1+t^3)^2}=\bigg(\frac{1-t^3}{1+t^3}\bigg)^2.$$The division only requires $$1+t^3\ne 0$$, which in turn trivially follows from $$x^3+y^3=2xy\ne 0$$.

• Based on the proof of J.G., I propose another solution – Blind Sep 23 at 21:25

My solution is based on the idea of J.G.

If $$x=0$$ then $$y=0$$ and so $$1-xy=1^2$$. Now we consider the case $$x\ne 0$$. Observe that

$$1-xy=1-\frac{x^3y}{x^2}=\frac{x^2-x^3y}{x^2}=\frac{x^2+y^4-2xy^2}{x^2}=\frac{(x-y^2)^2}{x^2}=\left(\frac{x-y^2}{x}\right)^2.$$

I think that this is the shortest solution.

• To be complete, needs something along the lines of "or $x=0$, which means $y=0$ and $1-xy=1$ which is a perfect square" – aschepler Sep 24 at 2:42