This question is asked in the mathematics contest for grade 9.

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_S Oct 3 at 20:17

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  • 2
    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
  • 1
    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
up vote 16 down vote accepted

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


I think that this is the shortest solution.

  • 1
    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

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