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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Carl Mummert, Theoretical Economist, GNUSupporter 8964民主女神 地下教會, D_S
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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

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

I think that this is the shortest solution.

  • 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.