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Prove that if $x, y$ are rational numbers and

$$ x^5 +y^5 = 2x^2y^2$$

then $1-xy$ is a perfect square.

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Welcome to MSE =) Have you tried something? Shooting questions like in math books is very arrogant if you do not show that you have tried something and only want us to give you the answer. If you just have no idea where to start then perhaps you could at least admit that, so that we could give tips/hints instead of providing a full answer. – Patrick Da Silva May 5 '12 at 19:19
@PatrickDaSilva Could you please edit the comment to remove the sentence containing the phrase "very arrogant". That is not a nice way to welcome new users. Please see the discussion here. – Bill Dubuque May 5 '12 at 19:36
I was not saying that Jona was arrogant, but reminding him that those types of questions on MSE are indeed arrogant. But I must admit I wasn't very smooth after re-reading my comment. Apologies here. No edits. – Patrick Da Silva May 5 '12 at 19:42
@Bill : Was I really this much arrogant? You speak like I did something really wrong, but I was suggesting Jona to comment about his question to begin interaction with OP, I never insulted him in any manner. And I don't see where I could start discussing with you in meta (I mean technically, I need to find the page of the discussion to discuss with you.) – Patrick Da Silva May 5 '12 at 19:49
@Patrick: I don't react to it quite so strongly as Bill, but your comment really is phrased pretty offensively. A simple What have you tried? would suffice. – Brian M. Scott May 5 '12 at 20:35

I would be so tempted to divide by $x^2 y^2$, so I would consider the following cases:

Case a:

if $x=0$ then that implies $y=0$.

Case b:

if $y=0$ then that would imply $x=0$

Case c:

$x \neq 0, y \neq 0$ Thus dividing by $x^2y^2$ will be legal

$$ \begin{align*} \frac{x^3}{y^2} - 2 + \frac{y^3}{x^2} = 0 \\ x\left(\frac{x}{y}\right)^2 - 2 +y\left(\frac{y}{x}\right)^2 = 0 \end{align*} $$

Substitute $u=\left(\frac{x}{y}\right)^2$

$$ \begin{align*} xu - 2 + \frac{y}{u} = 0\\ xu^2-2u+y=0 \end{align*} $$

This is a quadratic in $u$ and since $x$ and $y$ are rationals, $u$ is rational. The discriminant is $4(1-xy)$, which has to be a perfect square for $u$ to be rational.

Thus $1-xy$ is a perfect square.

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Very good! I like this answer. +1! – Patrick Da Silva May 5 '12 at 19:48
@PatrickDaSilva Thanks. – Kirthi Raman May 5 '12 at 19:49
from here ,it is clear that xy can't be more then 1 right?so it means only possible solutions are x=y=1 right? – dato datuashvili May 5 '12 at 20:09
or (x=y) and (x,y)<0,so infinity solution – dato datuashvili May 5 '12 at 20:10
@dato, the original question was prove $1-xy$ is a perfect square, so there is a proof required, not solutions to $x,y$. – Kirthi Raman May 5 '12 at 20:12

Hint $ $ Clear if $\rm y\!=\!0;$ else $\rm 0 = x^6\!-2x^3y^2\!+xy^5\! = (x^3\!-\!y^2)^2\!- y^4(1\!-\!xy)\Rightarrow1\!-\!xy = \smash[b]{\bigg(\!\!\dfrac{x^3}{y^2}\!-\!1\!\bigg)^2}$

Remark $\ $ Alternatively, instead of explicitly completing the square as I do above, one could, essentially equivalently, use the squareness of the discriminant from the quadratic formula

$$\rm f(X) = a\:X^2 + b\: X + c = 0\ \Rightarrow\ (2a\: X + b)^2 = b^2 - 4ac = discriminant(f) $$

Now $\rm\:f(X) = x\:X^2 - 2y^2\:X + y^5\in\mathbb Q[X]\:$ has root $\rm\:X = x^2\in\mathbb Q\:$ and the RHS of the above specializes precisely to ($4$ times) the middle equation in the hint. Thus, armed with the knowledge that a quadratic polynomial $\rm\in \mathbb Q[X]$ with rational root necessarily has square discriminant $\rm\in \mathbb Q^2,\:$ when seeking to prove that an expression is a square, it is natural to seek to represent it as a discriminant $\rm\:d,\:$ or $\rm\:d\:\! q^2.\:$ This is essentially what KV Raman does in his answer (which employs a scaled version of the above polynomial). Nice job KV.

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My +1 and Thanks. – Kirthi Raman May 6 '12 at 18:28

Here is a proof I really like: Using "polar coordinates", we have $x=r \cos\theta$ and $y=r\sin\theta$, where $r\geq 0$ and $\theta\in [0,2\pi)$. If $xy=0$, then $1-xy=1$ which is clearly a perfect square of a rational number. Therefore, we assume that $xy\neq 0$, which implies that $x^5 +y^5\neq 0$ thanks to $x^5 +y^5 = 2x^2y^2$. In particular, we have $r>0$ and $\cos^5\theta+\sin^5\theta\neq 0$. Therefore, from the equation $x^5 +y^5 = 2x^2y^2$, we get $$r=\frac{2\cos^2\theta \sin^2\theta}{\cos^5 \theta+\sin^5\theta}.$$ Hence $$1-xy=1-r^2\cos\theta\sin\theta=1-\left(\frac{2\cos^2\theta\sin^2\theta}{\cos^5\theta+\sin^5\theta}\right)^2\cos\theta\sin\theta=1-\frac{4\cos^5\theta\sin^5\theta}{(\cos^5\theta+\sin^5\theta)^2}=\frac{(\cos^5\theta-\sin^5\theta)^2}{(\cos^5\theta+\sin^5\theta)^2}=\left(\frac{\cos^5\theta-\sin^5\theta}{\cos^5\theta+\sin^5\theta}\right)^2 =\left(\frac{(r\cos\theta)^5-(r\sin\theta)^5}{(r\cos\theta)^5+(r\sin\theta)^5}\right)^2=\left(\frac{x^5-y^5}{x^5+y^5}\right)^2$$ which is a perfect square of a rational number, since $x$ and $y$ are rational by assumption.

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The initial equation can be manipulated such that one side is exactly $1-xy$. The other side ends up being a rather ugly fraction. I leave as an exercise for the reader the proof that it is a square.

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