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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.2012-05-05
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    Did you at least find trivial ones, like $x=y=0$ then $1-xy=1=1^2$ ?2012-05-05
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    @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.](http://meta.math.stackexchange.com/a/4114/242)2012-05-05
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    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.2012-05-05
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    @PatrickDaSilva Ok, I will open a meta discussion then. Comments like that have no place here.2012-05-05
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    @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.)2012-05-05
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    I think I do understand both your comments. Bill likes to welcome new users, and Patrick was right to mention that people who seek help should not just post a question without proper introduction. I think MSE should stop unregistered users to just post a question and vanish in thin air.2012-05-05
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    @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.2012-05-05
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    i am sure,only integers solutions are $x=y=1$ which satisfies our condition that $1-x*y=0$ is perfectly square,see this http://www.wolframalpha.com/input/?i=x%5E5%2By%5E5%3D2x%5E2*y%5E2 also $x=y=0$2012-05-05
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    How about $x=y=0$? And your answer does not look like a solution, you should post it as a comment next time. But nice remark, so $+1$!2012-05-05
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    yes it is also possible $x=y=0$2012-05-05
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    @dato You are sure???????2012-05-06
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    I will turn all this into comments to the question; it is not an answer.2012-05-06
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    @PatrickDaSilva I think you did not mean to be harsh to the OP, but you were by accident. Arrogant is a very strong and harsh word, and it does not fit in this situation. That's the main problem with your comment. If you were to change "very arrogant" to "considered poor form" then the comment is perfectly fine. I mean, you have a smily face in your comment, so hopefully the OP can see that you were not trying to be unkind, only trying to be helpful. But, sometimes people can get stuck on one word.2012-07-06

4 Answers 4

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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!2012-05-05
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    @PatrickDaSilva Thanks.2012-05-05
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    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?2012-05-05
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    or (x=y) and (x,y)<0,so infinity solution2012-05-05
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    @dato, the original question was prove $1-xy$ is a perfect square, so there is a proof required, not solutions to $x,y$.2012-05-05
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    aaa yes ,i see,just wanted to make sure in my thinking2012-05-05
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    Why is $u$ rational ?2012-07-22
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    @Belgi Because x is rational and y is rational, so will (x/y) be too.2014-02-24
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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.2012-05-06
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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.