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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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    @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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    @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.