If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square.

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

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

then $1-xy$ is a perfect square.

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.