Korean Mathematical Olympiad

Suppose that there are rational numbers $x$ and $y$ such that $x-\frac{1}{x}+y-\frac{1}{y}=4$. Since $\left(-\frac{1}{x},y\right)$, $\left(x,-\frac{1}{y}\right)$ and $\left(-\frac{1}{x},-\frac{1}{y}\right)$ are solutions of the equation, we can assume that $x>0$ and $y>0$.

Letting $u=xy$, we get $x+y=\frac{4xy}{xy-1}=\frac{4u}{u-1}$. Note that $u>0$ and $u\ne 1$. Now the quadratic equation $$T^2-\frac{4u}{u-1}T+u=0$$ has rational solutions $x$ and $y$. So the discriminant ${\left(\frac{4u}{u-1}\right)}^2-4u$ is a square of a rational number. Then $$4u^2-u(u-1{)}^2=u(6u-u^2-1)$$ is a square of a rational number. Let $u=\frac{q}{p}$ with distinct positive integers $p$ and $q$ such that $(p,q)=1$. From $$u(6u-u^2-1)=\frac{q(6pq-p^2-q^2)}{p^3}=\frac{1}{p^4}\cdot pq(6pq-p^2-q^2),$$ we know $pq(6pq-p^2-q^2)$ is a square of an integer. Because $p,q$ and $6pq-p^2-q^2$ are pairwise relatively prime, there are positive integers $s,t$ and $w$ such that $$p=s^2,q=t^2,\text{and}{w}^2=6pq-p^2-q^2.$$ Thus $w^2=6s^2{t}^2-s^4-t^4=(2st{)}^2-(s^2-t^2{)}^2$. Note that $(s,t)=1$ and $s\ne t$. Without loss of generality assume $s>t>0$. Assume that $(s,t,w)$ is the positive integer solution of $w^2=6s^2{t}^2-s^4-t^4=(2st{)}^2-(s^2-t^2{)}^2$ which makes $s+t$ minimum. Because $2st$ is even in the Pythagorean triple $\{w,s^2-t^2,2st\}$, we know $s^2-t^2$ is even and $s$ and $t$ are odd. So $\left(\frac{s^2-t^2}{2},st\right)=1$. Then $${\left(\frac{w}{2}\right)}^2+{\left(\frac{s^2-t^2}{2}\right)}^2=(st{)}^2$$ gives a primitive Pythagorean triple $\left(\frac{w}{2},\frac{s^2-t^2}{2},st\right)$. So there are positive integers $m$ and $n$ such that $(m,n)=1$, $s^2-t^2=4mn$ and $st=m^2+n^2$. We can assume that $m$ is even and $n$ is odd. From the equality $s^2-t^2=(s+t)(s-t)=4mn$, there are positive integers $A,B,C,D$ which are pairwise relatively prime satisfying $$s+t=2AB,s-t=2CD,m=AC,n=BD.$$ By plugging these to $st=m^2+n^2$, we have $2A^2{B}^2=(A^2+D^2)(B^2+C^2)$. Then we know $C$ is even from the assumption that $m$ is even. Note that $A,B,D$ are odd. Now from the equality $2A^2{B}^2=(A^2+D^2)(B^2+C^2)$, we get $A^2=B^2+C^2$, $2B^2=A^2+D^2$. Then from $A^2=B^2+C^2$, we get $A=a^2+b^2$, $B=a^2-b^2$, $C=2ab$ with

positive integers $a$ and $b$ such that $(a,b)=1$ and $a>b>0$. Then $2B^2=A^2+D^2$ becomes $a^4+b^4-6a^2{b}^2=D^2$. Now $(2D{)}^2=6(a+b{)}^2(a-b{)}^2-(a+b{)}^4-(a-b{)}^4$. We have a contradiction from the minimality of $s+t$ since $s+t=2AB=2B(a^2+b^2)>2a=(a+b)+(a-b)>0$. Therefore, there are no rational numbers $x$ and $y$ such that $x-\frac{1}{x}+y-\frac{1}{y}=4$.