I actually got this problem! Lemma 1. A rational number $t$ is in a form of $x-\frac{1}{x}$ with $x \in \mathbb{Q}$ if and only if $\sqrt{t^2+4}$ is rational. Proof of Lemma 1. If $t=x-\frac{1}{x}$, then $\sqrt{t^2+4} = x+\frac{1}{x}$, so we are done. If $\sqrt{t^2+4}$ is rational, then $t^2+4=q^2$, so $(q-t)(q+t)=4$, so $q+t=2x$ and $q-t=\frac{2}{x}$, we have $t=x-\frac{1}{x}$. Lemma 2. There are no solution in nonzero integers such that $x^4-y^4=z^2$. Proof of Lemma 2. Well-known lemma - just WLOG $x, y > 0$ and take the pair which has minimal value of $x+y$. Then use pythagorean triples to find a smaller pair. This is well-known so I will skip it here. Now back to the main problem. So by Lemma 1, it suffices to show that both $\sqrt{x^2+4}$ and $\sqrt{(4-x)^2+4}$ cannot be rational at the same time. It suffices to show that the midpoint of any rational points on the curve $y^2=x^2+4$ cannot have $2$ as its $x$-value. Now assume otherwise, and let $y=ax+b$ with $a, b \in \mathbb{Q}$ be the line passing through the two points. If $a=0$, we either have two points equal or reflection wrt $y$-axis, which forces the $x$ value of the midpoint to be $0$. We also check that $(2,2\sqrt{2})$ is not a rational point. Now by Vieta, $(ax+b)^2=x^2+4$ has roots adding up to $4$, so we have $b=-\frac{2(a^2-1)}{a}$. Plugging this in, we have $(x-2)^2 = \frac{8a^2-4}{a^2(a^2-1)}$. Clearly we can drop $a=1$. It now suffices to show that $\frac{2a^2-1}{a^2-1}$ cannot be a square of a rational. Let $a=\frac{p}{q}$, then we have $\frac{2p^2-q^2}{p^2-q^2}$, and the denominator and numerator are coprime. Now this forces $2p^2-q^2 = x^2$ and $p^2-q^2 = y^2$, so $p^2-y^2=q^2$ and $p^2+y^2=x^2$, giving $p^4-y^4=q^2x^2$. Since $q \not= 0$ and $p \not= 0$ and since $a \not= 1$, we now have $y \not = 0$. Since $\sqrt{2}$ is irrational, we also have $x \not=0$, and by our Lemma 2, we have a contradiction. If $2p^2-q^2=-x^2$ and $p^2-q^2=-y^2$, then we have $y^2-p^2=x^2$ and $y^2+p^2=q^2$, so $y^4-p^4=q^2x^2$. A similar contradiction follows, and we are done. $\blacksquare$

My bad,something completely wrong as pointed below

Here is my solution. WLOG, suppose that $x,y>0$ then let $x= \tfrac ab, y= \tfrac cd$ with $a,b,c,d \in \mathbb{N}, \; \gcd (a,b)= \gcd (c,d)=1$. The equation is equivalent to $$(x+y)(xy-1)=4xy \iff cd(a^2-b^2)+ab(c^2-d^2)=4abcd.$$From here, we obtain $ab \mid cd(a^2-b^2)$ but since $\gcd (a,b)=1$ so $ab \mid cd$. Similarly, we obtain $cd \mid ab$. Thus, $ab=cd$. Since $\gcd (a,b)= \gcd (c,d)=1$ so there exists pairwise relatively prime numbers $m,n,p,q$ such that $a=mp,c=mq, b=nq,d=np$. Therefore, the equation becomes $$\left( p^2+q^2 \right) \left( m^2-n^2 \right)= 4mnpq.$$If $2 \nmid mnpq$ then $8 \mid LHS$ but $RHS \equiv 4 \pmod{8}$, a contradiction. If $2 \mid m$ (or $2 \mid n$) then $2 \nmid npq$ (or $2 \nmid mpq$) so $LHS \equiv 2 \pmod{4}$ and $4 \mid RHS$, a contradiction. If $2 \mid pq$. WLOG, suppose that $2 \mid p$ then $2 \nmid qmn$. This follows that $2 \nmid p^2+q^2$. Also note that $\gcd (p^2+q^2,pq)= \gcd (m^2-n^2,mn)=1$ so $p^2+q^2=mn, m^2-n^2=4pq$. This follows $$\begin{aligned} 2 \left( \frac{m+n}{2} \right)^2 & = n^2+(p+q)^2, \\ 2x^2 &=z^2+w^2, \\ x^2 & = \left( \frac{z+w}{2} \right)^2+ \left( \frac{z-w}{2} \right)^2, \end{aligned}$$with $x= \tfrac{m+n}{2}, z=n, w=p+q$. Since $x, \tfrac{z+w}{2}, \tfrac{z-w}{2}$ are pairwise relatively prime numbers and $2 \nmid n$ so there exist $k,l \in \mathbb{N}$ with $k>l, \gcd (k,l)=1$ such that $$x= k^2+l^2, \frac{z+w}{2}=2kl, \frac{z-w}{2}=k^2-l^2 \; \text{or} \; \frac{z+w}{2}=k^2-l^2, \frac{z-w}{2}=2kl.$$We find $z=n= k^2-l^2+2kl$ and $\tfrac{m+n}{2}=x= k^2+l^2$ so $m= k^2+3l^2-2kl$. Hence, $$m^2-n^2=(k^2+3l^2-2kl)^2-(k^2-l^2+2kl)^2= 8l(l-k)(k^2+l^2) <0,$$a contradiction since $4pq=m^2-n^2>0$. Thus, there is no rational numbers $x,y$ such that $x- \frac 1x +y- \frac 1y=4$. $\blacksquare$

mathmaths wrote: $p^2+r^2=\frac {p^2q^2}{r}+q^2+4pq$ It should be $= \frac{p^2q^2}{r^{ { \color{red} 2}}}+q^2+4pq$. mathmaths wrote: $x\ge y$,then $p\ge r$,$RHS\ge q^2r+q^2+4pq=(r+1)q^2+4pq$,must be $p\ge q$,while $r\ge s$(else by the symmetric way we can get contradiction),but now $x+y<6,1/x+1/y>2,x+y-1/x-1/y<4$,contradiction If $p \ge q, r \ge s$ then shouldn't it be $\frac 1x+ \frac 1y <2$ instead of $>2$ ?

rkm0959 wrote: I actually got this problem! So by Lemma 1, it suffices to show that both $\sqrt{x^2+4}$ and $\sqrt{(4-x)^2+4}$ cannot be rational at the same time. It suffices to show that the midpoint of any rational points on the curve $y^2=x^2+4$ cannot have $2$ as its $x$-value. I'm confused here. What if there exists $x_0 \ne 4$ such that $\sqrt{x_0^2+4}$ and $\sqrt{(x_0-4)^2+4}$ are rational numbers ? The midpoints of those two points will be $\tfrac{x_0}{2}+\tfrac{x_0-4}{2}=x_0-2 \ne 2$ but they are still rational numbers. If the one above happens, then there exists $p,q$ such that $x_0= p- \tfrac 1p, x_0-4= q- \tfrac 1q$ so $p- \tfrac 1p+ \tfrac 1q - q =4$, which means $(x,y) = \left( p , \tfrac 1q \right)$.

Then $\sqrt{(4-x)^2+4}$ will be rational, so we can continue similarly.

Here's a brief sketch of my solution. The cool part is where we use infinite descent on a system of equations. (I mean, that's essentially rkm's Lemma 2, but actually, it's quite easy to see that it kills the problem. I'd also like to present my solution as its methodology revolves only around coprime numbers in factorizations.) Lemma 1. If $ab=cd$ and $(a,b)=(c,d)=1$, then there exist positive integers $p,q,u,v$ such that $a=pu$, $b=qv$ and $c=pv$, $d=qu$. (Proof: make $p,q,u,v$ equal to "$a\cap c$, $a\cap d$, $b\cap c$, $b\cap d$".) Lemma 2. If $a^2+b^2=c^2$, then one of $a,b$ is even, and if $b$ is even, then for coprime $m>n>0$, \[ a=m^2-n^2,\qquad b=2mn,\qquad c=m^2+n^2. \](Proof: see Wikipedia.) To finish, it's easy to see that by Lemma 1, a coprime numbers trick, parity discussion, and a $\frac{u+v}{2}$, $\frac{u-v}{2}$ substitution, the problem reduces to Lemma 3: Lemma 3. There don't exist positive integers $a,b,c,d$ satisfying $(a,b)=(c,d)=1$, $ab=cd$ and \[ a^2-b^2=c^2+d^2. \](Proof: use Lemma 1 to get $p^2(u^2-v^2)=q^2(u^2+v^2)$ for pairwise coprime $p,q,u,v$. It follows that $Ap^2=u^2+v^2$ and $Aq^2=u^2-v^2$ for $A\in\{1,2\}$ - by the way, the case $A=2$ corresponds to the case $A=1$ when switching $p,q$ and $u,v$. In both cases, we can use Lemma 2 to generate a similar quadruple $k,\ell,m,n$ whose product is smaller, so we get a contradiction by infinite descent.)

While I guess rkm's solution is unique, I feel like this can be solved with standard methods such as the one presented above pretty readily

This approach seems a little bit different than others, so... FSOC assume that the quantity is equal to $4$. Let $x = a/b$ and $y = c/d$ be reduced fractions. Then we have $$4 = \frac{a}{b} + \frac{c}{d} - \frac{b}{a} - \frac{d}{c} = \frac{a^2cd - b^2cd + c^2ab - d^2ab}{abcd} = 4.$$But then $a$ divides the numerator, i.e. it must divide $b^2cd$. But $\gcd(a,b) = 1$, so $a|cd$. Similarly, $b|cd$, $c|ab$, and $d|ab$. More specifically, we get $$ab|cd \quad \text{and} \quad cd|ab \implies ab = cd.$$By the Factor Lemma, write $a = k\ell, b = mn, c = kn, d = \ell m$ where $k,\ell, m, n$ are relatively prime. Then we get $$4 = \frac{k\ell}{mn} + \frac{kn}{\ell m} - \frac{mn}{k\ell} - \frac{\ell m}{kn} = \frac{(k^2 - m^2)(\ell^2 + n^2)}{k\ell mn}$$or $(k^2 - m^2)(\ell^2 + n^2) = 4k\ell mn$. Because $4 \not | \ell^2 + n^2$, we get $2|k^2-m^2$ or that $k$ and $m$ are both odd. But then $8|k^2-m^2$, so $2|mn$ and WLOG $\ell$ is even (so $n$ is therefore odd). We get in a similar fashion as before that $$4\ell n|k^2 - m^2 \quad \text{and} \quad km| \ell^2 + n^2$$which implies $$4\ell n = k^2 - m^2 \quad \text{and} \quad km = \ell^2 + n^2.$$Set $(k-m,k+m)=(2u,2v)$. Then we get $\ell n = uv$ and $u^2 - v^2 = \ell^2 + n^2$. Setting $\ell = AB, n = CD, u = AD, v = BC$ for relatively prime $A,B,C,D$ gives us $$A^2D^2 - B^2C^2 = A^2B^2 + C^2D^2 \implies (A^2 + C^2)(B^2 + D^2) = 2A^2D^2.$$In a similar fashion as before, we get WLOG $$A^2 + C^2 = 2D^2, \quad B^2 + D^2 = A^2.$$But then this implies that $C^2 + B^2 = D^2$ and $B^2 + D^2 = A^2$, which implies $$D^4 - B^4 = (D^2 - B^2)(D^2 + B^2) = (AC)^2$$which is impossible as it is well known $x^4 - y^4 = z^2$ has no solutions. Thus, we are done.

We will be using (without proof) the following well known lemma. Lemma: Suppose we have integers $a,b,c$ such that $a^2+b^2=c^2$ and $\gcd(a,b,c)=1$. Then there exist relatively prime integers $\alpha,\beta$, one even, one odd, such that \[\{a,b\}=\{2\alpha\beta,\alpha^2-\beta^2\}.\] Note that a rational number $a$ is of the form $x-1/x$ for some rational $x$ if and only if $a^2+4$ is the square of a rational. Suppose that we had rationals $x$ and $y$ satisfying the equation. Define \[\frac{m}{n}=x-\frac{1}{x}\]where $\gcd(m,m)=1$. By the above observation, we have that the quantities \begin{align*} m^2&+(2n)^2, \\ (4n-m)^2&+(2n)^2 \end{align*}are squares. Claim: We must have $m$ odd. Proof: Suppose for sake of contradiction that $m$ is even. Then, $\gcd(m/2,n)=\gcd(2n-m/2,n)=1$ and \begin{align*} (m/2)^2&+n^2, \\ (2n-m/2)^2&+n^2 \end{align*}are squares. Thus, by the lemma, we have integers $\alpha,\beta,\gamma,\delta$ satisfying $\gcd(\alpha,\beta)=\gcd(\gamma,\delta)=1$ and $\{\alpha,\beta\}\equiv\{\gamma,\delta\}\equiv\{0,1\}\pmod{2}$ and \[\frac{m}{2}=2\alpha\beta\quad\text{and}\quad n=\alpha^2-\beta^2\]and \[2n-\frac{m}{2}=2\gamma\delta\quad\text{and}\quad n=\gamma^2-\delta^2.\]This implies that \[\alpha^2-\beta^2=\gamma^2-\delta^2=\alpha\beta+\gamma\delta.\]Note that $\alpha\beta+\gamma\delta$ is even as both individual terms are even, but $\alpha^2-\beta^2$ is odd, so we have the desired contradiction. $\blacksquare$ Now since $m$ is odd, we have $\gcd(m,2n)=\gcd(4n-m,2n)=1$, so by the lemma, we have integers $\alpha,\beta,\gamma,\delta$ satisfying $\gcd(\alpha,\beta)=\gcd(\gamma,\delta)=1$ and $\{\alpha,\beta\}\equiv\{\gamma,\delta\}\equiv\{0,1\}\pmod{2}$ and \[m=\alpha^2-\beta^2\quad\text{and}\quad n=\alpha\beta\]and \[4n-m=\gamma^2-\delta^2\quad\text{and}\quad n=\gamma\delta.\]We will now show that there are no integers $\alpha,\beta,\gamma,\delta$ satisfying $\gcd(\alpha,\beta)=\gcd(\gamma,\delta)=1$ and $\{\alpha,\beta\}\equiv\{\gamma,\delta\}\equiv\{0,1\}\pmod{2}$, $\alpha\beta=\gamma\delta=n$, and \[\alpha^2-\beta^2+\gamma^2-\delta^2=4n.\]Indeed, under this formulation, the pairs $(\alpha,\beta)$ and $(\gamma,\delta)$ are symmetric, so by mod $4$ considerations, we may assume WLOG that $\alpha,\delta$ are odd and $\beta,\gamma$ are even. Let $p/q$ be the reduced form of $\alpha/\gamma=\delta/\beta$. Let $u=\alpha/p$ and $v=\delta/q$. Then, we have integers $p,q,u,v$ such that $\gcd(p,q)=\gcd(u,v)=1$, $p,u,v$ odd, $q$ even, and \[(p^2+q^2)(u^2-v^2)=4pquv.\]However $\gcd(4pq,p^2-q^2)=\gcd(u^2-v^2,uv)$, so in fact \[uv\mid p^2+q^2\]and \[4pq\mid u^2-v^2.\]Thus, for some $\varepsilon\in\{-1,1\}$, we have \[p^2+q^2=\varepsilon\cdot uv\]and \[u^2-v^2=\varepsilon\cdot 4pq.\]Letting $x=\tfrac{u+v}{2}$ and $y=\tfrac{u-v}{2}$, we have $\gcd(p,q)=\gcd(x,y)=1$, $q$ even, \[p^2+q^2=\varepsilon\cdot(x^2-y^2)\]and \[xy=\varepsilon\cdot pq.\]Replacing $x$ with $-x$ if need be, the problem reduces to the following claim. Claim: There do not exist integers $p,q,x,y$ such that $\gcd(p,q)=\gcd(x,y)=1$, $q$ even, \[p^2+q^2=x^2-y^2\]and \[xy=pq.\] Proof: It is easy to see that we have integers $A,B,C,D$ such that $\gcd(A,B)=\gcd(C,D)=1$, $D$ even, and $x=AC$, $p=BC$, $q=AD$, and $y=BD$. Plugging this in, we get \[(BC)^2+(AD)^2=(AC)^2-(BD)^2,\]or \[B^2(C^2+D^2)=A^2(C^2-D^2).\]We have $\gcd(A^2,B^2)=1$ and $\gcd(C^2-D^2,C^2+D^2)$, so $B^2=C^2-D^2$ and $A^2=C^2+D^2$. Thus, \[A^4-B^4=(2CD)^2,\]which has no solutions. $\blacksquare$ This completes the horrendous solution.

wrong....

Quote: Let $\left(\frac{1}{b}-\frac{b}{ac}\right)=k\in \{1,2,4,-1,-2,-4\}$ This quantity is not necessarily an integer.

Nice problem! Solved with Dimath27 Suppose that exists $x=\frac{p}{q}$ and $y=\frac{m}{n}$ with $p,q,m,n$ integers and $(p,q)=(m,n)=1$ so: $$\frac{p^2-q^2}{pq} + \frac{m^2-n^2}{mn}=4$$Both are fractions irreducibles, WLOG $p,q,m,n>0$ so $pq=mn \implies p=uv, q=wz, m=uw, n=vz$. Replacing we have: $$u^2v^2-w^2z^2+u^2w^2-v^2z^2=4uvwz \implies \frac{(u^2-z^2)(v^2+w^2)}{uvwz}=4$$WLOG $(w,v)=(u,z)=1$ so $uz \mid v^2 + w^2$ and $wv \mid u^2-z^2$: $$(\frac{v^2+w^2}{uz})(\frac{u^2-z^2}{wv})=4$$If $\frac{v^2+w^2}{uz}$ is even then $v,w$: odd so $uz$ is odd then $(\frac{u^2-z^2}{wv})$ is divisible by 4 this is contradiction. Then we have that $\frac{v^2+w^2}{uz}=1$ and $\frac{u^2-z^2}{wv}=4$, so $(\frac{u-z}{2})(\frac{u+z}{2})=wv$ exists $a,b,c,d$ such that: $$\frac{u-z}{2}=ab, \frac{u+z}{2}=cd, w=ac, v=bd$$$$\frac{v^2+w^2}{uz}=1 \implies b^2d^2+a^2c^2=(ab+cd)(cd-ab) \implies a^2(c^2+b^2)=d^2(c^2-b^2) \implies (\frac{a(c^2+b^2)}{d})^2=c^4-b^4$$So we have a solution of the equation $x^4-y^4=z^2$ this is a contradiction well-know. $\blacksquare$