$x-\frac1x+y-\frac1y=4\,$ has no rational solutions? Do there exist rational numbers $x,y$ such that $x-\frac1x+y-\frac1y=4?$
I think no, because the equation reduces to $\frac{x^2-1}{x}+\frac{y^2-1}{y}=4\implies x^2y+x(y^2-4y-1)-y=0$ in two variables is irreducible in the field $\mathbb{Q}$(Eisenstein) Is my reasoning right? Thanks beforehand.

Here is a sketch of a naive proof that it has no solutions (no elliptic curves).
We can assume x is positive, otherwise we would change x to -1/x. The same with y.
$x^2y+x(y^2-4y-1) -y =0$
$D=(y^2-4y-1)^2+4y^2$ should be a square of a rational number.
If $y=\frac{m}{n}$ (assuming m and n are co-prime), then multiplying by $n^4$ we get
$(m^2-4mn-n^2)^2 + 4m^2n^2 = v^2$, where $v$ is natural.


*

*If either $m$ or $n$ is even, than $2mn$ and $m^2-4mn-n^2$ are co-prime.
Hence 
$2mn=2xy$
and $m^2-4mn-n^2 = x^2-y^2$ for some natural $x$ and $y$.
Or the last one is $m^2-2mn +n^2-2n^2 = x^2+2xy + y^2-2y^2$ since $mn=xy$.


$(m-n)^2-2n^2=(x+y)^2-2y^2$, or $(m-n)^2 - (x+y)^2= 2 (n^2-y^2)$
$(m-n-x-y)(m-n+x+y)=2(n-y)(n+y)$.
Using the fact that numbers $m-n-x-y$ and $m-n+x+y$  are both odd or even and the same with numbers $n-y$ and $n+y$ we obtain that all of those numbers are even. 
Making similar step infinitely many times we obtain that both sides are divisible by infinite number of powers of 2, hence both sides are 0 which is impossible.
UPD: this part is wrong. I will try to change it later. Sorry.:(


*If both $m$ and $n$ are odd, then $mn=x^2-y^2$ and $m^2-4mn-n^2 = 8xy$ for some natural $x$ and $y$.
Then $5m^2-(2m+n)^2$ is divisible by 8 for some odd $m$ and $n$, which is impossible since odd square always has remainder 1.

