$x-\frac1x+y-\frac1y=4\,$ 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.
 A: Consider the equation 
$$x-\frac1x+y-\frac1y=4 \tag{1}$$
Let $a = x + y$ and let $b = -xy$. Then
$$x-\frac1x+y-\frac1y = a + \frac{a}{b}$$
$$(x - y)^2 = a^2 + 4b$$
It follows that equation (1) has a rational solution $(x,y)$ if and only if there exist rational numbers $a,b,u$ such that
$$a+\frac{a}{b} = 4 \tag{2}$$
$$a^2 + 4b = u^2 \tag{3}$$
Note that equation (2) implies $b \ne 0$.

Solving equation (2) for $a$ yields 
$$a = \frac{4b}{b+1}$$
which gives another restriction, namely $b \ne -1$.
Replacing $a$ by $((4b)/(b+1))$ in equation (3), and then simplifying the result, we get
$$v^2 = b^3 + 6b^2 + b \tag{4}$$
where $v = (u/2)(b+1)$.

It follows that equation (1) has a rational solution $(x,y)$ if and only if equation (4) has a rational solution $(b,v)$ with $b \ne 0$, $b \ne -1$.

Now equation (4) is the equation of an elliptic curve, so the question becomes, does the elliptic curve for equation (4) have a rational point $(b,v)$ with $b \ne 0$, $b \ne -1$?

I'll let the elliptic curve experts take it from here.
A: WLOG, suppose that $x,y>0$ (otherwise $x'=-\dfrac{1}{x}$),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 $$\dfrac{a}{b}-\dfrac{b}{a}+\dfrac{c}{d}-\dfrac{d}{c}=4 \iff cd(a^2-b^2)+ab(c^2-d^2)=4abcd.$$
From here, since $ab|(a^2-b^2)cd+(c^2-d^2)ab$,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 $x- \frac 1x +y- \frac 1y=4$.         $\blacksquare$
A: 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.

A: No, your reasoning is incorrect.
As a simple example, consider the equation $x^2 + y^2 - 1 = 0$. The LHS is irreducible over $\mathbb Q$, but the equation has lots of rational solutions (e.g., $(x,y) = (3/5,4/5)$).
As another example, take your equation and replace "4" by "3". With that "minor" change, you get the rational solution $(x,y) = (2,2)$, and yet
the corresponding polunomial in $x,y$ is irreducible over $\mathbb Q$.
As for your actual equation, I suspect there are no rational solutions, but if so, a proof will probably require methods from the theory of elliptic curves.
A: @quasi's answer establishes that the original question is basically equivalent to this one: "Are there any rational points on $v^2=b^3+6b^2+b$?" As quasi remarks, the solution $(b,v)=(0,0)$ doesn't lead to a rational solution to the original equation; neither (this is not mentioned at present in quasi's answer) does any solution with $b=-1$ because these correspond to $a=\infty$ in quasi's reparameterization of the original equation.
Well, this is an elliptic curve so let's ask an elliptic curve expert, in this case Sage.
In[2]: E = EllipticCurve([0,6,0,1,0]); E
Out[2]: Elliptic Curve defined by y^2 = c^3 + 6*x^2 + x over Rational Field

In[3]: E.analytic_rank_upper_bound()
Out[3]: 0

In[4]: E.prove_BSD()
Out[4]: []

In[5]: E.torsion_points()
Out[5]: [(-1 : -2 : 1), (-1 : 2 : 1), (0 : 0 : 1), (0 : 1 : 0)]

Here, Out[3] is telling us that the analytic rank of this curve (conditional on GRH) is definitely zero. In fact the curve is equivalent to Cremona's 32a4 which is known to have analytic rank 0, but I'm ignorant and don't know how to get Sage to report this. Out[4] is telling us that it can prove Birch-Swinnerton-Dyer for this curve with no exceptions, so the rank of the curve's group is also 0. And Out[5] is telling us that the only torsion points on the curve (hence, since the rank is zero, the only points) are the point at infinity, the point $(b,v)=(0,0)$, and the points $(b,v)=(-1,\pm2)$. None of these yields a rational solution to the original equation.
