$x-\frac1x+y-\frac1y=4\,$ has no rational solutions?

Question

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.

Answer 1

Doesn't solve the problem, but I thought this was an interesting equivalence.

Your question is equivalent to the the assertion:

There are no integers $a,b$ with $b\neq 0$ such that $a^2+b^2$ and $a^2+2b^2$ are both perfect squares.

Part 1: A solution to your equation gives a solution to mine

If $x-\frac{1}{x}+y-\frac{1}{y}=4$ for rational $x,y$, then there is a pair of nonzero integers $a,b$ such that $a^2+b^2$ and $a^2+2b^2$ are perfect squares.

If $x=\frac{u}{v}$ and $y=\frac{w}{z}$ then the equation is equivalent to:

$$(uz+wv)(uw-zv)=4uvwz$$

Now, if any prime power $p^k$ is a factor of $u$, then $p$ must be a factor of one or both of $uz+wv$ or $uw-zv$. Since $u,v$ are relatively prime, this means that $p$ must be a factor of $w$ or $z$, and, since $w,z$ are relatively prime, this means that $p^k$ must be a factor of $w$ or $z$. This means that we can write $u=u_0u_1,v=v_0v_1,w=u_0v_0,z=u_1v_1$ where $u_0=\gcd(u,w), u_1=\gcd(u,z),v_0=\gcd(v,w),v_1=\gcd(v,z)$. We get that $u_0,u_1,v_0,v_1$ pairwise relatively prime and:

$$u_0v_1(u_1^2+v_0^2)\cdot u_1v_0(u_0^2-v_1^2)=4u_0^2u_1^2v_0^2v_1^2$$

So you need a solution of $$(u_1^2+v_0^2)(u_0^2-v_1^2)=4u_0u_1v_0v_1$$

Now, since these are pairwise relatively prime, $u_1^2+v_0^2$ cannot be divisible by $4$. That means $u_0^2-v_1^2$ is even, so $u_0,v_1$ must both be odd, and thus $u_0^2-v_1^2$ is divisible by $8$, and thus one of $u_1,v_0$ must be even.

So we have that $\gcd(4u_1v_0,u_1^2+v_0^2)=1$ and $\gcd(u_0v_1,u_0^2-v_1^2)=1$, and hence that $$4u_1v_0=\pm(u_0^2-v_1^2)\\u_0v_1=\pm(u_1^2+v_0^2)$$

Now, if we have a solution to this equation, then we can find a solution with all the $u_i,v_i$ positive. So we want a solution to:

$$4u_1v_0=u_0^2-v_1^2\\u_0v_1=u_1^2+v_0^2$$

Substituting $v_1=\frac{u_1^2+v_0^2}{u_0}$ into the first, and expanding, we get:

$$4u_1v_0u_0^2=v_0^4-(u_1^2+v_0^2)^2$$

This is a quadratic equation in $u_0^2$ so we get:

$$u_0^2=2u_1v_0 +\sqrt{4u_1^2v_0^2+(u_1^2+v_0^2)^2}$$

Since one of $u_1,v_0$ is even, and they are relatively prime, we have that $(2u_1v_0,u_1^2+v_0^2)$ must be the legs of a primitive Pythagorean triple, and thus $2u_1v_0=2pq, u_1^2+v_0^2=p^2-q^2$ for some relatively prime pair $p,q$.

Since $u_1v_0=pq$, we can write $p=p_1p_2,q=q_1q_2,u_1=p_1q_1,v_0=p_2q_2$. Then you get:

$$u_1^2+v_0^2=p_1^2q_1^2+p_2^2q_2^2=p_1^2p_2^2-q_1^2q_2^2$$

And thus:

$$p_1^2(p_2^2-q_1^2)=q_2^2(p_2^2+q_1^2)$$

Now, $D=\gcd(p_2^2-q_1^2, p_2^2+q_1^2)=1$ or $2$.

If $D=1$ then $p_2^2-q_1^2=q_2^2$ and $p_2^2+q_1^2=p_1^2$, thus: $a=q_2$ and $b=q_1$ gives $a^2+b^2=p_2^2$ and $a^2+2b^2=p_1^2$.

If $D=2$ then $p_2^2-q_1^2=2q_2^2$ and $p_2^2+q_1^2=2p_1^2$. But this means that $p_2^2=p_1^2+q_2^2$ and $q_1^2=p_1^2-q_2^2$, and then $a=q_1,b=q_2$ gives the solution.

Part 2: A solution to my equations yields a solution to your equation

This is just reversing Part 1.

Assume $a^2+b^2=c^2$ and $a^2+2b^2=d^2$. Note that $$\begin{align}(dc-ab)(dc+ab)&=d^2c^2-a^2b^2\\ &=a^4+2a^2b^2 +2b^4\\ &=b^2(a^2+2b^2)+a^2(a^2+b^4)\\ &=b^2d^2+a^2c^2\end{align}\tag{1}$$

Let $$x=\frac{bd(cd+ab)}{ac(cd-ab)}\\y=\frac{ac(cd+ab)}{bd(cd-ab)}$$

Then you'll get, with help from (1) that: $$x+y=\frac{(cd+ab)^2}{abcd}$$

and $$\frac{1}{x}+\frac{1}{y}=\frac{(cd-ab)^2}{abcd}$$

So $$x-\frac{1}{x}+y-\frac{1}{y}=4$$

Part 3: Another equivalent

The requirement that $a^2+b^2$ and $a^2+2b^2$ is equivalent to the requirement that $c^4-b^4$ is a perfect square for some relatively prime $b,c$ with $b$ even (and $b\neq 0$ in both sides.)

If $c^4-b^4=k^2$ then $c^2-b^2$ and $c^2+b^2$ are relatively prime and hence squares, and hence $a^2=c^2-b^2$ and $c^2=a^2+b^2$ and $c^2+b^2=a^2+2b^2$ are both squares.

Answer 2

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.

Answer 3

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.

Answer 4

@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.

Answer 5

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 $x- \frac 1x +y- \frac 1y=4$. $\blacksquare$

Answer 6

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.

1. 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.:(

2. 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.