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.

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    $\begingroup$ I didn't downvote you, but you didn't really show any work. For example, you refer to an irreducible "resultant polynomial". What specific polynomial are talking about? Why don't you at least specify the polynomial and provide work to justify your claim of that it's irreducible? Also, how does irredicubility of whatever polynomial you're talking about relate to the existence or non-existence of rational solutions? $\endgroup$
    – quasi
    Jan 3, 2017 at 7:54
  • $\begingroup$ @quasi If a polynomial is irreducible in a field, then no element of the field can satisfy the equation, isnt it? $\endgroup$
    – vidyarthi
    Jan 3, 2017 at 7:57
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    $\begingroup$ In one variable, yes, (assuming degree greater than 1). You still didn't identify the polynomial you're talking about. $\endgroup$
    – quasi
    Jan 3, 2017 at 7:59
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    $\begingroup$ Eisenstein is applicable to polynomials in a single variable over a suitable field $K$. More often than not we use it with $K=\Bbb{Q}$. You can use it with $K=\Bbb{Q}(y)$ with indeterminate $y$, but it doesn't seem to apply directly to your quadratic over $\Bbb{Q}(y)$ (what is the irreducible element of $\Bbb{Q}[y]$ you use here?). Anyway, even though this polynomial is likely irreducible in $\Bbb{Q}[x,y]$ that does not help you towards the conclusion that there would be no rational solutions. This is because a polynomial in several variables can have zeros and also be irreducible. $\endgroup$ Jan 3, 2017 at 11:00
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    $\begingroup$ The title of this question is (soon, was) a really, really nasty clickbait. $\endgroup$
    – tomasz
    Jan 3, 2017 at 15:16

6 Answers 6


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.

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    $\begingroup$ No, I don't think it's simple. Feel free to prove me wrong. In any case, you asked whethet your reasoning was correct. Hopefully, from my reply, you see that it's not. $\endgroup$
    – quasi
    Jan 3, 2017 at 8:20
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    $\begingroup$ I didn't say I could solve it. I said that I suspect that there are no solutions, and that if so, it would probably require methods from the theory of elliptic curves. I don't claim to be able to work out the details. $\endgroup$
    – quasi
    Jan 3, 2017 at 8:26
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    $\begingroup$ @vidyarthi no, just that quasi doesn't know (nor do i) $\endgroup$ Jan 3, 2017 at 9:26
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    $\begingroup$ @vidyarthi As I indicated in my first reply, irreducibility of your bivariate polynomial doesn't imply there are no rational solutions. But in any case, you claimed that your polynomial is irreducible "By Eisenstein". I don't see how Eisenstein's criterion is applicable in this context. Could you show your work for that part of your claim? $\endgroup$
    – quasi
    Jan 3, 2017 at 10:34
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    $\begingroup$ @vidyarthi, your exact question was, "Is my reasoning right?" This answers your question fully. You should accept some answer, whether this or another one. $\endgroup$
    – Wildcard
    Jan 11, 2017 at 23:06

@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.

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    $\begingroup$ "...so let's ask an elliptic curve expert, in this case Sage". Hehe. $\endgroup$
    – quasi
    Jan 3, 2017 at 13:40
  • $\begingroup$ Well, it's certainly a lot more expert than I am. $\endgroup$ Jan 3, 2017 at 14:22

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.

  • $\begingroup$ Your rationale for $b \neq 0$ is, IMO, not good. The fact that $b \neq 0$ follows eventually from the implicit premise of the problem that $x \neq 0$ and $y \neq 0$. You don't explain the intermediate steps, but IMO it's more likely you implicitly use $b \neq 0$ in the derivation of (2). $\endgroup$
    – user14972
    Jan 3, 2017 at 12:34
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    $\begingroup$ The point $(b,v)=(-1,2)$ lies on your curve (4). But that leads to $u=\infty$ which is not very helpful. If I'm understanding correctly what my computer tells me, this has order 4 so there ought to be another possibly more interesting rational point... $\endgroup$ Jan 3, 2017 at 12:48
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    $\begingroup$ Hmm. It looks as if the square of that point is $(b,v)=(0,0)$ so $u=0$ which means $x=y$ so we would be trying to solve $2x-2/x=4$ which is equivalent to $x^2-2x+1=2$ whose solutions are not rational. And the inverse is $(b,v)=(-1,-2)$ which is obviously no better than $(-1,2)$. $\endgroup$ Jan 3, 2017 at 13:01
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    $\begingroup$ And the elliptic curve has rank 0 and torsion group cyclic of order 4, so these points plus the equally useless one at infinity are all there are. So indeed it seems that the original equation has no rational solutions. (And quasi's answer needs amending to note that as well as $b=0$ being useless, so is $b=-1$ because it leads to $a=\infty$.) I do wonder whether there's a more elementary solution for this particular curve, though. $\endgroup$ Jan 3, 2017 at 13:05
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    $\begingroup$ @Gareth: Thanks for pointing out the need to exclude $b = -1$. Also, thanks very much for the analysis of the elliptic curve. $\endgroup$
    – quasi
    Jan 3, 2017 at 13:23

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


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.:(

  1. 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.
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    $\begingroup$ This answer would be much better if you fixed the formatting and didn't re-use the variables $x,y$, which were originally rationals, but later integers. $\endgroup$ Jan 11, 2017 at 19:53

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:


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:


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


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:


And thus:


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.


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