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Does the Diophantine equation $2(x - \frac{1}{x}) = y - \frac{1}{y}$ have only trivial rational solutions, i.e, $x=\pm1, y = \pm1$?

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Looks like homework, voting to close. –  Igor Rivin Sep 30 '11 at 12:28
    
It is not a homework! If anybody can solve it in a simple way, please let me know? –  ksj03 Sep 30 '11 at 12:39
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4 Answers

Yes, these are the only points.

The numerator is $2 x^{2} y - x y^{2} + x - 2 y = 0$ which is genus 1 curve.

The Weierstrass model is: $y^2 = x^3 + 14x^2 - 4x - 56$ which turns out to be rank zero and the torsion points are your solutions.

To compute the rank from the Weierstrass model I used sage with the command E.gens().

I suppose sage uses "mwrank" for finding the generators and it can be used as standalone program.

Update One of the easiest ways to compute the Weierstrass model and the maps is the following Magma code that can the tried in Magma Calculator - I didn't use this for the answer.

  K<x,y>:=AffineSpace(Rationals(),2);
  C:=Curve(K,2*x^2*y - x*y^2 + x - 2*y);
  Genus(C);
  cl:=ProjectiveClosure(C);
  point:=cl!([1,1,1]);
  E,map1:=EllipticCurve(cl,point);
  mapinv:=Inverse(map1);
  aInvariants(E);
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Perhaps the OP (and the rest of us) would like to know how you computed the rank... –  Igor Rivin Sep 30 '11 at 12:58
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@ksj03 In sage I wrote "E.torsion_points()" then I mapped them to the original curve via the maps I computed first. –  joro Sep 30 '11 at 13:06
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As often happens with elliptic curves coming from natural-looking Diophantine equations, this curve happens to have a small enough conductor (namely $24$) that one can look it up in the Cremona or even Tingley ("Antwerp") tables. Indeed here the conductor is so small that the rank must vanish because the first case of positive rank is $37$. Even not knowing that, the presence of $2$-torsion (and vanishing of Sha[2]) means that Fermat's descent method suffices. But given the availability of tools like mwrank one rarely works out the details by hand nowadays. –  Noam D. Elkies Sep 30 '11 at 15:20
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Why are the Antwerp tables called Tingley ? –  Chandan Singh Dalawat Oct 1 '11 at 4:10
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@Chandan : The origins of the Antwerp tables is explained here : warwick.ac.uk/~masgaj/papers/ecdb_ants.pdf It seems that it was Swinnerton-Dyer who initiated the search. –  François Brunault Oct 1 '11 at 13:25
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Since at least one answer mentions the historical pedigree of this particular problem to some work in the 80's, I thought I would mention that the problem goes back much further. As various people have remarked in the comments, one first observes that

$$(y + 1/y)^2 = (y-1/y)^2 + 4 = 4(x^2 + 1/x^2 - 1).$$

In particular, if $u = x^2 + 1/x^2$, then $u-1$ is a square. Similarly, $u-2 = (x-1/x)^2$ and $u+2 = (x+1/x)^2$ are both squares. If follows that the product of $u-1$, $u-2$, and $u+2$ is also a square, leading to a point on the curve $$v^2 = (u-2)(u-1)(u+2).$$

Now Fermat proved many years ago that there do not exist four rational squares in a non-trivial arithmetic progression. This is the same as asking that $1$, $1+r$, $1+2r$, and $1+3r$ are all squares, leading to a point on the curve $$s^2 = (1+r)(1+2r)(1+3r).$$ Yet this is the same equation as the one above, with $u = 6r+4$ and $v = 6s$. Like in all cases of elliptic curves with rational $2$-torsion and trivial $2$-part of Sha, (edit and rank zero) there is an elementary proof that there are no rational points by infinite descent. However, these arguments, although classical, can sometimes be a little tricky to find. There are, however, several articles one can find on the web giving "elementary" proofs of Fermat's theorem, for example:

http://www.maths.mq.edu.au/~alf/SomeRecentPapers/183.pdf

gives an argument which involves finding pairs $A$, $B$ with $A^2+B^2$ and $A^2+4B^2$ are both squares, then using pythagorean triples and descent (something that came up in the comments). Google searches will reveal many more roughly equivalent arguments, which all essentially show that the elliptic curve above has rank zero, and all the rational torsion points correspond to "trivial" or "degenerate" solutions.

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Hint: Try expressing y in terms of x using the quadratic equation, and check if the solution will allow for integer values of y, when integer values of x other than $\pm1$ are substituted.

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What, no Weierstrass model? –  Igor Rivin Sep 30 '11 at 12:57
    
The discriminant of the quadratic is a 4th degree polynomial in $x$, and you are asking if this polynomial can ever be a square rational number. Is this really simpler than the original problem? –  J.C. Ottem Sep 30 '11 at 13:17
    
Is an argument of the following kind sufficient? $ y= x-\frac{1}{x} \pm \sqrt{x^2 + \frac{1}{x^2} -1}$ $ \Delta = x^2 + \frac{1}{x^2} -1 = \frac{(x^2 - \frac{1}{2})^2 + \frac{3}{4}}{x^2} = \frac{4(x^2 - \frac{1}{2})^2 + 3}{4x^2} = \frac{(2x^2 - 1)^2 + 3}{4x^2}$ where the numerator can only be a square if (2x^2 - 1)^2 = 1, which occurs only if $x = \pm 1$ so as to not divide by zero. –  user6204 Sep 30 '11 at 13:46
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Isn't the OP asking about rational solutions? –  joro Sep 30 '11 at 13:49
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it looks that appropriate descent should work for a^4-a^2b^2+b^4=c^2, similar to Euler's x^4+y^4=z^2. This descent has explanation via elliptic curves, so maybe opposite translation from elliptic curves to elementary language leads to some descent argument. –  Fedor Petrov Sep 30 '11 at 22:32
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This question is actually the first example, in a different formulation, of the following problem:

Find two integer right-angled triangles with a common base and altitudes in the integer ratio $N:1$,

which was considered by the late John Leech (in the 1980s I think).

We have to find integer solutions to

$B^2+A^2=C^2 \hspace{2cm} B^2+(NA)^2=D^2$

We can express $B=\alpha 2mn , A = \alpha (m^2-n^2)$ and $B=\beta 2pq , NA=\beta (p^2-q^2)$ giving

$\frac{NA}{B}=\frac{p^2-q^2}{2pq}=N\frac{A}{B}=N\frac{m^2-n^2}{2mn}$

and defining $X=m/n$ and $Y=p/q$ gives

$N(X-1/X )=(Y-1/Y)$

The original form of the problem is also just that of Euler's Concordant Numbers which is related to the elliptic curve

$V^2=U(U+1)(U+N^2)$

which has $3$ torsion points of order $2$ when $U=0,-1,-N^2$ and $4$ torsion points of order $4$ when $U=\pm N$, all of which give the trivial answer or lead to undefined quantities.

The first curve of rank greater than $0$ occurs for $N=7$ leading to $X=3/2, Y=6$ or $B=12, A=5$.

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