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

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.



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 $$(y + 1/y)^2 = (y1/y)^2 + 4 = 4(x^2 + 1/x^2  1).$$ In particular, if $u = x^2 + 1/x^2$, then $u1$ is a square. Similarly, $u2 = (x1/x)^2$ and $u+2 = (x+1/x)^2$ are both squares. If follows that the product of $u1$, $u2$, and $u+2$ is also a square, leading to a point on the curve $$v^2 = (u2)(u1)(u+2).$$ Now Fermat proved many years ago that there do not exist four rational squares in a nontrivial 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. 


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. 


This question is actually the first example, in a different formulation, of the following problem: Find two integer rightangled 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 $B=\alpha 2mn , A = \alpha (m^2n^2)$ and $B=\beta 2pq , NA=\beta (p^2q^2)$ giving $\frac{NA}{B}=\frac{p^2q^2}{2pq}=N\frac{A}{B}=N\frac{m^2n^2}{2mn}$ and defining $X=m/n$ and $Y=p/q$ gives $N(X1/X )=(Y1/Y)$ 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 $B=12, A=5$. 

