TITLE: A Diophantine Equation
QUESTION [8 upvotes]: Does the Diophantine equation $2(x - \frac{1}{x}) = y - \frac{1}{y}$ have only trivial rational solutions, i.e, $x=\pm1, y = \pm1$?

REPLY [7 votes]: 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.