What is the rational rank of the elliptic curve x^3 + y^3 = 2? In Unsolved Problems in Number Theory, 3rd edition, section D1, 2004, R. K. Guy says, "Andrew Bremner has computed the rational rank of the elliptic curve x^3 + y^3 = Taxicab(n) as equal to 2, 4, 5, 4 for n = 2, 3, 4, 5, respectively." Since Taxicab(1) = 2, my question can be restated:
What is the rational rank of the elliptic curve x^3 + y^3 = Taxicab(1)?
 A: To add to Abhinav's answer: the fact that $x^3+y^3=2$ has no solutions
other then $x=y=1$ is attributed by Dickson to Euler himself:
see Dickson's History of the Theory of Numbers  (1920) Vol.II, Chapter XXI
"Numbers the Sum of Two Rational Cubes", page 572.  The reference
(footnote 182) is
"Algebra, 2, 170, Art. 247; French transl., 2, 1774, pp. 355–60;
Opera Omnia, (1), I, 491".  In the next page Dickson also refers to
work of Legendre that includes this result (footnote 184: "Théorie
des nombres, Paris, 1798, 409; ...").
This result is actually easier than the $n=3$ case of Fermat,
because the curve has a $2$-torsion point, so only a $2$-descent
is required, and here one soon finds that there are only two
rational points (the second being the point at infinity
$(x:y:1) = (1:-1:0)$).  [It also happens that this curve has
conductor $36$, while it's known that the smallest conductor
of an elliptic curve of positive rank is $37$; but that's
using a huge modern cannon to dispatch a classical fly.]
An easy way to bring the curve $x^3+y^3=2$ into Weierstrass form
is to factor $x^3+y^3 = (x+y)(x^2-xy+y^2)$ and substitute
$(x,y)=(r+s,r-s)$ to get $r^3+3rs^2=1$.  Thus $s^2=(1-r^3)/(3r)$,
so $3r(1-r^3)$ is to be a square.  Now substitute $r=3/X$ to get
$s^2 = (X^3-27)/(3X)^2$, or equivalently $Y^2=X^3-27$ where $y=3sX$.
The solution $(x,y)=(1,1)$ corresponds to $(r,s)=(1,0)$ and then
$(X,Y)=(3,0)$, which we recognize as a 2-torsion point because $Y=0$.
A: It's isomorphic to $y^2 = x^3 - 27$, which has rank $0$.
