When is $(-1+\sqrt[3]{2})^n$ of the form $a+b\sqrt[3]{2}$? When is $(-1+\sqrt[3]{2})^n$ of the form $a+b\sqrt[3]{2}$ ($n$ being an integer) , i .e.,  when does $(-1+\sqrt[3]{2})^n$ not have a non-zero term in $\sqrt[3]{4}$. As you might have noticed, I'm interested in solving the diophantine equation $x^3-2y^3=1$ using this specific method. Is there any way I can use Skolem's p-adic method here?
 A: An account of Skolem’s method on that equation in $\mathbf Q_{31}$ (the first $\mathbf Q_p$ that contains  three cube roots of $2$) and in a finite extension of $\mathbf Q_3$ is here.
A: The only solutions of $x^3-2y^3=1$ are $(x,y)=(1,0)$ and $(x,y)=(-1,-1)$.
I don't know whether there's a nice Skolem-style proof, but here
this happens to be unnecessary because $(1,0)$ and $(-1,-1)$ are the 
only rational solutions and this can be proved by a Fermat-style descent:
the Weierstrass form is $Y^2 = X^3 - 27$, and there's a $2$-torsion point
at $(X,Y)=(3,0)$.  One could also use descent via $3$-isogeny to
$Y^2 = X^3+1$, which has $6$ rational points, at $\infty$, $(-1,0)$,
$(0,\pm1)$, and $(2,\pm3)$.
ADDED LATER: 1) As I already reported in a comment,
the result on $x^3 - 2y^3 = 1$ turns out to be due to Euler himself.
I found the reference in Dickson's History of the Theory of Numbers,
Vol II on page 572: it is Theorem 247 in Euler's Elements of Algebra,
see p.456 ff. of
this
English translation (Google Books scan of a Harvard library book from 1829).
It looks like Euler chose to use a 3-descent (presumably because
it was in the context of equations of the form $ax^3+bx^2+cx+d = y^3$),
even though a 2-descent was also available.
2) Meanwhile Rene Schoof notes that his book Catalan's Conjecture
reproduces a 3-adic proof using Skolem's method,
"from Bill McCallum's 1977 honours project at the University of Sydney".
See Proposition 4.1, pages 17-19.  [The $\root 3 \of 4$ coefficient of
$(\root 3 \of 2 - 1)^n$ is $0 \bmod 3$ iff $n = 3k$ or $n = 3k+1$,
and in both cases it vanishes mod $3^e$ iff $k$ does (each $e=1,2,3,\ldots$,
by induction on $e$), whence the known zeros for $k=0$ are the only ones.]
In the first paragraph of page 17, Schoof cites Euler's proof by descent, 
which he gives later in the book in an Appendix.
A: In section 4 of my book on the Catalan conjecture, I present such a proof (for $p=3$).  I took it from Bill McCallum's 1977 honours project at the University of Sydney
