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?
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$\begingroup$ Note that the coefficient on $\sqrt[3]{4}$ gives a linear recursive sequence. You should be able to do something with that. $\endgroup$– user13113Commented Jun 6, 2015 at 4:47
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1$\begingroup$ Have you seen Proposition 3.8 and Example 3.16 in esc.fnwi.uva.nl/thesis/centraal/files/f310232185.pdf ? $\endgroup$– so-called friend DonCommented Jun 6, 2015 at 5:53
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1$\begingroup$ See T. Nagell, Solution complète de quelques équations cubiques à deux indéterminées. J. Math. Pures Appl. 4 (1925), 209–270. I looked at the case n=2 in arxiv.org/pdf/1108.6218v2.pdf $\endgroup$– Franz LemmermeyerCommented Jun 6, 2015 at 6:28
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$\begingroup$ The sequence of coefficients is in the encyclopedia, though not with this context. oeis.org/A108369 $\endgroup$– Kevin O'BryantCommented Jun 7, 2015 at 5:01
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3 Answers
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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.
answered Jun 6, 2015 at 3:30
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5$\begingroup$ According to Dickson's History (Vol. II, p.572) Euler already proved in 1770 that if $x^3+y^3=2z^3$ (for integers $x,y,z$ with $z \neq 0$) then $x=y$. $\endgroup$ Commented Jun 6, 2015 at 4:06
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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
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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.
