Does the above Diophantine equation have other integer solutions besides $(x,y)=(1,2)$ and $(x, y) = (0, 1)$?
closed as offtopic by Alexey Ustinov, Vladimir Dotsenko, Wolfgang, Marco Golla, Stefan Kohl Dec 1 '15 at 11:03
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2$\begingroup$ This is a Thue equation which can be solved algorithmically. $\endgroup$ – Jeremy Rouse Dec 1 '15 at 0:40

$\begingroup$ Following up on Jeremy's comment, see the answers to mathoverflow.net/questions/115063/solvedcubicthueequation $\endgroup$ – socalled friend Don Dec 1 '15 at 1:03

1$\begingroup$ As a novice I would try factoring both sides of y^3  1 = 9x^3, or maybe 8x^3  1 = y^3  x^3, to see what that might say about x and y. Gerhard "Likes Doing Diophantine Equations Oldstyle" Paseman, 2015.11.30 $\endgroup$ – Gerhard Paseman Dec 1 '15 at 1:10
Theorem 6.4.30 in Cohen's Number Theory: Volume I asserts: For each nonzero integer $d$, there is at most one pair of integers $(X,Y)$ with $Y\ne 0$ and $X^3+dY^3=1$. Apply this with $X=y$, $Y=x$, and $d=9$, to see that that there are no more solutions. Theorem 6.4.30 is attributed to Skolem.

$\begingroup$ I think that settles it, thanks, will have a look at that ! $\endgroup$ – user83236 Dec 1 '15 at 1:25