Does the above Diophantine equation have other integer solutions besides $(x,y)=(1,2)$ and $(x, y) = (0, -1)$?
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2$\begingroup$ This is a Thue equation which can be solved algorithmically. $\endgroup$– Jeremy RouseDec 1, 2015 at 0:40
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$\begingroup$ Following up on Jeremy's comment, see the answers to mathoverflow.net/questions/115063/solved-cubic-thue-equation $\endgroup$– so-called friend DonDec 1, 2015 at 1:03
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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 Old-style" Paseman, 2015.11.30 $\endgroup$– Gerhard PasemanDec 1, 2015 at 1:10
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2 Answers
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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.
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$\begingroup$ I think that settles it, thanks, will have a look at that ! $\endgroup$– user83236Dec 1, 2015 at 1:25