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I'm not familiar with diophantine equations. At most my approaches doesn't give results. I need to solve the following equation $$x^3-2y^3=1$$ Where $x,y,z\in\mathbb{Z}$ I know $x=-1,y=-1.x=1,y=0,$ Aware of any other integer solutions. Prove

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Also, if $x$ is even, then $x^3$ is even, and $x^3 - 1$ is odd. But $x^3 - 1 = 2y^3$ and the RHS is even. So $x$ must be odd. –  user2468 Apr 19 '12 at 4:31
    
My first thought is to rewrite as $(x-y)(x^2+xy+y^2)=(y+1)(y^2-y+1)$ and think about factors, but I don't know if this will help. –  Ross Millikan Apr 19 '12 at 4:46
    
By Thue-Siegel-Roth, there are only finitely many solutions. Page 154, LeVeque, Topics in Number Theory, volume 2. –  Will Jagy Apr 19 '12 at 4:47

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up vote 7 down vote accepted

You have found the only solutions. These things are not so easy to prove. This one is done in Mordell's book, Diophantine Equations, in Chapter 15.

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