I know that ${x^3} - 8{y^3} = 12$ has no integer solutions but how can I prove it? If I had to sit down with someone and convince them (at least, fairly) rigorously that it has no integer solutions. How would I do it? Any help is appreciated.
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14$ x $ must be even, and hence $ 8 $ divides $ 12 $... – May 11 '16 at 11:46
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I don't understand why x must be even or the implications of that. – Ahmed May 11 '16 at 12:07
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2 divides $8y^3+12$, so it must divide $x^3$. Hence $x$ must be even. Hence 8 divides $x^3$ and hence also $x^3-8y^3$, so 8 divides 12. Contradiction. – almagest May 11 '16 at 12:11
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All even cubes are multiples of 8. Form the equation for $x^3$ and you will see it is even. – N.S.JOHN May 11 '16 at 12:13
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@almagest do you think this question must be down voted? I don't understand why. – N.S.JOHN May 11 '16 at 12:14
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1@N.S.JOHN I didn't downvote it. – almagest May 11 '16 at 12:16
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@the_hermit your comment should be added as an answer below so you get credit and so the question can be removed from unanswered queue. – JMoravitz May 11 '16 at 13:10
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Sure, added. Thanks. – May 11 '16 at 16:17