I have a pretty difficult math question that I have no idea even how to begin. Here it goes:

Find the nonzero integers $a$, $b$, $c$ such that $a^3 + b^3 + c^3 = 2$?

I would assume that at least one of the integers would be negative. Also, I got as far as $(a+b+c)^3 = 2^3$ but I'm not sure what good this will do.

(So, not to cower in shame I made a mistake above seeing as how $(a+b+c)^3 \ne 2^3$. I admit I failed there. But how to work this equation?)

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    $\begingroup$ How did you get to $(a+b+c)^3$?! $\endgroup$ – 5xum Apr 13 '15 at 8:50
  • $\begingroup$ Daw man, you're right. Sleep deprivation is causing my basic algebra skills to lack! Now, I'm even further behind that I 'believed' I was... $\endgroup$ – GreeNs Apr 13 '15 at 8:52
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    $\begingroup$ How did you get $(a+b+c)^3=8$? $\endgroup$ – Binary Geek Apr 13 '15 at 8:54
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    $\begingroup$ There are infinitely many non-zero solutions. $$(1+6x^3)^3 + (1-6x^3)^3 + (-6x^2)^3 = 2$$ The key is start from a known solution $(a,b,c) = (1,1,0)$ of the equation and play with it. $\endgroup$ – achille hui Apr 13 '15 at 8:55
  • $\begingroup$ Source?${}{}{}$ $\endgroup$ – Bart Michels Apr 13 '15 at 9:02

In $1908$, A.S. Werebrusov found the following parametrization of $x$, $y$, and $z$ for the sum of three cubes $x^3+y^3+z^3$ equal to $2$: $$ (1 + 6t^3)^3 + (1 − 6t^3)^3 + (−6t^2)^3 = 2. $$ A further references here is L.J. Mordell, On Sums of Three Cubes, Journal of the London Mathematical Society $17$ (1942), 139–144. There are several discussions about the sum of three integer cubes, see here and the references given therein.


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