How to rigourously prove that any integer divisible by $3$ can be written as a sum of four not necessarily posiitive cubes? I have been trying it for long


closed as off-topic by Travis, John B, Watson, user91500, JKnecht Apr 14 '16 at 10:47

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6 cannot be written as the sum of four non-negative cubes, so I assume you are allowing negative cubes.



  • 1
    $\begingroup$ How did you know that? I mean how did you get to those numbers. $\endgroup$ – TheRandomGuy Apr 14 '16 at 12:23
  • $\begingroup$ @Dhruv The first is well-known (and not hard to find). You can get the second by playing around on the basis that it might work for four linear expressions. $\endgroup$ – almagest Apr 14 '16 at 12:25

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