-1
$\begingroup$

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, John B, Watson, user91500, JKnecht
If this question can be reworded to fit the rules in the help center, please edit the question.

3
$\begingroup$

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

$6n=(n+1)^3+(n-1)^3+(-n)^3+(-n)^3$

$6n+3=n^3+(4-n)^3+(2n-5)^3+(4-2n)^3$

$\endgroup$
  • 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.