# How to rigourously prove that any integer divisible by $3$ can be written as a sum of four cubes? [closed]

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, JKnechtApr 14 '16 at 10:47

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## 1 Answer

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

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

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

• How did you know that? I mean how did you get to those numbers. – TheRandomGuy Apr 14 '16 at 12:23
• @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. – almagest Apr 14 '16 at 12:25