# 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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$6n=(n+1)^3+(n-1)^3+(-n)^3+(-n)^3$
$6n+3=n^3+(4-n)^3+(2n-5)^3+(4-2n)^3$