Multiple of 3 as Sum of 4 Cubes

Question

Prove that every multiple of 3 can be expressed as sum of four integer cubes.

When working with numbers of form $6n$, a very clear pattern emerged, which the equation below proves elegantly -

$\left(n-1\right)^3 + \left(-n\right)^3 + \left(n+1\right)^3 + \left(-n\right)^3 = 6n$.

This only leaves out the numbers of form $6n + 3$. But despite of lot of experimentation, I still couldn't work out on how to prove this. Please help.

Answer 1

Cubic Number (mathworld) has the following ((18) in that page):

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

By the way, it seems to me that the above page says that every number can be represented as a sum of four signed cubes except numbers of the form $9n\pm 4,108n\pm 38$.

Answer 2

Let me quote from here: $$ 6 x + 3 = x^{3} +(-x+4)^{3} +(2x-5)^{3} +(-2x+4)^{3}. $$

Answer 3

Reordering your equation $$A^3 - 2 (b^3) + c^3 = 6n,$$ and note the coefficients are the same as $1,2,1$ the coefficients of pascals #2.

Pascal wrote about these little Gems, which seem to be avoided by the modern world. A few hints are found in Dickson's 2nd History Dioiphantine Analysis.