# Can any integer can be written as the sum of 8 integer cubes?

Can anyone offer an elementary proof why: $$\forall n \in {\mathbb Z} \space \exists a, b, c, d, e, f, g, h \in {\mathbb Z}$$such that$$n=a^3+b^3+c^3+d^3+e^3+f^3+g^3+h^3{\rm ?}$$

In other words, why every integer is the sum of eight cubes integers.

My first thought was that since the gaps between $p^3$ and $(p+1)^3$ increase with $p,$ this would only hold for small integers. This is wrong; can anyone tell me why?

N.B. This is different to Waring's problem because $n,a,b,c...$ dot not have to be natural numbers, so we can have $27=3^3+(-1)^3+(-1)^3+(-1)^3+(-1)^3$

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$27=3^3+(-1)^3+(-1)^3+(-1)^3+(-1)^3$ is only true in base eight. –  Henning Makholm Oct 4 '12 at 20:32
Yes, sorry: that should be only two $(-1)^3$'s, of course. –  user43644 Oct 5 '12 at 16:56
Then it's true in base nine. :-) –  Henning Makholm Oct 5 '12 at 17:17

Allowing negative cubes, five suffice. It is suspected that four suffice but this is an open problem. I will see if I can find the argument for five, it is just one or two explicit formulas. By the way, these are called the "Easier" Waring problems.

Can't find them at this point, so here is section D5 form Richard K. Guy, Unsolved Problems in Number Theory. Note that he is allowing cubes to be positive, negative, or zero:

Is every number the sum of four cubes? This has been proved for all numbers except possibly those of the form $9n \pm 4$

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

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

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If you can find the argument, you can have an upvote :) –  Graphth Oct 4 '12 at 20:16
@Graphth, found it, not quite what I remembered but it works. –  Will Jagy Oct 4 '12 at 21:13
Great, thanks! . –  Graphth Oct 4 '12 at 21:16
Wow, that's neat! Thanks a lot. –  user43644 Oct 5 '12 at 16:56
1. Since $n - n^3 = n(1-n)(1+n)$ then $6 \mid (n - n^3)$ and we can write $n - n^3 = 6k$.

2. $6k = (k+1)^3 + (k-1)^3 - k^3 - k^3$.

3. Form 1 and 2 we derive that $n = n^3 + (k+1)^3 + (k-1)^3 - k^3 - k^3$ for some $k$.

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That is pretty! –  Erick Wong Nov 21 '14 at 16:34
(+1) That's really nice. –  aes Nov 21 '14 at 16:36