TITLE: Can any integer can be written as the sum of 8 integer cubes?
QUESTION [18 upvotes]: 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 $27=3^3+(-1)^3+(-1)^3+(-1)^3+(-1)^3$

REPLY [5 votes]: Numbers not of the form $9k \pm 4$ and $108k \pm 38$ can be written as the sum of 4 cubes (Demjanenko 1966, translated version Cohen 2004) using the following identities and complementary identities ($x \to -x$):
\begin{align}
6x &= (x − 1 )^3 + (−x )^3 + (−x )^3 + (x + 1 )^3 \\
6x + 3 &= x^3 + (−x + 4 )^3 + (2x − 5 )^3 + (−2x + 4 )^3 \\
18x + 1 &= (2x + 14 )^3 + (−2x − 23 )^3 + (−3x − 26 )^3 + (3x + 30 )^3 \\
18x + 7 &= (x + 2 )^3 + (6x − 1 )^3 + (8x − 2 )^3 + (−9x + 2 )^3 \\
18x + 8 &= (x − 5 )^3 + (−x + 14 )^3 + (−3x + 29 )^3 + (3x − 30 )^3 \\
54x + 2 &= (29484x^2 + 2211x + 43)^3 + (-29484x^2 - 2157x-41)^3 + (9828x^2+485x+4)^3 + (-9828x^2 -971x - 22)^3\\
54x + 20 &= (3x − 11 )^3 + (−3x + 10 )^3 + (x + 2 )^3 + (−x + 7 )^3 \\
216x-16 &= (14742x^2 - 2157x + 82)^3 + (-14742x^2 + 2211x - 86)^3 + (4914x^2-971x+44)^3 + (-4914x^2+485x-8)^3 \\
216x+92 &= (3x-164)^3 + (-3x+160)^3+(x-35)^3 + (-x+71)^3
\end{align}
More identities are listed here https://www.alpertron.com.ar/FCUBES.HTM