TITLE: Is every natural number representable as $\sum\limits_{k=1}^{n} \pm k^3$?
QUESTION [25 upvotes]: A well known identity is: $$(n+3)^2 - (n+2)^2-(n+1)^2 + n^2 = 4$$
and using this identity we prove that the set $\displaystyle \{\pm 1^2, \pm 1^2 \pm 2^2,\pm 1^2\pm 2^2\pm 3^2, \cdots\}$ contains every natural number greater than or equal to $1$.But for establishing this result we need to verify representation of the numbers $1,2,3,4$ in the form $\displaystyle \sum\limits_{k=1}^{n} \pm k^2$.
My question is can we say the same about the set of numbers $\displaystyle \sum\limits_{k=1}^{n} \pm k^3$, that is if every natural number greater than or equal to $1$ is representable in this form.
What about the general case $\displaystyle \sum\limits_{k=1}^{n} \pm k^s$ ?
This post shows how to compute the constant, $\displaystyle C_s = \sum\limits_{k=1}^{2^s} \pm k^{s} =  \large{2^{\frac{s(s-1)}{2}}s!}$ where, the signs $\pm$ are chosen as explained here.

REPLY [6 votes]: For 5th powers, there is an identity,
$$\sum\limits_{k=1}^{168}\pm(x+k)^5 = 480$$
analogous to the 3d powers mentioned by R. Millikan in his answer. 
What remains to be shown is that all $0\leq N<240$ can be decomposed into sums of fifth powers. See this post.
(P.S. The number of addends can be reduced to just $m=168$.)