"Alternating" sums of fourth powers.

Question

Problem: Show that any integer $n$ can be written as: $$n = \sum_{k = 0}^N \epsilon_k k^4$$ where $N$ is a nonnegative integer and $\epsilon_k$ is in $\{\pm 1\}$ for every $k\leq N$.

Answer 1

One can lower the degree by subtraction of consecutive temrs, that is $$k^4-(k+1)^4=-4k^3-6k^2-4k+1$$ is only of degree $3$. Next, $$(k^4-(k+1)^4)-((k+2)^4-(k+3)^4)=24k^2+72k+64$$ is only of degree $2$. Ultimately, one finds that 16 consecutive signs $+--+-++--++-+--+$ always produce the constant $1536$. So once you manage to find such sums for $n=0,1,\ldots,1535$, all remaining $n$ can be obtained from a sum to yield $n\bmod 1536$ followed by one or mor repetitions of that 16-sign sequence. (Is there possibly a shorter sign sequence leading to a - smaller - constant?)

EDIT: This paper shows the result (also for sums of cubes and reiterates the result for sums of squares squares) by exhibiting the above sign sequence that produces $1536$, and another sign sequence $ -++-+-+--+--++++---++--+$ to produce $2016$, and combine them to a very lengthy sign sequence that brings down the step size to $96$. So it remains to find specific solutions only for $n=0,\ldots,95$. In fact, special solutions only for $0,\ldots,48$ are enough as we can flip signs! (Oh, I just notice that the authors did not use this last reduction, tsk, tsk)