How many $n$th powers are required to partition any whole number?

Question

How many $n$th powers are required to partition every number into just that power? For example, only one number is required to do so for $1$st powers. Every number can be partitioned into at most $4$ squares. How many cubes do you need to partition every number? I know it's at least four. For the people who answered with Waring's Problem, I count negative cubes as well. For example, 12 could be 27-8-8+1. I know you cannot write numbers of the form 4 or 5 mod 9, with 3 cubes, so shouldn't you be able to write all integers as the sum of 4 cubes?

Answer 1

For general powers $n$ this is known as Waring's Problem. The problem is open for general $n$, but the presumed solution is $$g(n) := 2^n + \left\lfloor\left(\frac{3}{2}\right)^n\right\rfloor - 2 .$$ It's known that this is a lower bound for all $n$ and is a correct value for all $n \leq 4.7 \cdot 10^8$. The values of $g(n)$ are the content of OEIS A002804.

For $n = 3$ in particular, then, one needs $\leq 9$ positive cubes to write any positive integer (this was shown, before the above results were available, by Wieferer (1909) and Kempner (1912)). Dickson showed that the only numbers realizing this bound are $$23 = 2^3 + 2^3 + \underbrace{1^3 + \cdots + 1^3}_7 \qquad \textrm{and} \qquad 239 = 5^3 + 3^3 + 3^3 + 3^3 + 2^3 + 2^3 + 2^3 + 2^3 + 1^3.$$

Dickson's paper includes sobering remarks that make one thankful to live in the age of computational automation, including:

Table I gives, for each positive integer $N \leq 40\,000$, the least number $m$ such that $N$ is a sum of $m$ cubes. It was computed by R. D. von Sterneck by adding all cubes to each sum of $j$ but not fewer, cubes for $j = 1, 2, \ldots$. Any error in the $j$th step would introduce an increasingly large number of errors in the later steps. For this reason, my assistant, Miss Evelyn Garbe, computed in four weeks the table for $12\,000 \leq N \leq 40\,000$ by my improved method described later. The only error found is $$32\,822 = 3^3 + 3^3 + 32^3,$$ whose $m$ was given to be $4$ by von Sterneck.

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Dickson, L.E., All integers except 23 and 239 are sums of eight cubes [pdf] Bull. Amer. Math. Soc. 45 (1939), 588-591.