Hilbert-Waring Theorem/Variant Form/Particular Cases/3

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Particular Case of the Hilbert-Waring Theorem -- Variant Form: $k = 3$

The Hilbert-Waring Theorem -- Variant Form states that:


For each $k \in \Z: k \ge 2$, there exists a positive integer $G \left({k}\right)$ such that every sufficiently large positive integer can be expressed as a sum of at most $G \left({k}\right)$ $k$th powers.


The case where $k = 3$ is:

Every sufficiently large positive integer can be expressed as the sum of a number of positive cubes.


The exact number is the subject of ongoing research, but at the time of writing ($11$th February $2017$) it is known that it is between $4$ and $7$.


That is:

$4 \le \map G 3 \le 7$


Progress

From Cube Modulo 9, a cube is congruent to $0$, $1$, or $2$ modulo $9$.

So it takes at least $4$ cubes to make an integer $k \equiv \pm 4 \pmod 9$.

Hence $\map G 3$ is at least $4$.


For numbers less than $1.3 \times 10^9$, it has been demonstrated that $1 \, 290 \, 740$ is the highest number to require $6$ cubes.

It is also known that the number of numbers between $n$ and $2 n$ requiring $5$ cubes falls with increasing $n$. Melvyn Bernard Nathanson has expressed a belief that this decrease is rapid enough to suggest that $\map G 3 = 4$.

From the work of Jean-Marc Deshouillers, François Hennecart, Bernard Landreau and I. Gusti Putu Purnaba, $7 \, 373 \, 170 \, 279 \, 850$ is the largest number known that is not the sum of $4$ cubes. They offer an argument that this may indeed be the largest possible.


Historical Note

It had been established by $1912$ that it takes no more than $9$ positive cubes to represent any positive integer as a sum.

The value of $G \left({3}\right)$ is still the subject of research.

Yuri Vladimirovich Linnik showed in $1943$ that the upper bound of $G \left({3}\right)$ is $7$.

Recent research, however, suggests that $G \left({3}\right)$ may indeed be $4$.


Sources

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