---
tag: 7VWGA444K3
kind: problem
name: Waring's problem
msc: [11P05, 11P55]
uses: [0C4JS9RF5M, ZQFBEX62D0]
about: [0KYYSMEVHH]
sources:
  - url: https://en.wikipedia.org/wiki/Waring's_problem
same_as:
  wikidata: Q657903
  wikipedia: Waring's problem
history:
  posed_by: "Edward Waring"
  first_stated: "1770"
---

## Problem

Let $k \ge 2$ be an [[integer|0C4JS9RF5M]].
Let $g(k)$ be the least $s$ such that every positive integer is a sum of at most $s$ $k$-th powers of positive integers.
Let $G(k)$ be the least integer $s$ such that every sufficiently large integer is a sum of at most $s$ $k$-th powers of positive integers.
Waring's problem asks:

1. whether $g(k)$ exists for every $k \ge 2$;
2. what the values of $g(k)$ and $G(k)$ are.

## Remarks

Hilbert proved in 1909 that $g(k)$ exists for every $k$ (the Hilbert–Waring theorem).
By definition $G(k) \le g(k)$.
[[Lagrange's four-square theorem|0KYYSMEVHH]] gives $g(2) = 4$, and $G(2) = 4$ because no integer $n \equiv 7 \pmod 8$ is a sum of three squares.
Known values include $g(3) = 9$ and $g(4) = 19$; the numbers $23$ and $79$ need $9$ cubes and $19$ fourth powers.
The lower bound $g(k) \ge 2^k + \floor{(3/2)^k} - 2$ comes from the integer $n = 2^k \floor{(3/2)^k} - 1$, which is less than $3^k$ and so must be written with the powers $1^k$ and $2^k$ only.
Equality is known for all $k$ outside a possible finite set, and the possible exceptions have been excluded far beyond any small $k$.
For $G(k)$ much less is known: $G(4) = 16$ (Davenport), and $4 \le G(3) \le 7$, the upper bound being due to Linnik.
The [[Hardy–Littlewood circle method|ZQFBEX62D0]] gives an asymptotic formula for the number of representations when the number of powers is large compared with $k$.

## Explanations

### L1 human

Every positive integer is a sum of four squares, for example $7 = 4 + 1 + 1 + 1$.
Waring asked whether the same is true for cubes, fourth powers and so on, with some fixed number of terms.
The answer is yes: for instance, every positive integer is a sum of at most nine cubes.
The hard part is to find the smallest number of terms that always suffices, or that suffices for all large numbers.