Waring’s Problem

M. Ram Murty, FRSC, FNA, FNASc

Queen’s Research Chair

Queen’s University

Lagrange’s theorem

In 1770, Lagrange proved that every natural number can be written as a sum of four squares.

This was first conjectured by Bachet in 1621 who verified the conjecture for every number less than 326.

Joseph Louis Lagrange (1736-1813)

Fermat and Euler Fermat claimed a proof of the

four square theorem. But the first documented proof

of a fundamental step in the proof was taken by Euler.

Pierre de Fermat (1601-1665)

Leonhard Euler (1707-1783)

Edward Waring and Meditationes Algebraicae In 1770, Waring wrote in his

book Meditationes Algebraicae that every natural number can be written as a sum of four squares, as a sum of nine cubes, as a sum of 19 fourth powers and so on.

This is called Waring’s problem.

Edward Waring (1736-1798)

The problem with cubes

Can every natural number be expressed as a sum of 9 cubes?

This was first proved in 1908 by Arthur Wieferich (1884-1954)

Wieferich was a high school teacher and wrote only five papers in his entire life.

But all of these papers were of high quality. A small error in Wieferich’s paper was

corrected by A.J. Kempner in 1912.

What about fourth powers?

Theorem (Balasubramanian, Deshouillers, Dress, 1986) Every number can be written as a sum 19 fourth powers.

R. Balasubramanian

J.-M. Deshouillers

So what exactly is Waring’s problem? For each natural number k, there is a number

g= g(k) such that every number can be written as a sum of g kth powers.

Moreover g(2)=4, g(3)=9, g(4)=19 and so on. The first question is if g(k) exists. The second is what is the formula (if there is

one) for g(k)?

Hilbert’s Theorem

David Hilbert (1862-1943)

Theorem (Hilbert, 1909) For each k, there is a g=g(k) such that every number can be written as a sum of g kth powers.

What is g(k)? J.A. Euler (the son of L. Euler) conjectured in 1772 that g(k) = 2k + [(3/2)k] – 2.

The number 2k[(3/2)k]-1 <3k can only use 1’s and 2k’s when we try to write it as a sum of kth powers.

The most frugal choice is [(3/2)k] -1 2k’s followed by 1’s.

This gives g(k) ≥ 2k + [(3/2)k] – 2. Thus, g(1)=1, g(2)=4, g(3)=9,

g(4)=19. g(5)=37 (J. Chen, 1964) g(6)=73 (S. Pillai, 1940)

J. Chen (1933-1996)

S.S. Pillai (1901-1950)

Pillai’s Theorem

Write 3k = 2kq + r, with 0 < r < 2k.

If r+q ≤2k, then g(k) = 2k + [(3/2)k] – 2. Equivalent formulation: if {(3/2)k}≤1-(3/4)k,

then g(k) = 2k + [(3/2)k] – 2. Mahler (1957) proved that this condition holds

for all k sufficiently large. However, his proof was ineffective since it uses Roth’s theorem in Diophantine approximation which is ineffective.

The circle method

In his letter to Hardy written in 1912, Ramanujan alluded to a new method called the circle method. S. Ramanujan (1887-1920)

This was developed by Hardy and Littlewood in several papers and the methodis now called the Hardy-Littlewood method.

The function G(k)

Define G(k) as follows. For each k there is an no(k) such that every n≥ no(k) can be written as a sum of G(k) kth powers.

Clearly G(k) ≤ g(k). Note that g(k) = 2k + [(3/2)k] – 2 implies that g(k) has

exponential growth. Using the circle method, Vinogradov in 1947 showed

that G(k)≤k(2log k + 11) Hardy & Littlewood conjectured that G(k)<4k and this is

still an open problem.

Schnirelman’s theorem Let A be an infinite set and set

A(n) be the number of elements of A less than or equal to n.

If B is another infinite set, then what can we say about the set A+B = {a+b: a ε A, b ε B}?

Here we allow for the empty choice.

For example, is there a relation between A(n), B(n), and (A+B)(n)?

L. Schnirelman (1905-1938)

Schnirelman’s density

Define the density of A as δ(A) = infn≥1 A(n)/n. Thus, A(n)≥δn for all values of n. Note the density of even numbers is zero

according to this definition since A(1)=0. The density of odd numbers is ½.

δ(A)=1 if and only if A is the set all natural numbers.

Theorem (Schnirelman, 1931): δ(A+B)≥δ(A)+δ(B)-δ(A)δ(B).

An elementary approach Let A(n)=s, B(n)=t. Write a1 < a2 < ... < as ≤n.

Let ri = B(ai+1 – ai -1) and write these numbers as b1 < b2 < ... < bri

Then ai < ai + b1 < ai + b2 < ... < ai +bri< ai+1

Therefore, (A+B)(n) is at least A(n) + r1 + r2 + ... + rs-1 + B(n-as) +B(a1-1) A(n)+ δ(B)((a1-1) + (a2-a1-1) + ... + (as-as-1-1) + (n-as) ) = A(n) + δ(B)(n-s)= (1-δ(B))A(n)+δ(B)n ≥(1-δ(B))δ(A)n + δ(B)n.

An application of induction

So we have δ(A+B) ≥1-(1-δ(A))(1-δ(B)). By induction, we have δ(A1+ ... + As) ≥1 – (1-δ(A1))...(1-δ(As)) Notation: 2A = A+A, 3A = A+A+A, etc. Corollary. If δ(A)>0, then for some t, we have δ(tA) >

½. Proof. By the above, δ(tA) ≥1-(1-δ(A))t. If δ(A)=1, we are done. Suppose 0<δ(A)<1. Then, (1-δ(A))t tends to zero as t tends to infinity.

What happens if δ(A)>1/2?

Then A(n) > n/2. This is the size of A = {a ε A, a≤n}. Consider the set B= {n – a: a ε A, a ≤ n}. This set has size A(n) > n/2. If B and A are disjoint, we get more than n

numbers which are ≤n, a contradiction. Thus, every number can be written as a sum

of two elements of A.

Consequence of Schnirelman’s Theorem If A has positive Schnirelman density, then

for some t, we have tA is the set of natural numbers.

In other words, every number can be written as a sum of at most t elements from the set A.

Let us apply this observation to count the number of numbers ≤n which can be written as a sum of t kth powers.

X1k + X2

k + ... + Xtk ≤ n

Let A be the set of numbers that can be written as a sum of t kth powers.

Key lemma: The number of solutions is at most A(n)nt/k-1.

On the other hand, a lower bound is given by [(n/t)1/k]t .

This gives that A(n) has positive Schnirelman density.

U.V. Linnik (1915-1972)

Some open problems Can every sufficiently large number be

written as a sum of 6 cubes? (Unknown) The conjecture is that G(3)=4. What we

know is that 4≤G(3)≤7. Hypothesis K: (Hardy and Littlewood) Let

rg,k(n) be the number of ways of writing n as a sum of g kth powers. Then, rk,k(n) =O(nε) for any ε>0.

G(k)=max(k+1, γ(k)) where γ(k) is the smallest value of g predicted by “local” obstructions.

THANK YOU!