Download Institute Colloquium on Waring`s Problem

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

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.
David Hilbert (1862-1943)
What is g(k)?



J. Chen (1933-1996)




S.S. Pillai (1901-1950)
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)
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.
This was developed by Hardy and
Littlewood in several papers and
the method
is now called the Hardy-Littlewood
method.
S. Ramanujan (1887-1920)
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.
k
X1




+
k
X2
+ ... +
k
Xt
≤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!