Title: Waring
1Warings Problem
- M. Ram Murty, FRSC, FNA, FNASc
- Queens Research Chair
- Queens University
2Lagranges 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)
3Fermat 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)
4Edward 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 Warings problem.
Edward Waring (1736-1798)
5The 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 Wieferichs paper was corrected
by A.J. Kempner in 1912.
6What about fourth powers?
- Theorem (Balasubramanian, Deshouillers, Dress,
1986) Every number can be written as a sum 19
fourth powers.
R. Balasubramanian
J.-M. Deshouillers
7So what exactly is Warings 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)?
8Hilberts Theorem
- Theorem (Hilbert, 1909) For each k, there is a
gg(k) such that every number can be written as a
sum of g kth powers.
David Hilbert (1862-1943)
9What 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 lt3k can only use 1s and
2ks when we try to write it as a sum of kth
powers. - The most frugal choice is (3/2)k -1 2ks
followed by 1s. - 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)
10Pillais Theorem
- Write 3k 2kq r, with 0 lt r lt 2k.
- If rq 2k, then g(k) 2k (3/2)k 2.
- Equivalent formulation if (3/2)k1-(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 Roths theorem in
Diophantine approximation which is ineffective.
11The 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 method is now called the
Hardy-Littlewood method.
12The 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)lt4k and
this is still an open problem.
13Schnirelmans 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 AB ab a e A, b e B? - Here we allow for the empty choice.
- For example, is there a relation between A(n),
B(n), and (AB)(n)?
L. Schnirelman (1905-1938)
14Schnirelmans density
- Define the density of A as d(A) infn1 A(n)/n.
- Thus, A(n)dn 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. - d(A)1 if and only if A is the set all natural
numbers. - Theorem (Schnirelman, 1931)
- d(AB)d(A)d(B)-d(A)d(B).
15An elementary approach
- Let A(n)s, B(n)t. Write a1 lt a2 lt ... lt as n.
- Let ri B(ai1 ai -1) and write these numbers
as b1 lt b2 lt ... lt bri - Then ai lt ai b1 lt ai b2 lt ... lt ai brilt
ai1 - Therefore, (AB)(n) is at least
- A(n) r1 r2 ... rs-1 B(n-as) B(a1-1)
- A(n)
- d(B)((a1-1) (a2-a1-1) ... (as-as-1-1)
(n-as) ) - A(n) d(B)(n-s) (1-d(B))A(n)d(B)n
- (1-d(B))d(A)n d(B)n.
16An application of induction
- So we have d(AB) 1-(1-d(A))(1-d(B)).
- By induction, we have
- d(A1 ... As) 1 (1-d(A1))...(1-d(As))
- Notation 2A AA, 3A AAA, etc.
- Corollary. If d(A)gt0, then for some t, we have
d(tA) gt ½. - Proof. By the above, d(tA) 1-(1-d(A))t.
- If d(A)1, we are done. Suppose 0ltd(A)lt1.
- Then, (1-d(A))t tends to zero as t tends to
infinity.
17What happens if d(A)gt1/2?
- Then A(n) gt n/2.
- This is the size of A a e A, an.
- Consider the set B n a a e A, a n.
- This set has size A(n) gt 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.
18Consequence of Schnirelmans 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.
19X1k X2k ... 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/kt . - This gives that A(n) has positive Schnirelman
density.
U.V. Linnik (1915-1972)
20Some 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 4G(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(ne) for any egt0. - G(k)max(k1, ?(k)) where ?(k) is the smallest
value of g predicted by local obstructions.
21 THANK
YOU!