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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.

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Waring’s Problem

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  1. Waring’s Problem M. Ram Murty, FRSC, FNA, FNASc Queen’s Research Chair Queen’s University

  2. 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)

  3. 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)

  4. 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)

  5. 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.

  6. 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

  7. 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)?

  8. 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)

  9. 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)

  10. 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.

  11. 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 method is now called the Hardy-Littlewood method.

  12. 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.

  13. 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)

  14. 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).

  15. 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.

  16. 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.

  17. 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.

  18. 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.

  19. X1k + 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/k]t . • This gives that A(n) has positive Schnirelman density. U.V. Linnik (1915-1972)

  20. 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.

  21. THANK YOU!

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