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

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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 ( )

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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 ( ) Leonhard Euler ( )

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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 ( )

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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 ( ) 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.

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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

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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 k th 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)?

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

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What is g(k)? J.A. Euler (the son of L. Euler) conjectured in 1772 that g(k) = 2 k + [(3/2) k ] – 2. The number 2 k [(3/2) k ]-1 <3 k can only use 1’s and 2 k ’s when we try to write it as a sum of k th powers. The most frugal choice is [(3/2) k ] -1 2 k ’s followed by 1’s. This gives g(k) ≥ 2 k + [(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 ( ) S.S. Pillai ( )

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Pillai’s Theorem Write 3 k = 2 k q + r, with 0 < r < 2 k. If r+q ≤2 k, then g(k) = 2 k + [(3/2) k ] – 2. Equivalent formulation: if {(3/2) k }≤1-(3/4) k, then g(k) = 2 k + [(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.

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The circle method In his letter to Hardy written in 1912, Ramanujan alluded to a new method called the circle method. S. Ramanujan ( ) This was developed by Hardy and Littlewood in several papers and the method is now called the Hardy-Littlewood method.

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The function G(k) Define G(k) as follows. For each k there is an n o (k) such that every n≥ n o (k) can be written as a sum of G(k) k th powers. Clearly G(k) ≤ g(k). Note that g(k) = 2 k + [(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.

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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 ( )

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Schnirelman’s density Define the density of A as δ(A) = inf n≥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).

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An elementary approach Let A(n)=s, B(n)=t. Write a 1 < a 2 <... < a s ≤n. Let r i = B(a i+1 – a i -1) and write these numbers as b 1 < b 2 <... < b r i Then a i < a i + b 1 < a i + b 2 <... < a i +b r i < a i+1 Therefore, (A+B)(n) is at least A(n) + r 1 + r r s-1 + B(n-a s ) +B(a 1 -1) A(n)+ δ(B)((a 1 -1) + (a 2 -a 1 -1) (a s -a s-1 -1) + (n-a s ) ) = A(n) + δ(B)(n-s)= (1-δ(B))A(n)+δ(B)n ≥(1-δ(B))δ(A)n + δ(B)n.

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An application of induction So we have δ(A+B) ≥1-(1-δ(A))(1-δ(B)). By induction, we have δ(A A s ) ≥1 – (1-δ(A 1 ))...(1-δ(A s )) 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.

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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.

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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 k th powers.

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X 1 k + X 2 k X t k ≤ n Let A be the set of numbers that can be written as a sum of t k th powers. Key lemma: The number of solutions is at most A(n)n t/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 ( )

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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 r g,k (n) be the number of ways of writing n as a sum of g k th powers. Then, r k,k (n) =O(n ε ) for any ε>0. G(k)=max(k+1, γ (k)) where γ( k) is the smallest value of g predicted by “local” obstructions.

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