J- UIt-[0,h(k-2)+t(b+1)]
1.7 Application: Sumsets and powers of 2
Let n > 1, and let B' be the set of all multiples of 3 contained in the interval [ 1, n).
Then I B I < n/3, and every sum of elements of B' is divisible by 3. Certainly, no such sum is a power of 2. This set B' is the extremal case: We shall prove that if B is any subset of [1, n] such that I BI > n13, then some power of 2 can be written as the sum of at most four (not necessarily distinct) elements of B.
Lemma 1.5 Let m > 1, and let C be a subset of [0, m] such that m
ICI
2+1.
Then some power of 2 is either an element of C or the sum of two distinct elements of C.
32 1. Simple inverse theorems
Proof. The proof is by induction on m. It is easy to check that the result is true for m - 1, 2, 3, and 4. Let m > 4, and assume that the result holds for all positive integers m' < m. Choose s > 2 such that
2'<m<2'+I
and let
r-m-2'E[O,2'-1].
Let
C'-Cfl[0,2'-r-1]
and
C"-Cfl[2'-r,2'+r].
Then C is the disjoint union of C' and C", and ICI - IC'I + IC"I.
Suppose that the lemma is false for the set C. Then ICI > m/2+ 1, but no power of 2 either belongs to C or is the sum of two distinct elements of C. It follows that 2' ¢ C" and, f o r each i - 1, ... , r, the set C" contains at most one of the two integers 2' - i, 2' + i. Therefore,
IC"I<r.
If m - 2` - 1, then r - 2' - I and C' c (0); thus IC'I<1.
It follows that
2+1<ICI<1+r-2'-m21
which is impossible.
Similarly, if 2' <m <2'+1 -1,then 0<r <2'-Iand m'-2'-r-1
Since the set C contains C', it follows that no power of 2 either belongs to C' or is the sum of two distinct elements of C'. By the induction hypothesis, we have
IC'I< 2+1-2-2-1+1 r
m
2''-r-1 m+l
2
+ 1 <ICI - ICI+ICI <
2+1+r--2
which is also impossible.
Theorem 1.18 Let n > 1, and let B be a set of integers contained in the interval [ 1, n J. If I B I > n /3, then there is a power of 2 that can be written as the sum of at most four elements of B.
Proof. Since B C [1, n], we have I B I
> n/3 > max(B)/3. Thus, we can
assume that max(B) - n. Let d be the greatest common divisor of the elements ofB. The number of multiples of d in the interval [1, n] is [n/d], so
n n
3<IBI<d
Therefore, d = I or2.
If d = 2, we can consider the set
B'a{b/2:EB}c[1,2].
The greatest common divisor of the elements of B' is 1. The set B' also satisfies the hypotheses of the theorem. If the theorem holds in the case d = 1, then there exists h < 4 and there exist integers b,, ... , bh E B' such that b, + + bh = 2S.
It follows that 2b,, ..., 2b' E B and 2b, +
+ 2bh = 2'+'. Therefore, we can assume that d = 1.Let A = {0} U B. Then d(A) - 1, max(A) = max(B) = n, and
k=IAI=CBI+l> 3+1.
nIt follows from Theorem 1.15 that
12A I > min(3k - 3, k + n) > n + 1.
Since 2A C [0. 2n], we can apply Lemma 1.5 with C = 2A and m = 2n. Since 2C = 4A, it follows that some power of 2 can be written as the sum of at most four elements of A. This completes the proof.
In Exercise 19, we construct examples of finite sets B C [ 1, n ] such that I B I >
n/3, but no power of 2 can be written as the sum of three elements of B. This shows that Theorem 1.18 is best possible.
1.8 Notes
The principal result in this chapter is Theorem 1.16, which was proved by Frei- man [49, 54, 55]. Freiman [52] has extended this to sumsets of the form A + B.
Steinig [1211 has an expanded version of Freiman's proof. In Chapter 4 1 give a different proof, discovered by Lev and Smeliansky [81], that uses a theorem of Kneser on sumsets in abelian groups. Freiman's monograph Foundations of a Structural Theory of Set Addition [54] is devoted to Freiman's work on inverse theorems.
Theorem 1.1 is due to Nathanson [91 J. For some related results, see Lev [80].
The simple inverse theorem for hA (Theorem 1.5) is probably ancient, but I have not seen it in print. The inverse theorems for the sets h ^ A are due to Nathanson (951.
34 1. Simple inverse theorems
Very little is known about Erdos's conjecture that IE2(A)I »E k2-`. Erd6s and Szemeredi [44] have shown that there exists a real number S > 0 such that
IE2(A)I >> k'+a and Nathanson [97] proved that
IE2(A)I > ck3213i
where c - 0.00028.... Ford [47] has improved the exponent to 16/15. The proof of Lemma 1.4 is due to Erd6s and Pomerance (personal communication). Us- ing a theorem of Vinogradov and Linnik [125], Nathanson and Tenenbaum [99]
strengthened this result: They proved that if Q is an arithmetic progression of length 1, then dQ(m) << 12(logl)3 for all m E Q2. This implies that if IAI = k and 12A1 < 3k - 4, then IA21 >> k2/(logk)3
Erd6s and Freud had conjectured that if A C [ 1, n] and I A I > n/3, then some power of 2 can be written as the sum of distinct elements of A. This was proved by Erdos and Freiman [39] (with an unbounded nun]ber of summands), and Nathanson and Sarkozy [98] (with a bounded number of summands). Theorem 1.18, due to Lev [79], improves results of Nathanson and Sark6zy [98] and of Freiman [58].
Closely related to the "structural" inverse problems is another class of inverse problems in additive number theory that we can call recognition problems or de- composition problems. We write A B if A and B are sets of integers that coincide from some point on. If we are given a finite or infinite set B of integers, can we determine whether B is a sumset or even asymptotically a sumset? This means
the following: Let h > 2. Does there exist a set A such that hA
B? More generally, do there exist sets A,, ... , Ah such that IAi I > 2 for i = 1, 2, ... , h and A, + + Ah @ B? Do there exist sets At , ... , Ah such that IA; I > 2 fori - 1, 2, ... , h and A, +
+ Ah ^- B? Ostmann [ 100] introduced this class of inverse problems.Sets of integers that decompose into sumsets are rare. Let us associate to each set A of nonnegative integers the real number
E2-a-' E [0, 1].
aeA
Wirsing [128] proved that the Lebesgue measure of the set of real numbers that correspond to sets A such that A B + C for some B and C is zero.
An important decomposition problem is the following: Do there exist infinite sets A and B of nonnegative integers such that the sumset A + B and the set P of odd prime numbers eventually coincide, that is,
P^-A+B.
(1.26)The answer is almost surely no, but there is no proof. Hornfeck [69, 70] proved that (1.26) is impossible if the set A is finite and I A I > 2.
There are other kinds of recognition problems: Does the set B contain a sumset?
Given sets A and B, does there exist a set C such that B + C C A? The twin prime
conjecture is a special case of this inverse problem: Let P be the set of odd prime numbers. Does there exist an infinite set A such that
A + {0, 2} c P?
Practically nothing is known about these questions.
We do not consider "partition problems" in this book. A partition of a posi- tive integer N is a representation of N as the sum of an unrestricted number of elements taken from a fixed set of positive integers. A good reference for the clas- sical approach to partitions is Andrews's monograph The Theory of Partitions [4].
For interesting examples of inverse theorems for partitions, see the papers of Cas- sels [15], Erdos [35], Erdos, Gordon, Rubel, and Straus [41], and Folkman [46].
An early version of this chapter appeared in Nathanson [92].
In this book we investigate only h-fold sumsets of finite sets. A subsequent volume (Nathanson [90]) will examine sums of infinite sets of integers in additive number theory. For example, it includes a deep and beautiful inverse theorem of Kneser [76] concerning the asymptotic density of sumsets, and an important recent improvement of this due to Bilu [10).
For a comprehensive treatment of many of the most important results on War- ing's problem and the Goldbach conjecture, see Nathanson, Additive Number The- ory: The Classical Bases [96]. There is no other recent book on additive number theory.