INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A05

SETS WITH MORE SUMS THAN DIFFERENCES

Melvyn B. Nathanson^{1}

Lehman College (CUNY), Bronx, New York 10468

                      melvyn.nathanson@lehman.cuny.edu

Received: 1/6/07, Accepted: 1/29/07, Published: 2/3/07

Abstract

Let A be a finite subset of the integers or, more generally, of any abelian group, written additively. The set A has more sums than differences if |A + A| > |A − A|. A set with this property is called an MSTD set. This paper gives explicit constructions of families of MSTD sets of integers.

Subject class: Primary 11B05, 11B13,11B25,11B75.

Keywords: Sumsets, difference sets, MSTD sets

1. MSTD Sets

Let G be an abelian group, written additively, and let A and B be subsets of G. We define the sumset

A + B = {a + b : a ∈ A, b ∈ B}

and the difference set

A − B = {a − b : a ∈ A, b ∈ B}.

In particular, if A = B, then

A + A = {a + a′ : a, a′ ∈ A}

and

A − A = {a − a′ : a, a′ ∈ A}.

1This is the text of a lecture at the SIAM Conference on Discrete Mathematics in Victoria, British Columbia, on June 27, 2006. This work was supported in part by grants from the NSA Mathematical Sciences Program and the PSC-CUNY Research Award Program.

Let |A| denote the cardinality of the set A.

Since G is abelian, for every pair a, b of distinct elements of G we have

a + b = b + a

but

a − b ≠ b − a

unless a − b has order 2. This suggests that if A is finite, then we should “usually” have |A − A| ≥ |A + A|. Thus, sets with more sums than differences (MSTD) should be rare. On the other hand, elements are omitted from A − A in pairs: If c ∉ A − A, then −c ∉ A − A. This provides an opportunity to construct sets with more sums than differences. We shall call a finite set A with |A + A| > |A − A| an MSTD set.

Let Z denote the ring of integers. For a, b ∈ Z, let [a, b] denote the interval of integers {a, a + 1, a + 2, …, b}. The set

A_{1} = {0, 2, 3, 4, 7, 11, 12, 14}

is an MSTD set with A_{1} + A_{1} = [0, 28] \ {1, 20, 27} and A_{1} − A_{1} = [−14, 14] \ {±6, ±13}, hence |A_{1} + A_{1}| = 26 and |A_{1} − A_{1}| = 25. The set

A_{2} = {0, 2, 3, 4, 7, 9, 13, 14, 16}

is also an MSTD set, with A_{2} +A_{2} = [0, 32]\{1, 24, 31} and A_{2} −A_{2} = [−16, 16]\{±8, ±15}, hence |A_{2} + A_{2}| = 30 and |A_{2} − A_{2}| = 29.

For every set A of integers and every integer x, we define the dilation x∗A = {xa : a ∈ A}. If A is an MSTD set of integers, then, for all integers x ≠ 0 and y, the set

x ∗ A + {y} = {xa + y : a ∈ A}

is also an MSTD set, and so the property of having MSTD is an affine invariant of the ring of integers.

In this paper we provide the first explicit constructions of infinite families of MSTD sets of integers. We also use the probability method to prove the existence of MSTD sets in certain finite abelian groups, and give an algorithm to construct MSTD sets of integers from MSTD sets in finite abelian groups. This answers a question posed by Nathanson [6].

The first examples of MSTD sets of integers appear in Marica [4] and Freiman-Pigarev [1]. Stein [11] and Hennecart-Robert-Yudin [3] also considered MSTD sets. Ruzsa [8, 9, 10] used probability methods to show the existence of MSTD sets. Hegarty [2] proved that there is no MSTD set of cardinality less than 8, and that the set A_{1} is, up to affine equivalence, the unique MSTD set of cardinality 8. Roesler [7] showed that, in the collection of all sets of size k in {1, 2, …, n}, the average number of sums does not exceed the average number of differences. On the other hand, Martin and O’Bryant [5] recently proved that a positive proportion of the sets in {1, 2, …, n} have more sums than differences.

2. A Construction of MSTD Sets of Integers

The subset A of an abelian group G is called symmetric with respect to a^{∗} if a^{∗} ∈ G and

A = a^{∗} − A.

If A is symmetric with respect to a^{∗}, then A has the same number of sums as differences, since

|A + A| = |A + (a^{∗} − A)| = |a^{∗} + (A − A)| = |A − A|.

For example, the arithmetic progression {a, a + m, a + 2m, …, a + (k − 1)m} is symmetric with respect to a^{∗} = 2a+(k−1)m. For every d ≥ 1, the d-dimensional generalized arithmetic progression

L = {a + x_{1}m_{1} + ⋯ + x_{d}m_{d} : ℓ_{i} ≤ x_{i} ≤ ℓ_{i} + k_{i} − 1 for i = 1, …, d}

in the group G is symmetric with respect to

a^{∗} = 2a + ∑_{i=1}^{d} (2ℓ_{i} + k_{i} − 1) m_{i}.

If A is a finite set of integers that is symmetric with respect to a^{∗}, then a^{∗} = min(A)+max(A).

In this section we shall construct infinite families of MSTD sets of integers. Each MSTD set will be constructed by adjoining a single element to a symmetric set that is a small perturbation of a generalized arithmetic progression.

We begin with an examination of the set A_{1} = {0, 2, 3, 4, 7, 11, 12, 14}. Observe that A_{1} \ {4} is symmetric with respect to 14, and that

A_{1} = {0, 2, 3, 4, 7, 11, 12, 14}
= {0, 2, 3, 7, 11, 12, 14} ∪ {4}
= {0, 2} ∪ {3, 7, 11} ∪ (14 − {0, 2}) ∪ {4}.

Hegarty and Roesler (personal communication) observed that this set belongs to the following infinite sequence of MSTD sets: For k ≥ 3, let

A^{∗} = {0, 2} ∪ {3, 7, 11, …, 4k − 1} ∪ {4k, 4k + 2}

and

A = A^{∗} ∪ {4}.

To see that the set A is an MSTD set of integers, we note that A^{∗} is symmetric and so |A^{∗}+A^{∗}| = |A^{∗}−A^{∗}|. Since 8 ∈ (A+A)\(A^{∗}+A^{∗}), it suffices to prove that |A−A| = |A^{∗}−A^{∗}|.

For this, we only need to show that A^{∗} − {4} ⊆ A^{∗} − A^{∗}, and this is true because

1 = 3 − 2 ∈ A^{∗} − A^{∗}
4 = 7 − 3 ∈ A^{∗} − A^{∗}
4k − 4 = (4k − 1) − 3 ∈ A^{∗} − A^{∗}
4k − 2 = (4k) − 2 ∈ A^{∗} − A^{∗}.

We can generalize this construction as follows.

Theorem 1 Let m, d, and k be integers such that m ≥ 4, 1 ≤ d ≤ m − 1, d ≠ m/2, and k ≥ 3 if d < m/2 and k ≥ 4 if d > m/2. Let

B = [0, m − 1] \ {d}
L = {m − d, 2m − d, …, km − d}
a^{∗} = (k + 1)m − 2d
A^{∗} = B ∪ L ∪ (a^{∗} − B)
A = A^{∗} ∪ {m}.

The set A is an MSTD set of integers.

Proof. The set A^{∗} is symmetric with respect to a^{∗}, and so

|A^{∗} + A^{∗}| = |A^{∗} − A^{∗}|.

Also, 2m ∈ A + A. We shall prove that 2m ∉ A^{∗} + A^{∗}.

Assume that 2m = a + a′ ∈ A^{∗} + A^{∗} with a ≤ a′. Then m ≤ a′ ≤ 2m and so a′ ∉ B. Also,

min(a^{∗} − B) ≥ a^{∗} − (m − 1) = km − 2d + 1 > 2m

if (k − 2)m ≥ 2d, and this inequality holds if k ≥ 4 or if k ≥ 3 and d ≤ m/2. Thus, a′ ∉ a^{∗} − B.

Since 3m − d > 2m, we have a′ ∈ L only if a′ = 2m − d. Therefore, if a′ ∈ A^{∗}, then a′ = 2m − d and a = d. Since d ∉ B and m − d ≠ d, it follows that a ∉ A^{∗} and so 2m ∉ A^{∗} + A^{∗}. This proves that

2m ∈ (A + A) \ (A^{∗} + A^{∗})

and so

|A + A| > |A^{∗} + A^{∗}|.

We shall prove that A − A = A^{∗} − A^{∗}. It suffices to show that A^{∗} − {m} ⊆ A^{∗} − A^{∗}. Note that 0 ∈ A^{∗} implies that A^{∗} ⊆ A^{∗} − A^{∗}.

Since m = (2m − d) − (m − d) ∈ L − L, it follows that

{m} − B = {1, 2, …, m − 1, m} \ {m − d} ⊆ B ∪ (L − L) ⊆ A^{∗} − A^{∗}

and also B − {m} ⊆ A^{∗} − A^{∗}. Similarly,

L − {m} = {−d, m − d, …, (k − 1)m − d} ⊆ L ∪ {−d}.

If 1 < d < m − 1, then {1, d + 1} ⊆ B ∪ L ⊆ A^{∗} and d ∈ A^{∗} − A^{∗}. If d = 1, then {2, 3} ⊆ B ∪ L and 1 ∈ A^{∗} − A^{∗}. If d = m − 1, then m + 1 = 2m − d ∈ L, 2 ∈ B, and so m − 1 ∈ B − L ⊆ A^{∗} − A^{∗}. Therefore,

L − {m} ⊆ A^{∗} − A^{∗}.

Finally, we must prove that

a^{∗} − B − {m} = {a^{∗} − b − m : b ∈ B} ⊆ A^{∗} − A^{∗}.

If d = m − 1, then a^{∗} = (k − 1)m + 2 and

A^{∗} = [0, m − 2] ∪ {1, m + 1, 2m + 1, …, (k − 1)m + 1} ∪ (a^{∗} − [0, m − 2]).

If b ∈ [0, m − 4], then

a^{∗} − b − m = a^{∗} − (b + 2) − (m − 2) ∈ A^{∗} − A^{∗}.

If b = m − 3, then

a^{∗} − b − m = a^{∗} − (m − 4) − (m + 1) ∈ A^{∗} − A^{∗}.

If b = m − 2, then

a^{∗} − b − m = a^{∗} − (m − 3) − (m + 1) ∈ A^{∗} − A^{∗}.

Suppose that 1 ≤ d ≤ m − 2. For b ∉ {d − 1, m − 1}, we have b + 1 ∈ B and so a^{∗} − (b + 1) ∈ a^{∗} − B. Since m − 1 ∈ B, we have

a^{∗} − b − m = (a^{∗} − (b + 1)) − (m − 1) ∈ A^{∗} − A^{∗}.

If b = m − 1, then

a^{∗} − b − m = (k − 1)m − 2d + 1
= ((k − 1)m − d) − (d − 1)
∈ L − B ⊆ A^{∗} − A^{∗}.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A05 6

If b = d − 1 and d ≠ m − 2, then 2 ≤ d ≤ m − 3 and

a^{∗} − b − m = a^{∗} − (d − 1) − m
= (a^{∗} − (d + 1)) − (m − 2)
∈ A^{∗} − A^{∗}.

The last case is b = d − 1 and d = m − 2. Then m ≥ 5, k ≥ 4, and

a^{∗} = (k + 1)m − 2d = (k − 1)m + 4

and

a^{∗} − b − m = (k − 1)m + 4 − (d − 1) − m
= (k − 3)m + 7
= (k − 2)m + 2 − (m − 5)
∈ L − B ⊆ A^{∗} − A^{∗}.

Therefore,

a^{∗} − B − m ⊆ A^{∗} − A^{∗}

and so A − A = A^{∗} − A^{∗}. This completes the proof. □ The MSTD set A_{1} is the case m = 4, d = 1, and k = 3 of Theorem 1. Next we consider the MSTD set A_{2} = {0, 2, 3, 4, 7, 9, 13, 14, 16}. This set is a perturbation

of a two-dimensional generalized arithmetic progression:

A_{2} = {0, 2, 3, 4, 7, 9, 13, 14, 16}
= ({0, 2} ∪ {3, 7} ∪ {9, 13} ∪ {14, 16}) ∪ {4}
= {0, 2} ∪ {3 + 6x_{1} + 4x_{2} : 0 ≤ x_{1} ≤ 1 and 0 ≤ x_{2} ≤ 1} ∪ (16 − {0, 2}) ∪ {4}.

The following sequence of sets extends this example.

Theorem 2 For k ≥ 2, let

A^{∗} = {0, 2} ∪ {3, 7, 11, …, 4k − 1} ∪ {9, 13, 17, …, 4k + 5} ∪ {4k + 6, 4k + 8}
= {0, 2} ∪ {3 + 6x_{1} + 4x_{2} : 0 ≤ x_{1} ≤ 1 and 0 ≤ x_{2} ≤ k − 1} ∪ (4k + 8 − {0, 2})

and

A = A^{∗} ∪ {4}.

Then A is an MSTD set of integers.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A05 7

Proof. It suffices to show that 4 − A^{∗} ⊆ A^{∗} − A^{∗}, and this follows from

1 = 3 − 2 ∈ A^{∗} − A^{∗}
4 = 7 − 3 ∈ A^{∗} − A^{∗}
5 = 7 − 2 ∈ A^{∗} − A^{∗}
4k + 2 = (4k + 5) − 3 ∈ A^{∗} − A^{∗}
4k + 4 = (4k + 6) − 2 ∈ A^{∗} − A^{∗}.

This completes the proof. □ Lemma 1 Let m ≥ 4 and let r and s be integers such that

r ≥ 1 and r + 1 ≤ s ≤ m − 1.

Let

B = [0, r − 1] ∪ [s, m − 1].

If

s ≤ 2r − 1 and 2s ≤ m + r − 1  (1)

then

B + B = [0, 2m − 2].

If

s ≤ 2r − 1 or 2s ≤ m + r − 1  (2)

then

B − B = [1 − m, m − 1].

Let m ≥ 4 and 1 ≤ r ≤ m − 2 and let

B = [0, m − 1] \ {r}.

Then

B − B = [−(m − 1), m − 1].

If 2 ≤ r ≤ m − 3, then

B + B = [0, 2m − 2].

If r = 1, then

B + B = [0, 2m − 2] \ {1}.

If r = m − 2, then

B + B = [0, 2m − 2] \ {2m − 3}.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A05 8

Proof. This is a straightforward calculation. If s ≤ 2r − 1 and 2s ≤ m + r − 1, then

2A = [0, 2r − 2] ∪ [s, m + d − 2] ∪ [2s, 2m − 2] = [0, 2m − 2].

If s ≤ 2r − 1 or 2s ≤ m + r − 1, then

(A − A) ∩ [0, m − 1] = [0, r − 1] ∪ [s − r + 1, m − 1] ∪ [0, m − 1 − s] = [0, m − 1] and so A − A = [1 − m, m − 1].

If s = r + 1, then 1 ≤ r ≤ m − 1 and B = [0, m − 1] \ {r}. Condition (2) is always satisfied, and condition (1) is equivalent to 2 ≤ r ≤ m − 3. The proofs for the cases r = 1 and r = m − 2 are similar. □

The following theorem produces explicit families of MSTD sets constructed from 2dimensional arithmetic progressions.

Theorem 3 Let m and d be integers such that

(i) m ≥ 4 (ii) 1 ≤ d ≤ m − 2 and d ≠ m/2 (iii) if d < m/2, then d ≠ m/3 (iv) if d > m/2, then d ≠ 2m/3.

For k ≥ 3, let

a^{∗} = (k + 2)m
B = [0, m − 1] \ {d}
L = {2m − d, 3m − d, …, km − d} ∪ {2m + d, 3m + d, …, km + d}
A^{∗} = B ∪ L ∪ ({a^{∗}} − B)
A = A^{∗} ∪ {m}.

Then A is an MSTD set of integers.

Proof. The set A^{∗} is symmetric with respect to a^{∗} and so

|A^{∗} + A^{∗}| = |A^{∗} − A^{∗}|.

Since

A^{∗} ∩ [0, 2m] = B ∪ {2m − d}

and d ∉ B, we have 2m ∉ A^{∗} + A^{∗} and so 2m ∈ (A + A) \ (A^{∗} + A^{∗}). Thus, it suffices to show that A − A = A^{∗} − A^{∗}. We need only prove that

A^{∗} − {m} ⊆ A^{∗} − A^{∗}.

Since d ∈ B − B by Lemma 1 and since m = (3m − d) − (2m − d) ∈ L − L, it follows that

{m} − B = [1, m] \ {m − d} ⊆ (B − B) ∪ (L − L) ⊆ A^{∗} − A^{∗}

and so

B − {m} ⊆ A^{∗} − A^{∗}.

The inequalities 1 ≤ d < m/2 and d ≠ m/3 imply that m − 2d ≠ d, m − 2d ∈ B, and so

m + d = (2m − d) − (m − 2d) ∈ L − B.

The inequalities m/2 < d ≤ m − 2 and d ≠ 2m/3 imply that 2m − 2d ≠ d, 2m − 2d ∈ B, and so

m + d = (3m − d) − (2m − 2d) ∈ L − B.

Also, m − d ∈ B. Therefore,

L − {m} ⊆ B ∪ L ∪ (L − B) ⊆ A^{∗} − A^{∗}.

Suppose that 1 ≤ d ≤ m − 3. If b ∈ B and b ≠ m − 1, then Lemma 1 implies there exist b_{1}, b_{2} ∈ B such that m + b = b_{1} + b_{2}, and so

a^{∗} − b − m = (k + 1)m − b = ((k + 2)m − b_{1}) − b_{2} ∈ ({a^{∗}} − B) − B.

If d = m − 2 and b = m − 3, then 2m − 3 ∉ B + B but m ≥ 5 and

a^{∗} − b − m = km + 3 = (km + d) − (d − 3) ∈ (a^{∗} − B) − B.

For all d ∈ [1, m − 2] we have

a^{∗} − (m − 1) − m = km + 1 = (km + d) − (d − 1) ∈ L − B

and so

{a^{∗}} − B − {m} ⊆ (({a^{∗}} − B) − B) ∪ (L − B) ⊆ A^{∗} − A^{∗}.

This completes the proof. □

The next theorem gives a method to construct MSTD sets of integers from generalized arithmetic progressions of dimension d for every d ≥ 1.

Theorem 4 Let m ≥ 4 and let B be a subset of [0, m − 1] such that

B + B = [0, 2m − 2]

and

B − B = [−m + 1, m − 1].

Let L^{∗} be a (d − 1)-dimensional arithmetic progression contained in [0, m − 1] \ B such that min(L^{∗}) − 1 ∈ B. Let k ≥ 2 and let L be the d-dimensional arithmetic progression

L = (m − L^{∗}) + m ∗ [1, k].

Let

a^{∗} = min(L) + max(L) = (k + 3)m − min(L^{∗}) − max(L^{∗})

and

A^{∗} = B ∪ L ∪ ({a^{∗}} − B).

Then A = A^{∗} ∪ {m} is an MSTD set of integers.

Proof. The set A^{∗} is symmetric with respect to a^{∗} and so

|A^{∗} + A^{∗}| = |A^{∗} − A^{∗}|.

Since B ∩ L^{∗} = ∅ and

A^{∗} ∩ [0, 2m] = B ∪ ({2m} − L^{∗})

we have 2m ∉ A^{∗} + A^{∗} and so 2m ∈ (A + A) \ (A^{∗} + A^{∗}). The set A will be an MSTD set if A − A = A^{∗} − A^{∗}, and for this it suffices to show that

A^{∗} − {m} ⊆ A^{∗} − A^{∗}.

Note that the condition B + B = [0, 2m − 2] implies that 0, m − 1 ∈ B. In particular, A^{∗} ⊆ A^{∗} − A^{∗}.

Since m ∈ L − L, we have

{m} − B ⊆ [1, m] ⊆ (B − B) ∪ (L − L) ⊆ A^{∗} − A^{∗}

and so

B − {m} ⊆ A^{∗} − A^{∗}.

Also,

L − {m} = ({m} − L^{∗}) ∪ ((m − L^{∗}) + m ∗ [1, k − 1])
⊆ (B − B) ∪ L
⊆ A^{∗} − A^{∗}.

Finally, let a^{∗} − b − m ∈ a^{∗} − B − {m}. If b ≠ m − 1, then there exist b_{1}, b_{2} ∈ B such that m + b = b_{1} + b_{2}, and so

a^{∗} − b − m = (a^{∗} − b_{1}) − b_{2} ∈ (a^{∗} − B) − B ∈ A^{∗} − A^{∗}.

If b = m − 1, then

a^{∗} − b − m = a^{∗} − 2m + 1
= (k + 3)m − min(L^{∗}) − max(L^{∗}) − 2m + 1
= (k + 1)m − max(L^{∗}) − (min(L^{∗}) − 1)
∈ L − B
⊆ A^{∗} − A^{∗}

because (m − max(L^{∗})) + km ∈ L and min(L^{∗}) − 1 ∈ B. This completes the proof. □

Theorem 4 is a powerful tool for constructing MSTD sets. For example, let P be any (d − 1)-dimensional arithmetic progression of nonnegative integers with min(P) = 0 and max(P) = M. Choose integers r and s so that

r ≥ M + 2

and

r + M + 1 ≤ s ≤ 2r − 1.

Choose an integer m such that

2s ≤ m + r − 1.

Let

B = [0, r − 1] ∪ [s, m − 1]

and

L^{∗} = {r} + P.

Then L^{∗} is a (d − 1)-dimensional arithmetic progression contained in [0, m − 1] \ B, and min(L^{∗})−1 = r−1 ∈ B. By Lemma 1, we have B+B = [0, m−2] and B−B = [1−m, m−1]. The sets B and L^{∗} satisfy the conditions of Theorem 4.

A slight variation of the proof of Theorem 4 gives the following result.

Theorem 5 Let m ≥ 4 and let B be a subset of [0, m − 1] such that

B + B = [0, 2m − 2]

and

B − B = [−m + 1, m − 1].

Let L^{∗} be a (d − 1)-dimensional arithmetic progression contained in [0, m − 1] \ B such that min(L^{∗}) − 1 ∈ B and m ∉ L^{∗} + L^{∗}. Let k ≥ 2 and let L be the d-dimensional arithmetic progression

L = (m − L^{∗}) + m ∗ [0, k].

Let

a^{∗} = min(L) + max(L) = (k + 3)m − min(L^{∗}) − max(L^{∗})

and

A^{∗} = B ∪ L ∪ ({a^{∗}} − B).

Then A = A^{∗} ∪ {m} is an MSTD set of integers.

3. Finite Abelian Groups and Lattices

Let A be a nonempty subset of an abelian group. For all integers h ≥ 0, we define the h-fold sumset hA inductively:

0A = {0}
1A = A
2A = A + A
hA = (h − 1)A + A for all h ≥ 3.

For nonnegative integers h and k we define the generalized sum-difference set

hA − kA = {a_{1} + ⋯ + a_{h} − a′_{1} − ⋯ − a′_{k} : a_{j} ∈ A
for j = 1…. , h and a′_{ℓ} ∈ A for ℓ = 1, …, k}.

Let G be a finite abelian group. Since every finite abelian group is isomorphic to a direct product of cyclic groups, we can assume that

G = Z/m_{1}Z × ⋯ × Z/m_{d}Z,

where m_{1}, …, m_{d} are integers greater than 1. Then

|G| = m_{1} ⋯ m_{d}.

Let Z^{d} denote the standard d-dimensional integer lattice. The sublattice Λ_{G} of Z^{d} determined by the group G is

Λ_{G} = {(q_{1}m_{1}, …, q_{d}m_{d}) ∈ Z^{d} : q_{i} ∈ Z for i = 1, …, d}.

The fundamental integer parallelepiped in Z^{d} defined by G is the set

P_{G} = {(r_{1}, …, r_{d}) ∈ Z^{d} : 0 ≤ r_{i} ≤ m_{i} − 1 for i = 1…. , d}

Every lattice point in Z^{d} is uniquely the sum of an element in P_{G} and an element in Λ_{G}, that is,

Z^{d} = P_{G} ⊕ Λ_{G}.  (3)

We define the canonical embedding φ of G into the fundamental integer parallelepiped P_{G} of the lattice Λ_{G} by

φ(u_{1} + m_{1}Z, …, u_{d} + m_{d}Z) = (r_{1}, …, r_{d})  (4)

where r_{i} is the unique integer such that

r_{i} ≡ u_{i} (mod m_{i})  (5)

and

0 ≤ r_{i} ≤ m_{i} − 1.  (6)

We define a map

π : Z^{d} → P_{G}

by

(x_{1}, …, x_{d}) ↦ (r_{1}, …, r_{d})

where

r_{i} ≡ x_{i} (mod m_{i})

and

0 ≤ r_{i} ≤ m_{i} − 1.

For integers s ≤ t, we define

Λ_{G}(s, t) = {(q_{1}m_{1}, …, q_{d}m_{d}) ∈ Λ_{G} : s ≤ q_{i} < t for i = 1, …, d}.

Then

|Λ_{G}(s, t)| = (t − s)^{d}  (7)

Λ_{G}(s_{1}, t_{1}) + Λ_{G}(s_{2}, t_{2}) = Λ_{G}(s_{1} + s_{2}, t_{1} + t_{2} − 1)  (8)

and

Λ_{G}(s_{1}, t_{1}) − Λ_{G}(s_{2}, t_{2}) = Λ_{G}(s_{1} − t_{2} + 1, t_{1} − s_{2}).  (9)

In particular, we have the generalized sum-difference relation

hΛ_{G}(0, t) − kΛ_{G}(0, t) = Λ_{G}(−kt + k, ht − h + 1).  (10)

Lemma 2 Let A be a nonempty subset of the finite abelian group G = Z/m_{1}Z×⋯×Z/m_{d}Z. For all integers h ≥ 1 and k ≥ 0,

(i)

φ(hA − kA) = π(hφ(A) − kφ(A))

(ii)

hφ(A) − kφ(A) ⊆ φ(hA − kA) + Λ_{G}(−k, h).

(iii)

φ(hA − kA) ⊆ hφ(A) − kφ(A) + Λ_{G}(−h + 1, k + 1).

Proof. (i) If (s_{1}, …, s_{d}) ∈ φ(hA−kA), then the group G contains elements a_{1}, …, a_{h}, a′_{1}, …, a′_{k} such that

(s_{1}, …, s_{d}) = φ(a_{1} + ⋯ a_{h} − a′_{1} − ⋯ − a′_{k}).

For j = 1, …, h and i = 1, …, d, we can write

a_{j} = (r_{1j} + m_{1}Z, …, r_{dj} + m_{d}Z)  (11)

with

0 ≤ r_{ij} ≤ m_{i} − 1.  (12)

Similarly, for ℓ = 1, …, k and i = 1, …, d, we have

a′_{ℓ} = (r′_{1ℓ} + m_{1}Z, …, r′_{dℓ} + m_{d}Z)  (13)

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A05 15

with

0 ≤ r′_{iℓ} ≤ m_{i} − 1.  (14)

There exist unique integers q_{i} and s′_{i} such that 0 ≤ s′_{i} ≤ m_{i} − 1 and

∑_{j=1}^{h} r_{ij} − ∑_{ℓ=1}^{k} r′_{iℓ} = q_{i}m_{i} + s′_{i}.

The ith component of the group element

a_{1} + ⋯ + a_{h} − a′_{1} − ⋯ − a′_{ℓ} ∈ hA − kA

is

(∑_{j=1}^{h} r_{ij} − ∑_{ℓ=1}^{k} r′_{iℓ}) + m_{i}Z = s′_{i} + m_{i}Z

and so

(s_{1}, …, s_{d}) = φ (a_{1} + ⋯ + a_{h} − a′_{1} − ⋯ − a′_{ℓ}) = (s′_{1}, …, s′_{d}).

Therefore, s_{i} = s′_{i} for i = 1, …, d and

∑_{j=1}^{h} r_{ij} − ∑_{ℓ=1}^{k} r′_{iℓ} = q_{i}m_{i} + s_{i}.  (15)

Since

φ(a_{j}) = (r_{1j}, …, r_{dj}) ∈ φ(A) for j = 1, …, h

and

φ(a′_{ℓ}) = (r′_{1ℓ}, …, r′_{dℓ}) ∈ φ(A) for ℓ = 1, …, k

it follows that

φ(a_{1}) + ⋯ + φ(a_{h}) − φ(a′_{1}) − ⋯ − φ(a′_{k})

= (∑_{j=1}^{h} r_{1j} − ∑_{ℓ=1}^{k} r′_{1ℓ}, …, ∑_{j=1}^{h} r_{dj} − ∑_{ℓ=1}^{k} r′_{dℓ})

= (q_{1}m_{1} + s_{1}, …, q_{d}m_{d} + s_{d})

∈ hφ(A) − kφ(A)

and so

π (φ(a_{1}) + ⋯ + φ(a_{h}) − φ(a′_{1}) − ⋯ − φ(a′_{k})) = π(q_{1}m_{1} + s_{1}, …, q_{d}m_{d} + s_{d})
= (s_{1}, …, s_{d})
∈ π(hφ(A) − kφ(A)).

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A05 16

Therefore,

φ(hA − kA) ⊆ π(hφ(A) − kφ(A)).

Similarly,

π(hφ(A) − kφ(A)) ⊆ φ(hA − kA).

This proves (i).

(ii) Every lattice point in the generalized sum-difference set hφ(A)−kφ(A) is of the form

φ(a_{1}) + ⋯ + φ(a_{h}) − φ(a′_{1}) − ⋯ − φ(a′_{k})

with a_{1}, …, a_{h}, a′_{1}, …, a′_{k} ∈ G. Using the same notation as above, we see that the ith coordinate of this lattice point is

∑_{j=1}^{h} r_{ij} − ∑_{ℓ=1}^{k} r′_{iℓ} = q_{i}m_{i} + s_{i}  (16)

where

0 ≤ s_{i} ≤ m_{i} − 1

and

(s_{1}, …, s_{d}) ∈ φ(hA − kA).

Moreover,

−km_{i} ≤ −k(m_{i} − 1) ≤ q_{i}m_{i} + s_{i} ≤ h(m_{i} − 1) < hm_{i}

and so

−k ≤ q_{i} < h  (17)

for i = 1, …, d. Therefore,

(q_{1}m_{1}, …, q_{d}m_{d}) ∈ Λ_{G}(−k, h)

and

φ(a_{1}) + ⋯ +φ(a_{h}) − φ(a′_{1}) − ⋯ − φ(a′_{k})
= (q_{1}m_{1} + s_{1}, …, q_{d}m_{d} + s_{d})
= (s_{1}, …, s_{d}) + (q_{1}m_{1}, …, q_{d}m_{d})
∈ φ(hA − kA) + Λ_{G}(−k, h).

(iii) If (s_{1}, …, s_{d}) ∈ φ(hA − kA), then there exist elements a_{1}, …, a_{h}, a′_{1}, …, a′_{k} ∈ G such that

(s_{1}, …, s_{d}) = φ(a_{1} + ⋯ + a_{h} − a′_{1} − ⋯ − a′_{k}).

Since the ith component of φ(a_{1}) + ⋯ + φ(a_{h}) − φ(a′_{1}) − ⋯ − φ(a′_{k}) is (16), we have

φ(a_{1}) + ⋯ +φ(a_{h}) − φ(a′_{1}) − ⋯ − φ(a′_{k})
= (q_{1}m_{1} + s_{1}, …, q_{d}m_{d} + s_{d})
= (s_{1}, …, s_{d}) + (q_{1}m_{1}, …, q_{d}m_{d}).

Equivalently,

(s_{1}, …, s_{d}) = φ(a_{1}) + ⋯ + φ(a_{h}) − φ(a′_{1}) − ⋯ − φ(a′_{k}) − (q_{1}m_{1}, …, q_{d}m_{d})
∈ hφ(A) − kφ(A) − Λ_{G}(−k, h)
= hφ(A) − kφ(A) + Λ_{G}(−h + 1, k + 1).

This completes the proof. □

Theorem 6 Let A be a nonempty subset of the finite abelian group G = Z/m_{1}Z × ⋯ × Z/m_{d}Z. Let φ : G → P_{G} be the canonical embedding defined by (4), (5), and (6). Let

B_{t} = φ(A) + Λ_{G}(0, t) ⊆ Z^{d}.

Then

hB_{t} − kB_{t} ⊆ φ(hA − kA) + Λ_{G}(−kt, ht)

and

hB_{t} − kB_{t} ⊇ φ(hA − kA) + Λ_{G}(−kt + h + k − 1, ht − h − k + 1).

Moreover,

|hB_{t} − kB_{t}| ≤ |hA − kA|((h + k)t)^{d}

and

|hB_{t} − kB_{t}| ≥ |hA − kA|((h + k)t − 2(h + k − 1))^{d}.

Proof. By Lemma 2 (ii),

hφ(A) − kφ(A) ⊆ φ(hA − kA) + Λ_{G}(−k, h).

It follows from (8)–(10) that

hB_{t} − kB_{t} = h (φ(A) + Λ_{G}(0, t)) − k (φ(A) + Λ_{G}(0, t))
= hφ(A) − kφ(A) + hΛ_{G}(0, t) − kΛ_{G}(0, t)
⊆ φ(hA − kA) + Λ_{G}(−k, h) + Λ_{G}(−kt + k, ht − h + 1)
= φ(hA − kA) + Λ_{G}(−kt, ht).

Applying (3) and (7), we obtain

|hB_{t} − kB_{t}| ≤ |φ(hA − kA) + Λ_{G}(−kt, ht)|
= |φ(hA − kA)||Λ_{G}(−kt, ht)|
= |hA − kA|((h + k)t)^{d}.

Similarly, by Lemma 2 (iii),

φ(hA − kA) ⊆ hφ(A) − kφ(A) + Λ_{G}(−h + 1, k + 1).

and so

hB_{t} − kB_{t} = h (φ(A) + Λ_{G}(0, t)) − k (φ(A) + Λ_{G}(0, t))
= hφ(A) − kφ(A) + Λ_{G}(−kt + k, ht − h + 1)
= hφ(A) − kφ(A) + Λ_{G}(−h + 1, k + 1) + Λ_{G}(−kt + h + k − 1, ht − h − k + 1)
⊇ φ(hA − kA) + Λ_{G}(−kt + h + k − 1, ht − h − k + 1).

Therefore,

|hB_{t} − kB_{t}| ≥ |hA − kA|((h + k)t − 2(h + k − 1))^{d}.
□

Theorem 7 Let G = Z/m_{1}Z × ⋯ × Z/m_{d}Z be a finite abelian group, and let φ be the canonical embedding of G into the fundamental integer parallelepiped P_{G} of the lattice Λ_{G}. Let A be a nonempty subset G and let h_{1}, h_{2}, k_{1}, k_{2} be nonnegative integers such that h_{1} ≥ 1, h_{2} ≥ 1,

h_{1} + k_{1} = h_{2} + k_{2}

and

|h_{1}A − k_{1}A| > |h_{2}A − k_{2}A|.

For every sufficiently large integer t, the set of lattice points

B_{t} = φ(A) + Λ_{G}(0, t)

satisfies the inequality

|h_{1}B_{t} − k_{1}B_{t}| > |h_{2}B_{t} − k_{2}B_{t}|.

Proof. Let h_{1} + k_{1} = h_{2} + k_{2} = c. By Theorem 6,

\frac{|h_{1}B_{t} − k_{1}B_{t}|}{|h_{2}B_{t} − k_{2}B_{t}|} ≥ \frac{|h_{1}A − k_{1}A|((h_{1} + k_{1})t − 2(h_{2} + k_{2} − 1))^{d}}{|h_{2}A − k_{2}A|((h_{2} + k_{2})t)^{d}}

≥ \frac{|h_{1}A − k_{1}A|}{|h_{2}A − k_{2}A|} \frac{(ct − 2c + 2)^{d}}{(ct)^{d}}

= \frac{|h_{1}A − k_{1}A|}{|h_{2}A − k_{2}A|} (1 − \frac{2c − 2}{ct})^{d}

Since

lim_{t→∞} (1 − \frac{2c − 2)}{ct})^{d} = 1

it follows that

|h_{1}B_{t} − k_{1}B_{t}| > |h_{2}B_{t} − k_{2}B_{t}|

for all sufficiently large t. This completes the proof. □

Theorem 8 Let G = Z/m_{1}Z × ⋯ × Z/m_{d}Z be a finite abelian group, and let φ be the canonical embedding of G into the fundamental integer parallelepiped P_{G} of the lattice Λ_{G}. Let A be a nonempty subset G such that

|A + A| > |A − A|.

For every sufficiently large integer t, the set of lattice points

B_{t} = φ(A) + Λ_{G}(0, t)

satisfies the inequality

|B_{t} + B_{t}| > |B_{t} − B_{t}|.

Proof. Apply Theorem 7 with (h_{1}, k_{1}) = (2, 0) and (h_{2}, k_{2}) = (1, 1). □

The final step is to embed a finite set of lattice points in Z^{d} into the integers. For a = (a_{1}, …, a_{d}) ∈ Z^{d}, define

∥a∥ = max(|a_{i}| : i = 1, …, d).

Lemma 3 Let m be a positive integer, and consider the group homomorphism ψ : Z^{d} → Z defined by

ψ(a) = ∑_{i=1}^{d} a_{i}m^{i−1}

for all a = (a_{1}, …, a_{d}) ∈ Z^{d}. If ∥a∥ < m and ψ(a) = 0, then a = 0.

Proof. Suppose that a ∈ Z^{d} satisfies ∥a∥ < m and ψ(a) = 0. The inequality ∥a∥ < m implies that |a_{i}| ≤ m − 1 for all i = 1, …, d. If a ≠ 0, let r be the largest integer such that a_{r} ≠ 0. Then ψ(a) = 0 implies that

a_{r}m^{r−1} = − ∑_{i=1}^{r−1} a_{i}m^{i−1}

but this is impossible since

|a_{r}m^{r−1}| ≥ m^{r−1}

and

|− ∑_{i=1}^{r−1} a_{i}m^{i−1}| ≤ m^{r−1} − 1.

This completes the proof. □

Theorem 9 Let L be a positive integer, and let A be a nonempty finite subset of Z^{d}. Let m be an integer such that

m > 2L max(∥a∥ : a ∈ A).

Define ψ : Z^{d} → Z by

ψ(a_{1}, …, a_{d}) = ∑_{i=1}^{d} a_{i}m^{i−1}.

If h and k are nonnegative integers such that

h + k ≤ L

then

|hA − kA| = |hψ(A) − kψ(A)|.

Proof. Since ψ is a homomorphism, we have

ψ(hA − kA) = hψ(A) − kψ(A).

We must show that the map ψ : hA − kA → Z is one-to-one. Let u, v ∈ hA − kA such that ψ(u) = ψ(v). The set A contains lattice points a_{j} = (a_{1,j}, …, a_{d,j}) and b_{j} = (b_{1,j}, …, b_{d,j}) for j = 1, …, h, and lattice points a′_{ℓ} = (a′_{1,ℓ}, …, a′_{d,ℓ}) and b′_{ℓ} = (b′_{1,ℓ}, …, b′_{d,ℓ}) for ℓ = 1, …, k such that

u = a_{1} + ⋯ + a_{h} − a′_{1} − ⋯ − a′_{k}

and

v = b_{1} + ⋯ + b_{h} − b′_{1} − ⋯ − b′_{k}.

The ith coordinate of u is ∑_{j=1}^{h} a_{i,j} − ∑_{ℓ=1}^{k} a′_{i,ℓ}, and so

∥u∥ ≤ ∑_{j=1}^{h} ∥a_{j}∥ + ∑_{ℓ=1}^{k} ∥a′_{k}∥

≤ (h + k) max(∥a∥ : a ∈ A)

< \frac{m}{2}.

Similarly,

∥v∥ ≤ ∑_{j=1}^{h} ∥b_{j}∥ + ∑_{ℓ=1}^{k} ∥b′_{k}∥ < \frac{m}{2}

and so

∥u − v∥ < m.

If ψ(u) = ψ(v), then ψ(u − v) = 0. By Lemma 3, it follows that u = v and the map ψ is one-to-one. □

4. A Counting Argument

In this section we prove that the finite abelian group

G = Z/nZ × Z/2Z

contains MSTD sets.

Let Ω be the set of all subsets of G of the form

A = {(i + nZ, ε_{i} + 2Z) : i = 0, 1, …, n − 1 and ε_{i} ∈ {0, 1}}.

Then |Ω| = 2^{n} and |A| = n for all A ∈ Ω. If g = (b + nZ, δ + 2Z) is in the difference set A − A, then there exist a = (i + nZ, ε_{i} + 2Z) ∈ A and a′ = (j + nZ, ε_{j} + 2Z) ∈ A such that g = a − a′, or, equivalently,

(b + nZ, δ + 2Z) = (i + nZ, ε_{i} + 2Z) − (j + nZ, ε_{j} + 2Z)
= (i − j + nZ, ε_{i} − ε_{j} + 2Z).

If b + nZ = nZ, then i ≡ j (mod n) and so i = j. It follows that ε_{i} = ε_{j} and δ ≡ 0 (mod 2). This implies that for every set A ∈ Ω, if (nZ, δ + 2Z) ∈ A − A, then δ ≡ 0 (mod 2), that is,

(nZ, 1 + 2Z) ∉ A − A

and so

|A − A| ≤ |G| − 1 = 2n − 1.

We shall prove that there exists A ∈ Ω such that A + A = G, that is,

|A + A| = 2n

and so A is an MSTD set. Indeed, we shall prove that A + A = G for almost all sets A ∈ Ω.

Theorem 10 Let G = Z/nZ × Z/2Z and let Ω be the set of all subsets of G of the form

A = {(i + nZ, ε_{i} + 2Z) : i = 0, 1, …, n − 1 and ε_{i} ∈ {0, 1}}.

Let Ψ(G) denote the number of A ∈ Ω such that A + A = G. Then

Ψ(G) ≥ \begin{cases}2^{n} (1 − \frac{\sqrt{2}n}{2^{n/2}}) & if n is odd \\ 2^{n} (1 − \frac{2n}{2^{n/2}} ) & if n is even. \end{cases}

Proof. Let g = (b + nZ, δ + 2Z) ∈ G and let ϕ(g) denote the number of A ∈ Ω such that g ∉ A + A. Then

|Ω| − Ψ(G) ≤ ∑_{g∈G} ϕ(g).

Consider first the case that n is odd. There is a unique congruence class a_{0} + nZ ∈ Z/nZ such that 2a_{0} ≡ b (mod n), and (a_{0} + nZ, ε + 2Z) ∈ A for some ε ∈ {0, 1}. If δ ≡ 0 (mod 2), then

g = (b + nZ, 2Z) = (a_{0} + nZ, ε + 2Z) + (a_{0} + nZ, ε + 2Z) ∈ A + A

and ϕ(g) = 0.

If δ ≡ 1 (mod 2), then (a_{0} +nZ, ε+2Z)+(a_{0} +nZ, ε+2Z) ≠ g. If a ≢ a_{0} (mod n), then b − a ≢ a (mod n). It follows that Z/nZ \ {a_{0} + nZ} is the union of (n − 1)/2 disjoint sets of the form {a_{j} + nZ, b − a_{j} + nZ}, for j = 1, …, (n − 1)/2. Let {ε_{j}}_{j=0}^{(n−1)/2} be any sequence of 0’s and 1’s. We define the sequence {ε′_{j}}_{j=1}^{(n−1)/2} by

ε′_{j} = δ − ε_{j} + 1.

Defining the set A ∈ Ω by A = {(a_{j} + nZ, ε_{j} + 2Z}_{j=0}^{(n−1)/2} ∪ {(b − a_{j} + nZ, ε′_{j} + 2Z}_{j=1}^{(n−1)/2} we see that g ∉ A + A. It follows that

ϕ(g) = \begin{cases}0 & if δ ≡ 0 (mod 2) \\ 2^{(n+1)/2} & if δ ≡ 1 (mod 2) \end{cases}

and so

∑_{g∈G} ϕ(g) = (\frac{|G|}{2}) 2^{(n+1)/2} = n2^{(n+1)/2}

and

Ψ(G) ≥ |Ω| − ∑_{g∈G} ϕ(g)

= 2^{n} − n2^{(n+1)/2}

= 2^{n} (1 − \frac{\sqrt{2}n}{2^{n/2}}) .

There is a similar argument in the case that n is even. If b is odd, then the congruence 2x ≡ b (mod n) has no solution, and Z/nZ can be partitioned into n/2 pairs of elements that sum to b. It follows that ϕ(g) = 2^{n/2}. There are exactly n/2 elements in the group Z/nZ with odd b, and so n elements in the group G with odd b. Therefore,

∑_{\begin{matrix}g∈G \\ b odd \end{matrix}} ϕ(g) = n2^{n/2}.

If b is even, then there is an integer a_{0} such that

b + nZ = 2(a_{0} + nZ) = 2(a_{0} + n/2 + nZ).

If δ ≡ 0 (mod 2), then ϕ(g) = 0. If δ ≡ 1 (mod 2), then Z/nZ \ {a_{0} + nZ, a_{0} + n/2 + nZ} can be partitioned into (n − 2)/2 pairs of elements that sums to b. It follows that ϕ(g) = 4 · 2^{(n−2)/2} = 2^{(n+2)/2} and

∑_{\begin{matrix}g∈G \\ b even \end{matrix}} ϕ(g) = (\frac{n}{2}) 2^{(n+2)/2} = n2^{n/2}.

Therefore, ∑_{g∈G} ϕ(g) = 2n2^{n/2} and

Ψ(G) ≥ |Ω| − ∑_{g∈G} ϕ(g)

= 2^{n} − n2^{(n+2)/2}

= 2^{n} (1 − \frac{2n}{2^{n/2}} ) .

This completes the proof. □

5. Problems

It would be interesting to classify the structure of all MSTD sets of integers. We can stratify this problem in the following way. Let F(Z) denote the set of all finite sets of integers. Define the function n : F(Z) → Z by

n(A) = |A + A| − |A − A|.

What is the range of this function? Is it possible to construct, for every integer t, a set A ∈ F(Z) with n(A) = t? Can we describe all sets A with n(A) = 1?

Acknowledgements. The idea of constructing of MSTD sets of lattice points and integers from MSTD subsets of finite abelian groups is due to Terence Tao [12], and I thank him for his permission to include it in this paper. I also thank Boris Bukh and Peter Hegarty for helpful discussions.

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