1. Introduction. In combinatorics, for a finite sequence

LXXXVII.2 (1998)

On sums of distinct representatives

by

Hui-Qin Cao and Zhi-Wei Sun (Nanjing)

1. Introduction. In combinatorics, for a finite sequence

(1) {A

}

of sets, a sequence

(2) {a

}

of elements is called a system of distinct representatives (abbreviated to SDR) of (1) if a

∈ A

, . . . , a

∈ A

and a

6= a

for all 1 ≤ i < j ≤ n. A celebrated theorem of P. Hall [H] says that (1) has an SDR if and only if (3)

   [

i∈I

A

 ≥ |I| for all I ⊆ {1, . . . , n}.

Clearly (1) has an SDR provided that |A

| ≥ i for all i = 1, . . . , n, in particular an SDR of (1) exists if |A

| = . . . = |A

| ≥ n or 0 < |A

| < . . .

< |A

|.

Let G be an additive abelian group and A

, . . . , A

its subsets. We as- sociate any SDR (2) of (1) with the sum P

i=1

a

and set S({A

}

) = S(A

, . . . , A

)

(4)

= {a

+ . . . + a

: {a

}

forms an SDR of {A

}

}.

Of course, S(A

, . . . , A

) 6= ∅ if and only if (3) holds. A fascinating and challenging problem is to give a sharp lower bound for |S({A

}

)| and determine when the bound can be reached.

Let p be a prime. In 1964 P. Erd˝os and H. Heilbronn (cf. [EH] and [G]) conjectured that for each nonempty subset A of Z

= Z/pZ there are at least min{p, 2|A| − 3} elements of Z

that can be written as the sum of two distinct elements of A. With the help of Grassmann spaces this was

1991 Mathematics Subject Classification: Primary 11B75; Secondary 05A05.

The research is supported by the Return-from-abroad Foundations of the Chinese Educational Committee and Nanjing City.

[159]

confirmed by J. A. Dias da Silva and Y. O. Hamidoune [DH] in 1994, in fact they proved the following generalization for n-fold sums: If A ⊆ Z

then (5) |n

A| ≥ min{p, n|A| − n

+ 1}

where n

A denotes the set of sums of n distinct elements of A, i.e. n

A = S(A, . . . , A) with A repeated n times on the right hand side. In 1995 and 1996, N. Alon, M. B. Nathanson and I. Z. Ruzsa [ANR1, ANR2] introduced an ingenious polynomial method and obtained the following result by con- tradiction: Let F be any field of characteristic p and A

, . . . , A

its subsets with 0 < |A

| < . . . < |A

| < ∞. Then

(6) |S(A

, . . . , A

)| ≥ min

 p,

X

|A

| − n(n + 1)

2 + 1



providing N < p where N = P

i=1

|A

| − n(n + 1)/2. We mention that (6) also holds in the case N ≥ p. In fact, let s be the smallest positive integer with P

i=1

(|A

| − i) > N − p and choose A

⊆ A

, . . . , A

⊆ A

so that

|A

| = i for i < s, A

= A

for i > s, and

|A

| = s − 1 + X

(|A

| − i) − (N − p) ≤ |A

| − 1.

Then |A

| < . . . < |A

|, N

= P

i=1

|A

| − n(n + 1)/2 = p − 1 and so

|S(A

, . . . , A

)| ≥ |S(A

, . . . , A

)| ≥ N

+ 1 = p = min{p, N }.

Alon, Nathanson and Ruzsa [ANR2] posed the question when the lower bound in (6) can be reached and considered it to be interesting.

In view of the fundamental theorem on finitely generated abelian groups (cf. [J]), if a finite addition theorem holds in Z then it holds in any torsion- free abelian groups. So, without any loss of generality, we may work within Z.

For a finite subset A of Z, in 1995 Nathanson [N] showed the inequality

|n

A| ≥ n|A| − n

+ 1 and proved that if equality holds then A must be an arithmetic progression providing 2 ≤ n < |A| − 2. The same result was independently obtained by Y. Bilu [B].

Let A

, . . . , A

be finite subsets of Z with 0 < |A

| < . . . < |A

|. Take a sufficiently large prime p greater than P

i=1

|A

| − n(n + 1)/2 and the largest element of S(A

, . . . , A

). Applying the Alon–Nathanson–Ruzsa re- sult stated above, we have the inequality

(7) |S(A

, . . . , A

)| ≥ X

|A

| − n(n + 1)

2 + 1 =

X

(|A

| − i) + 1.

In this paper we will make a new approach to sums of distinct repre-

sentatives. The method allows us to give a somewhat constructive proof of

(7) provided that A

, . . . , A

are finite nonempty subsets of Z with distinct

cardinalities. Furthermore we are able to make key progress in the equality case.

Let us first look at two examples.

Example 1. Let a ∈ Z and d ∈ Z \ {0}. Let k ≥ n ≥ 1, A = {a + jd : j = 0, 1, . . . , k − 1}, and A

, . . . , A

be subsets of A with |A

| = k − n + i for every i = 1, . . . , n. Obviously S(A

, . . . , A

) ⊆ n

A. If S ⊆ A and |S| = n, then for each i = 1, . . . , n at least i elements of S lie in A

since |S \ A

| ≤

|A \ A

| = n − i, therefore we can write S in the form {a

, . . . , a

} where a

∈ A

, . . . , a

∈ A

. So n

A ⊆ S(A

, . . . , A

). Let X = {j

+ . . . + j

: 0 ≤ j

< . . . < j

< k}. For each j = 0, 1, . . . , n(k − n), there exist 0 ≤ u < n and 0 ≤ v ≤ k − n such that j = u(k − n) + v, hence

n(n − 1) 2 + j =

X

(i − 1) + u(k − n) + v

= X

0<i<n−u

(i − 1) + (n − u − 1 + v) + X

n−u<i≤n

(k − n + i − 1) belongs to X. Thus {n(n − 1)/2 + j : 0 ≤ j ≤ n(k − n)} ⊆ X. Apparently the least and the largest elements of X are 0 + 1 + . . . + (n − 1) = n(n − 1)/2 and (k − n) + . . . + (k − 1) = n(n − 1)/2 + n(k − n) respectively. So, by the above

S(A

, . . . , A

) = n

A = n X

i=1

(a + j

d) : 0 ≤ j

< . . . < j

< k o

= {na + xd : x ∈ X}

=

 na +

 n(n − 1)

2 + j



d : 0 ≤ j ≤ n(k − n)



and hence

|S(A

, . . . , A

)| = |n

A| = n(|A| − n) + 1 = X

|A

| − n(n + 1) 2 + 1.

Example 2 (cf. [N]). Let a

, a

, a

, a

∈ Z, a

< a

< a

< a

and a

− a

= a

− a

(but a

− a

may be different from a

− a

). Let A

= {a

, a

, a

} and A

= {a

, a

, a

, a

}. Then

S(A

, A

) = {a

+ a

, a

+ a

, a

+ a

= a

+ a

, a

+ a

, a

+ a

}.

Note that in this example |A

| = 3 < |A

| = 4 < |S(A

, A

)| = 5 =

|A

| + |A

| − 2(2 + 1)/2 + 1.

Now we introduce some notations to be used throughout the paper. For a

subset A of Z, −A refers to {−a : a ∈ A}, min A and max A denote the least

and the largest elements of A respectively. If there exist a ∈ Z, d ∈ Z \ {0}

and a positive integer k such that

A = {a + jd : 0 ≤ j < k}, then we call A an arithmetic progression (for short, AP).

In this paper, by a novel method we obtain the following

Theorem. Let A

, . . . , A

be subsets of Z with 0 < |A

| < . . . <

|A

| < ∞. Then inequality (7) holds. Moreover , in the equality case we have S

i=1

A

= A

for every m ∈ M = {1 ≤ j < n : |A

| > |A

| + 1} ∪ {n}, and A

forms an AP unless n = 1 or |A

| ≤ 3.

The result of Nathanson and Bilu stated above actually follows from the Theorem. For i = 1, . . . , n let A

⊆ A and |A

| = |A| − (n − i). Obviously 0 < |A

| < . . . < |A

| < ∞. It follows from Example 1 and the Theorem that

|n

A| = |S(A

, . . . , A

)| ≥ X

|A

| − n(n + 1)

2 + 1 = n|A| − n

+ 1.

When |n

A| = n|A| − n

+ 1, we have

|S(A

, . . . , A

)| = X

i=1

|A

| − n(n + 1) 2 + 1,

hence by the Theorem if 2 ≤ n ≤ |A| − 3 (i.e., n ≥ 2 and |A

| ≥ 4) then A = A

is an AP.

In the next section we shall provide two lemmas. The proof of the The- orem will be given in Section 3.

2. Auxiliary results

Lemma 1. Let G be an additive abelian group, and A

, . . . , A

its fi- nite subsets. Let r ∈ {1, . . . , n} and suppose that {a

}

forms an SDR of {A

}

. Then

(i) There exists a J ⊆ {1, . . . , n} containing r such that if J ⊆ I ⊆ {1, . . . , n} then

S

 [

j∈J

A



⊆ S({A

}

), n

i ∈ I \ {r} : a

∈ [

j∈J

A

o

= J \ {r}

and hence  

S

 [

j∈J

A

   =

   [

j∈J

A

 − |J| + 1, where for any subset A of G we let

(8) S

(A) = n X

i∈I

a

: a

∈ A \ {a

: i ∈ I \ {r}}

o

.

(ii) Let k

= |A

| < . . . < k

= |A

|. For any J described in (i) and I ⊆ {1, . . . , n} containing J, we have

(9)

S

 [

j∈J

A

 

 ≥ k

− r + 1,

and equality holds if and only if there exists an l ∈ {r, . . . , n} for which J = {1, . . . , l}, S

j=1

A

= A

and k

= k

+ (l − r).

P r o o f. (i) Let J be the class of those J ⊆ {1, . . . , n} containing r such that if J ⊆ I ⊆ {1, . . . , n} then for each j ∈ J there exists a one- to-one mapping σ

: I \ {r} → I \ {j} for which a

∈ A

for all i ∈ I \ {r}. Obviously J is nonempty (for, {r} belongs to J ) and finite.

Let J be any maximal set in J with respect to the semiorder ⊆, and let J ⊆ I ⊆ {1, . . . , n}.

Set A = S

j∈J

A

. Apparently J

= {r} ∪ {i ∈ I \ {r} : a

∈ A} contains J. Let J

⊆ I

⊆ {1, . . . , n}. Since J ∈ J and J ⊆ I

, for j ∈ J there is a one-to-one mapping σ

: I

\ {r} → I

\ {j} such that a

∈ A

for all i ∈ I

\ {r}. For j

∈ J

\ J, there is a j ∈ J with a

∈ A

. Since J ∈ J and I

= I

\ {j

} ⊇ J, there also exists a one-to-one mapping σ

: I

\ {r} → I

\ {j} such that a

∈ A

for i ∈ I

\ {r}. Obviously by letting j

∈ I

\ {r} correspond to j ∈ I

\ {j

} we can extend σ

to a one-to-one mapping σ

: I

\ {r} → I

\ {j

} for which a

∈ A

for all i ∈ I

\ {r}. Thus J

∈ J . As J ⊆ J

and J is a maximal set in J , we must have J

= J, i.e., {i ∈ I \ {r} : a

∈ S

j∈J

A

} = J \ {r}.

If j ∈ J and x

∈ A

\ {a

: i ∈ I \ {r}}, then x

+ P

i∈I\{r}

a

∈ S({A

}

) because a

∈ A

for i ∈ I \ {r}. So S

(A) ⊆ S ({A

}

).

Note that

|S

(A)| = |A \ {a

: i ∈ I \ {r}}| = |A| − |{i ∈ I \ {r} : a

∈ A}|

= |A| − |J \ {r}| = |A| − |J| + 1.

This proves part (i).

(ii) Let J be as described in (i), A = S

j∈J

A

and J ⊆ I ⊆ {1, . . . , n}.

If |J| < r, then

|A| − |J| ≥ |A

| − |J| > k

− r.

When |J| ≥ r, clearly max J ≥ r and k

−k

= P

r<i≤max J

(k

−k

) ≥ max J − r, therefore

|A| − |J| ≥ |A

| − |J| ≥ k

+ max J − r − |J| ≥ k

− r, and |A| − |J| = k

− r if and only if

A = A

, k

= k

+ max J − r, max J = |J|,

i.e.,

J = {1, . . . , |J|} , A = A

, k

= k

+ |J| − r.

This together with the equality |S

(A)| = |A| − |J| + 1 yields the second part.

Lemma 2. Let A and B be finite subsets of Z with 4 ≤ k = |A| < l = |B|, A ⊆ B, min A = min B, max A 6= max B and |S(A, B)| = k + l − 2. Then B is an AP.

P r o o f. Let A = {a

, . . . , a

} and B = {b

, . . . , b

} where a

< . . . < a

and b

< . . . < b

. Put C = {a

+ b

, . . . , a

+ b

, a

+ b

, . . . , a

+ b

}.

Clearly C ⊆ S(A, B) and |C| = k + l − 2. As |S(A, B)| = k + l − 2, S(A, B) coincides with C. Since A ⊆ B and a

6= b

, for i = 2, . . . , k we may suppose that a

= b

where i ≤ f (i) < l. Because

S(A, B) ⊇ {a

+ b

: 2 ≤ j < f (i)} ∪ {a

+ b

: j 6= f (i)}

∪ {a

+ b

: i < j ≤ k}, we have

k + l − 2 = |S(A, B)| ≥ (f (i) − 2) + (l − 1) + (k − i) = k + l + f (i) − i − 3, i.e. a

∈ {b

, b

}. Observe that a

< a

≤ b

. Since

a

+ b

< a

+ b

< a

+ b

< a

+ b

, a

+ b

= a

+ b

and a

+ b

= a

+ b

, it follows that

a

− a

= b

− b

= a

− a

.

If a

6= b

, then a

= b

, a

= b

, b

− b

< a

− b

= a

− a

, a

+ b

<

b

+ a

< a

+ b

= a

+ b

; this contradicts the fact a

+ b

∈ S(A, B) = C.

So a

= b

. As a

+ b

∈ C we must have a

+ b

= a

+ b

, similarly a

+ b

= a

+ b

, . . . , a

+ b

= a

+ b

. Thus

b

− b

= . . . = b

− b

= a

− a

= b

− b

= a

− b

.

If a

6= b

, then a

= b

and hence b

= b

−a

+b

= b

, which is impossible.

So a

= b

and B forms an AP.

3. Proof of Theorem. The case n = 1 is trivial. Below we let n ≥ 2 and assume the statement holds for smaller values of n.

Put k

= |A

| for i = 1, . . . , n. Set a = min S

i=1

A

, I = {1 ≤ i ≤ n : a ∈ A

}, r = min I and t = max I. For i ∈ I let

A

=

 A

\ {a} if i 6= r,

{a} if i = r;

and for i ∈ ¯ I = {1, . . . , n} \ I put A

=

 A

\ {a

} if r < i < t and i 6∈ M , A

otherwise,

where a

is an arbitrary element of A

. Apparently all the A

are finite, nonempty and contained in Z, also |S(A

, . . . , A

)| = |S({A

}

)|. Let k

= |A

| for i = 1, . . . , n. Observe that k

< k

if 1 ≤ i < j ≤ n and i, j 6= r.

By the induction hypothesis,

|S({A

}

)| ≥ X

i6=r

k

− (n − 1)(n − 1 + 1)

2 + 1 > 0.

Suppose that max S({A

}

) = P

i6=r

a

where {a

}

is an SDR of {A

}

. By Lemma 1 there exists a J ⊆ {1, . . . , n} containing r for which

J \ {r} = {i 6= r : a

∈ A}, S

(A) ⊆ S({A

}

), |S

(A)| ≥ k

− r + 1, where A = S

j∈J

A

and S

(A) = { P

i=1

a

: a

∈ A \ {a

: i 6= r}}. As S(A

, . . . , A

) ⊇ S(A

, . . . , A

) ∪ S

(A) and

max S(A

, . . . , A

) = a + X

i6=r

a

= min S

(A), we have

|S(A

, . . . , A

)| ≥ |S(A

, . . . , A

)| + |S

(A)| − 1

≥ X

i6=r

k

− n(n − 1)

2 + 1 + (k

− r + 1) − 1

≥ X

i6=r

k

− (t − r) − n(n − 1)

2 + 1 + k

− r

= X

k

− n(n + 1)

2 + n − t + 1

≥ X

k

− n(n + 1) 2 + 1.

From now on we assume that (10) |S(A

, . . . , A

)| =

X

k

− n(n + 1) 2 + 1.

The above deduction yields (11) |S({A

}

)| = X

i6=r

k

− (t − r) − n(n − 1)

2 + 1,

(12) r < i < t ⇒ i 6∈ ¯ I ∩ M,

|S

(A)| = k

− r + 1, (13)

t = n.

(14)

By (13) and Lemma 1 there is an l ∈ {r, . . . , n} for which J = {1, . . . , l}, S

j=1

A

= A

, k

= k

+ (l − r) and

(15) {1, . . . , l} \ {r} = J \ {r} = {i 6= r : a

∈ A = A

}.

(14) indicates that a ∈ A

. Let b = max S

i=1

A

. Clearly a 6= b, for otherwise each A

contains exactly one element, which contradicts the inequality k

<

k

. As −b = min S

i=1

(−A

) and |S(−A

, . . . , −A

)| = |S(A

, . . . , A

)| = P

i=1

|−A

| − n(n + 1)/2 + 1, similarly we have −b ∈ −A

. So b ∈ A

= A

\ {a}. Choose the smallest s ≤ n such that b ∈ A

.

Let m ∈ M . We now show that S

i=1

A

= A

, i.e. A

contains both S

i=1,i6=r

A

and A

.

If m = r, then r ∈ M , hence l = r = m and A

= S

i=1

A

⊇ S

i=1,i6=r

A

.

Since A

= A

for all i < r, by (11), (12) and the induction hypothesis, if m < r then S

i=1

A

= S

i=1,i6=r

A

= A

.

Assume r < n. Clearly b ∈ {a

: i 6= r} (otherwise P

i6=r,n

a

+ b ∈ S({A

}

) would be greater than P

i6=r

a

= max S({A

}

)). Suppose that b = a

where j 6= r. In view of (15), b ∈ A

if and only if j ≤ l. Since b = a

∈ A

⊆ A

, we have j ≥ s. If l = s, then b ∈ A

, s ≤ j ≤ l = s, s = j 6= r.

Now suppose that r < m ≤ n. If m < t = n, then m ∈ I by (12), and k

− k

≥ (k

− 1) − (k

− 1) > 1. By (11), (12) and the induction hypothesis S

i=1,i6=r

A

= A

. If 1 ≤ i < r, then A

= A

⊆ A

⊆ A

; if r < i ≤ m and i ∈ I, then A

= A

∪ {a} ⊆ A

∪ {a} = A

; if r < i ≤ m but i 6∈ I, then |A

| ≥ k

+ 1 ≥ 2 and hence for any given x

∈ A

by taking a

∈ A

different from x

at the beginning we find that x

∈ A

\ {a

} = A

⊆ A

⊆ A

. So S

i=1,i6=r

A

⊆ A

.

Since s is the smallest index such that −A

contains min S

i=1

(−A

) =

−b, by analogy S

i=1,i6=s

(−A

) ⊆ −A

. Thus, if r 6= s then −A

⊆ −A

, i.e. A

⊆ A

. If r = s, then by the above l 6= s = r, also l ≤ m since k

= k

+ l − r, therefore A

⊆ S

j=1

A

= A

⊆ A

. So S

i=1

A

= A

. Now let us check that A

is an AP except the case k

≤ 3.

If r = 1 then min{k

: i 6= r} = k

≥ k

− 1 ≥ k

, and if r > 1 then min{k

: i 6= r} = k

= k

. So min{k

: i 6= r} ≥ k

. Below we assume that k

≥ 4.

Suppose n > 2. By (11), (12) and the induction hypothesis, if r < n then

A

= A

\ {a} is an AP. Similarly, if s < n then −A

\ {−b} is an AP and

hence so is A

\ {b}. Thus, if r < n and s < n then A

= {a} ∪ (A

\ {a}) =

(A

\ {b}) ∪ {b} forms an AP. (Note that |A

| = k

> k

≥ 4.)

Now consider the case r = s = 1 < n = 2. By the above l = 2 (since l 6= s = 1) and k

= k

+ 1. Let a = b

< . . . < b

= b be all the elements of A

. If 1 ≤ i < j ≤ k

, then either b

or b

belongs to A

because |A

\A

| = 1, therefore b

+ b

∈ S(A

, A

). So

S(A

, A

) ⊇ S(A

\ {b}, A

)

⊇ C = {b

+ b

, . . . , b

+ b

, b

+ b

, . . . , b

+ b

}.

As |S(A

, A

)| = k

+ k

− 2 = 2k

− 3 = |C|, we have S(A

, A

) = C = S(A

\{b}, A

). Clearly |A

\{b}| = k

−1 = k

≥ 4, min(A

\{b}) = min A

and max(A

\ {b}) 6= max A

. Applying Lemma 2 we find that A

= A

forms an AP.

With respect to the case r = n we make the following remarks:

(i) Since A

= S

j=1

A

, we have A

⊂ A

. If n > 2, then by (11), (12) and the induction hypothesis A

= S

i=1

A

forms an AP, i.e. S

i=1

A

= A

is an AP. Set

A

= {x ∈ A

: x ≤ max A

} and A

= {x ∈ A

: min A

≤ x}.

Whether n = 2 or n > 2 we always have S

i=1

A

= A

⊆ A

∩ A

and hence |A

∩ A

| ≥ k

≥ k

≥ 4. Among A

, . . . , A

, A

, the index r

of the first one containing min( S

i=1

A

∪ A

) = a is identical with r while the index s

of the first one containing max( S

i=1

A

∪ A

) = max A

is less than n. Similarly, among A

, . . . , A

, A

, the index r

of the first one containing min( S

i=1

A

∪ A

) = min A

is less than n while the index s

of the first one containing max( S

i=1

A

∪ A

) = b is equal to s.

(ii) Suppose that A

6= A

. Then |A

| > k

> . . . > k

. According to the previous reasoning,

|S(A

, . . . , A

, A

)| ≥

n−1

X

i=1

k

+ |A

| − n(n + 1)

2 + 1 > 0.

Observe that

max S(A

, . . . , A

, A

) ≤ max S(A

, . . . , A

) + max A

< max S({a

}, . . . , {a

}, A

\ A

), and that |S({a

}, . . . , {a

}, A

\ A

)| = |A

\ A

|. So

|S(A

, . . . , A

)| ≥ |S(A

, . . . , A

, A

)| + |S({a

}, . . . , {a

}, A

\ A

)|

≥

n−1

X

i=1

k

+ |A

| − n(n + 1)

2 + 1 + (k

− |A

|)

= X

k

− n(n + 1)

2 + 1.

Since (10) holds, we must have (16) |S(A

, . . . , A

, A

)| =

n−1

X

i=1

k

+ |A

| − n(n + 1) 2 + 1.

(iii) By analogy, when −A

= {−x ∈ −A

: −x ≤ max(−A

)} 6=

−A

(i.e. A

6= A

), we have

|S(A

, . . . , A

, A

)| = |S(−A

, . . . , −A

, −A

)|

(17)

=

n−1

X

i=1

k

+ |A

| − n(n + 1) 2 + 1.

Assume that s < r = n. Then both r

and s

= s are less than n.

If A

6= A

, then (17) holds and hence A

forms an AP by previous arguments. If n > 2, then A

is an AP anyway and so is A

\ {b} by the above, therefore A

forms an AP. If n = 2, then s = 1, min(−A

) = min(−A

), max(−A

) 6= max(−A

) (since r = 2), and |S(−A

, −A

)| =

|S(A

, A

)| = k

+ k

− 2, hence −A

is an AP by Lemma 2, thus A

= A

forms an AP.

In the case r < s = n, by applying the above result to the subsets

−A

, . . . , −A

instead of A

, . . . , A

, we see that −A

forms an AP, i.e., A

is an AP.

Finally, we handle the remaining case r = s = n. Since r

< s

= s = n and s

< r

= r = n, by the above A

forms an AP if A

6= A

, and A

forms an AP if A

6= A

. Providing n > 2, both A

and A

are APs, therefore A

forms an AP. When n = 2, if A

= A

is not an AP, then A

or A

coincides with A

, hence min A

= min A

= min A

and max A

6= max A

(since s = 2), or min(−A

) = min(−A

) = min(−A

) and max(−A

) 6= max(−A

) (since r = 2), thus A

forms an AP by Lemma 2, which leads a contradiction. So, whether n > 2 or n = 2, A

always forms an AP.

The induction step is now completed and the proof of the Theorem is finished.

References

[ANR1] N. A l o n, M. B. N a t h a n s o n and I. Z. R u z s a, Adding distinct congruence classes modulo a prime, Amer. Math. Monthly 102 (1995), 250–255.

[ANR2] —, —, —, The polynomial method and restricted sums of congruence classes, J. Number Theory 56 (1996), 404–417.

[B] Y. B i l u, Addition of sets of integers of positive density, ibid. 64 (1997), 233–275.

[DH] J. A. D i a s d a S i l v a and Y. O. H a m i d o u n e, Cyclic space for Grassmann

derivatives and additive theory, Bull. London Math. Soc. 26 (1994), 140–146.

[EH] P. E r d ˝o s and H. H e i l b r o n n, On the addition of residue classes mod p, Acta Arith. 9 (1964), 149–159.

[G] R. K. G u y, Unsolved Problems in Number Theory, 2nd ed., Springer, New York, 1994, 129–131.

[H] P. H a l l, On representatives of subsets, J. London Math. Soc. 10 (1935), 26–30.

[J] N. J a c o b s o n, Basic Algebra I , 2nd ed., W. H. Freeman and Company, New York, 1985.

[N] M. B. N a t h a n s o n, Inverse theorems for subset sums, Trans. Amer. Math. Soc.

347 (1995), 1409–1418.

Hui-Qin Cao

The Fundamental Division Nanjing Auditing College Nanjing 210029

The People’s Republic of China

Zhi-Wei Sun Department of Mathematics Nanjing University Nanjing 210093 The People’s Republic of China E-mail: zwsun@netra.nju.edu.cn

Received on 1.3.1998

and in revised form on 9.8.1998 (3346)