Sums of distinct squares by

LXVII.4 (1994)

Sums of distinct squares

by

Paul T. Bateman (Urbana, Ill.), Adolf J. Hildebrand (Urbana, Ill.) and George B. Purdy (Cincinnati, Ohio)

1. Introduction. Throughout this paper we shall suppose that s is an integer ≥ 5. Then order of magnitude considerations show that every sufficiently large integer is expressible as a sum of s distinct non-zero squares.

In fact, E. M. Wright [Wr] proved that, if s ≥ 5, then for large n we can essentially prescribe the ratios of the squares in expressing n as a sum of s squares. Thus, for each s ≥ 5 there exists a largest integer N (s) which is not expressible as a sum of s distinct non-zero squares. In this paper we shall obtain asymptotic estimates for N (s).

In a recent paper [HK], Halter-Koch considered representations of inte- gers as sums of s distinct non-zero coprime squares, and he proved among other things the following results.

Theorem 0 (Halter-Koch). The largest odd integer not expressible as a sum of 4 distinct non-zero squares with greatest common divisor 1 is 157.

Moreover , if N

(s) denotes the largest integer not expressible as a sum of s distinct non-zero squares with greatest common divisor 1, then N

(5) = 245, N

(6) = 333, N

(7) = 330, N

(8) = 462, N

(9) = 539, N

(10) = 647, N

(11) = 888, and N

(12) = 1036.

Halter-Koch also proved a number of related results. For example, he showed that for s ≥ 5,

N

(s + 1) ≤ 2( p

N

(s) + 2)

,

which enables one to derive an explicit (but rather crude) bound for N

(s).

Of the two quantities N (s) and N

(s), the former is the more natural one, and we shall express our results in terms of N (s). Trivially, we have N

(s) ≥ N (s) for all s ≥ 5, and we shall show in Theorem 5 that the two functions are in fact identical. Thus, the coprimality condition in the definition of N

(s) does not affect the results in any way.

Research of the second author supported in part by an NSF Grant.

[349]

Since any sum of s distinct positive squares must be greater than or equal to the sum of the first s positive squares, namely

P (s) = X

i

= s(s + 1)(2s + 1)/6,

we have the trivial lower bound N (s) ≥ P (s) − 1. In fact, N (s) must be strictly larger than P (s) since, for example, P (s) + 1 is not expressible as a sum of s distinct squares. Our principal result (Theorem 1) shows among other things that N (s) is asymptotically equal to this lower bound P (s) and gives a fairly precise estimate for the difference

R(s) = N (s) − P (s).

In order to state this main theorem, we define λ

≥ 0 by λ

= 2 max(k √

2sk, k √

2s − 1/2k),

where k · k denotes the distance to the nearest integer. It is easy to see that

(1.1) λ

= 1/2 + k √

8s − 1/2k.

We further set, for any non-negative real number x,

L

= log log max(x, e

), t

= bL

/ log 2c, f (x) =

t_x

X

i=0

x

.

Theorem 1. (i) We have the asymptotic formula

(1.2) R(s) = 2s{ √

2s + λ

(2s)

+ O(s

)}.

(ii) We have the upper estimate

(1.3) R(s) ≤ 2s{f ( √

2s) + O(L

)}.

(iii) The bound (1.3) is best possible in the sense that there exists an increasing sequence {s

} of positive integers such that

(1.4) R(s

) ≥ 2s

{f ( √

2s

) + O(L

)}.

An example of such a sequence is given by taking s

= 1 and s

= 2s

+ s

for k ≥ 2.

The estimate (1.2) shows in particular that R(s)  p

P (s). The second main term on the right-hand side of (1.2) involves the oscillatory quantity λ

, which depends on how 8s is situated relative to the sequence of squares.

From the representation (1.1) of λ

it is clear that 1/ √

2 ≤ λ

≤ 1 and that these bounds are best possible. Specifically, λ

will be near its maximal value 1 if 8s is close to a square; and λ

will be near its minimal value 1/ √

2 when

8s is roughly midway between two consecutive squares, for example, when

s has the form m(8m ± 1). Thus, R(s) = N (s) − P (s) oscillates between

the limits (2s)

+ (2s)

/ √

2 and (2s)

+ (2s)

, up to an error term O(s

).

The inequality (1.3) gives a universal upper bound for R(s) which sharp- ens that of (1.2) when 8s is close to a square and which by (1.4) is best possible.

The remainder of this paper is organized as follows. In Section 2 we give an explicit polynomial upper bound for N (s), namely N (s) < (s − 1)

for s ≥ 5 (Theorem 2), which will be needed as a basis for the subsequent arguments. In Section 3 we reformulate the problem of determining N (s) and state a result (Theorem 3) about a related extremal problem. This problem concerns the minimum Q(m) of P

i=1

a

for all representations of the integer m in the form m = P

i=1

ε

a

, where ε

= ±1 for all i and a

, . . . , a

are distinct positive integers. Theorem 3 gives estimates for Q(m) parallel to those of Theorem 1 and forms the principal ingredient in the proof of that theorem, but is also of some interest for its own sake. In Sections 4 and 5 we prove Theorem 3, and in Sections 6 and 7 we prove Theorem 1. In Section 8, we give the explicit upper bound (Theorem 4)

(1.5) N (s) < P (s) + 2s √

2s + 44s

+ 108s (s ≥ 166),

which is useful for various purposes. In particular, we use (1.5) to show that the function N (s) is monotonic for s ≥ 7; this answers a question of Erd˝os.

(Note, however, that the function R(s) = N (s) − P (s) is not monotonic, since (1.2) gives R(8m

) > R(8m

+ m) for all large m.) In Section 9, we prove the above remark that N (s) = N

(s) for every s ≥ 5; in fact, we show (Theorem 5) that if a positive integer is expressible as a sum of s ≥ 5 distinct non-zero squares then it is also expressible as a sum of s distinct non-zero squares with greatest common divisor 1. In Section 10 we make some remarks on the more general problem of expressing an integer as a sum of s distinct positive kth powers. Using the results of Hardy and Littlewood on Waring’s problem, we show (Theorem 6) that if N

(s) denotes the largest integer not expressible in this form, then

N

(s) = s

k + 1 + O(s

).

In the final section, we discuss the computation of N (s) and we give two tables of numerical data.

2. An initial upper bound. Using the result of Halter-Koch on four squares mentioned in the preceding section, we obtain the rough bound N (s) < (s − 1)

, which will be needed later on.

Theorem 2. If s ≥ 5 and if n ≥ (s − 1)

, then n is expressible as a

sum of s distinct non-zero squares.

P r o o f. It is convenient to prove the assertion of the theorem under the slightly weaker assumption n ≥ (s − 1)

(s − 3). For i = 1, 2, . . . , s − 5 we put a

= b p

n/(s − 3)c + i; we also put a

= b p

n/(s − 3)c + s − 4 + δ, where δ ∈ {0, 1} is chosen so that r = n − a

− a

− . . . − a

is odd. (When s = 5, only a

is needed.) Then a

> n/(s − 3) for each i and thus r < n/(s − 3).

Moreover, r ≥ n −

s−5

X

i=1

r n

s − 3 + i



−

r n

s − 3 + s − 3



= f

(n),

say. A simple calculation gives f

(n) = n

s − 3 − (s

− 7s + 14)

r n

s − 3 − (2s

− 21s

+ 85s − 126)/6.

Clearly f

(n) is an increasing function of n provided p

n/(s − 3) > (s

− 7s + 14)/2. This condition is satisfied if s ≥ 5 and n ≥ (s − 1)

(s − 3). Thus, if n ≥ (s − 1)

(s − 3), we have

r ≥ f

(n) ≥ f

((s − 1)

(s − 3)) = 14

3 s

− 39

2 s

+ 101 6 s + 8.

The polynomial on the right-hand side here is an increasing function of s for s ≥ 5 and hence

r ≥ 14

3 5

− 39

2 5

+ 101

6 5 + 8 = 188.

Since r is odd and greater than 157, Theorem 0 shows that r is expressible as a sum of four distinct non-zero squares. Since each of these four squares is less than

r < n/(s − 3) < a

< a

< . . . < a

and since n = r + a

+ a

+ . . . + a

, the assertion of the theorem follows.

3. An extremal problem. In this section we rephrase the problem of estimating N (s) in a form which is more suitable when dealing with integers that are close to P (s), and we state a result (Theorem 3), which will form the principal ingredient in the proof of Theorem 1. The underlying idea is that if n is an integer close to P (s) = P

i=1

i

which has a representation n = P

i=1

a

as a sum of s distinct squares, then the set {a

: i ≤ s} can be expected to be “close” to the set {i : i ≤ s}.

To make this idea precise, we note that any set {a

: i ≤ s} of distinct

positive integers can be obtained from the set {i : i ≤ s} by replacing some

of the integers i ≤ s, say s − h

, i ≤ t, by distinct integers > s, say s + k

,

i ≤ t. The associated representation n = P

i=1

a

can then be written as n =

X

i

− X

i=1

(s − h

)

+ X

(s + k

)

(3.1)

= P (s) + 2s X

(h

+ k

) + X

(k

− h

), where the numbers h

and k

satisfy

h

distinct, 0 ≤ h

< s, (3.2)

k

distinct, k

≥ 1.

(3.3)

Conversely, any integer n expressible in the form (3.1) with the conditions (3.2) and (3.3) is a sum of s distinct positive squares. Therefore, R(s) = N (s) − P (s) is the largest integer r not expressible in the form

(3.4) r = 2s

X

(h

+ k

) + X

(k

− h

) with integers h

and k

satisfying (3.2) and (3.3).

The above formulation leads naturally to the problem of minimizing the sum P

i=1

(h

+ k

), subject to the conditions (3.2) and (3.3), while holding the sum P

i=1

(k

− h

) fixed. However, this extremal problem is somewhat awkward to deal with directly, as the conditions (3.2) and (3.3) are not symmetrical and depend on the parameter s. We therefore consider the following related, but simpler and more natural problem, which is sufficient for the application to the proof of Theorem 1 and also is of some intrinsic interest. For m 6= 0 set

(3.5) Q(m) = min

n X

i=1

a

: X

ε

a

= m o

,

where the minimum is taken over all sets {a

: i ≤ t} of distinct positive integers satisfying P

i=1

ε

a

= m with suitable numbers ε

∈ {±1}, and define Q(0) = 0. The quantity Q(m) may be viewed as a measure for how

“economically” m can be represented as a difference of sums of distinct squares. The following result gives precise upper and lower bounds for Q(m) that are largely parallel to those of Theorem 1. Since m = ((m + 10)/2)

− ((m + 8)/2)

− 3

for even positive integers m and m = ((m + 17)/2)

− ((m + 15)/2)

− 4

for odd positive integers m, every non-zero integer m has indeed a representation m = P

i=1

ε

a

of the above form, so that Q(m) is well-defined. Halter-Koch’s result that N

(5) = 245, along with Schwarz’s inequality, shows that trivially Q(m) ≤ √

5m for m ≥ 246.

Theorem 3. (i) We have the asymptotic formula

(3.6) Q(m) = p

|m| + p

2θ

|m|

+ O(|m|

), where θ

= k √

xk.

(ii) We have the upper estimate

(3.7) Q(m) ≤ f ( p

|m| ) + O(L

), where f (x) and L

are defined as in Theorem 1.

(iii) The inequality (3.7) is best possible in the sense that if the sequence {m

} is defined by m

= 1 and m

= m

+ m

for k ≥ 1, then we have

(3.8) Q(m

) ≥ f ( √

m

) + O(L

).

(iv) The upper bounds in (3.6) and (3.7) remain valid if in the definition (3.5) of Q(m), t is restricted by the condition

(3.9) t ≤ CL

,

where C is a suitable absolute constant.

4. Proof of Theorem 3; upper bounds. Call a representation m = P

i=1

ε

a

admissible if ε

∈ {±1} and the numbers a

are distinct positive integers. To obtain the upper bounds of Theorem 3 (in the stronger form claimed in the last part of Theorem 3), we need to construct an admissible representation with t ≤ CL

for which the sum P

i=1

a

is bounded by the right-hand sides of (3.6) and (3.7). Our construction is essentially that obtained by the greedy algorithm, supplemented by a direct argument for the first few values of m. We first dispose of the case of small m with the following lemma.

Lemma 4.1. If 0 < |m| ≤ 37, then m has an admissible representation m = P

i=1

ε

a

such that a

≤ 5 for all i.

P r o o f. The identities 1 = 1

, 2 = 4

− 3

− 2

− 1

, 3 = 2

− 1

, 4 = 2

, 5 = 2

+ 1

, 6 = 3

− 2

+ 1

, 7 = 4

− 3

, 8 = 3

− 1

, 9 = 3

, 10 = 3

+ 1

, 11 = 4

− 2

− 1

, 12 = 4

− 2

, and 13 = 3

+ 2

show that every m with 0 < m ≤ 13 has a representation of the required form with a

≤ 4. Replacing ε

by −ε

in each of these representations, we see that the same is true for

−13 ≤ m < 0. In the remaining range 13 < |m| ≤ 37 the result follows by writing m = ε5

+ m

with ε ∈ {±1} and |m

| ≤ 12 and representing m

in the above form using squares a

with a

≤ 4.

The lemma shows that for 0 < |m| ≤ 37, Q(m) is well-defined and satisfies the bounds (3.6) and (3.7) trivially, provided the O-constants are suitably chosen. The same is true for m = 0, since by definition Q(0) = 0. To deal with the general case, we begin with the following observation. Given an arbitrary integer m, let q = b p

|m|c, so that q

≤ |m| ≤ q

+ 2q, and

set a = h p

|m|i, where hxi denotes the nearest integer to x. (Note that, since p

|m| cannot be half an odd integer, there is no ambiguity in the definition of h p

|m|i.) Then a = q if q

≤ |m| ≤ q

+ q, a = q + 1 if q

+ q + 1 ≤ |m| ≤ q

+ 2q, and in either case we have m = εa

+ r with ε = sign(m) (with the convention sign(0) = 1) and |r| ≤ q = b p

|m|c.

Iterating this procedure, we obtain, for any given integer m, sequences of integers {a

} and {r

} defined by

(4.1) r

= m, a

= h p

|r

|i, ε

= sign(r

), r

= ε

a

+r

(i ≥ 1), such that

(4.2) |r

| ≤ b p

|r

|c (i ≥ 1).

We then have for any k ≥ 1 the representation

(4.3) m =

X

ε

a

+ r

.

In fact, for sufficiently large k we have the exact representation m = P

i=1

ε

a

, since it is easily seen that the sequence {r

} must be eventu- ally zero; however, in order to ensure that the numbers a

are distinct, we need to work with the truncated version (4.3) in which the term r

is not necessarily 0.

Assume now that |m| = |r

| > 37. Then a

= h p

|m|i ≥ h √

37i ≥ 6.

Moreover, if i ≥ 2 and a

≥ 3 then (4.2) and (4.1) imply that 3 ≤ a

= h p

|r

|i ≤ h|r

|

i < h p

|r

|i = a

,

since any real number x with hxi ≥ 3 must be at least equal to 5/2 and hence satisfies x < x

− 1 and hxi < hx

i. Therefore, defining k to be the maximal index such that a

≥ 6, we have

a

> a

> . . . > a

≥ 6 > a

. Furthermore, by (4.1) we have h p

|r

|i = a

≤ 5, so that |r

| ≤ (5 + 1/2)

< 36. If r

= 0, then (4.3) gives an admissible representation of m.

Otherwise we have 0 < |r

| < 36 and we can therefore apply Lemma 4.1 to represent r

in the form

r

= X

ε

a

, 5 ≥ a

> . . . > a

≥ 1.

Combining this representation with (4.3) we obtain again an admissible

representation of m involving t ≤ k + 5 squares. In either case we obtain

the inequality

(4.4) Q(m) ≤ X

a

≤ X

a

+ X

i=1

i = X

a

+ 15.

To bound the sum P

i=1

a

, we first observe that by (4.2) and induction we have for each i ≥ 1,

|r

| ≤ |r

|

= |m|

. Together with (4.1), this implies

(4.5) a

= h p

|r

|i ≤ |m|

+ 1/2 and, in particular,

6 ≤ a

≤ |m|

+ 1/2.

The last estimate implies

(4.6) k ≤ 1

log 2 L

,

which in view of the inequality t ≤ k + 5 shows that the representation constructed above satisfies the additional restriction (3.9) stated in part (iv) of the theorem. Moreover, (4.5) and (4.6) yield

X

a

≤ X

(|m|

+ 1/2) ≤ f ( p

|m|) + O(L

).

In view of (4.4) this establishes the bound (3.7).

To prove the upper bound in (3.6), we observe that if p

|m| = a + ϑ with

|ϑ| ≤ 1/2, then we have a = h p

|m|i, |ϑ| = θ

and

|r

| = ||m| − a

| = |(a + ϑ)

− a

| = 2a|ϑ| + O(1) = 2θ

p

|m| + O(1).

Using this estimate together with (4.4), (4.5), and (4.6), we obtain Q(m) ≤

X

a

+ O(1) ≤ p

|r

| + p

|r

| + X

(|m|

+ 1/2) + O(1)

= p

|m| + p

2θ

|m|

+ O(|m|

), which is the desired estimate.

5. Proof of Theorem 3; lower bounds. We begin with a lemma which supplies the key step in the proof.

Lemma 5.1. (i) For any integer m, we have Q(m) = Q(|m|) ≥ p

|m|.

(ii) If m is a sufficiently large positive integer , then we have (5.1) Q(m) = min 

q + Q(m − q

), q + 1 + Q(m − (q + 1)

) , where q = b p

|m|c.

P r o o f. (i) The identity Q(m) = Q(|m|) follows immediately from the definition of Q(m). The bound Q(m) ≥ p

|m| holds trivially for m = 0, since Q(0) = 0. If m 6= 0, then any representation of the form

(5.2) m =

X

ε

a

, ε

∈ {±1}, a

> a

> . . . > a

≥ 1,

satisfies

X

a

≥

 X

i=1

a



≥   

X

ε

a

= p

|m|.

By the definition of Q(m) this implies Q(m) ≥ p

|m|.

(ii) We first show that Q(m) is bounded from below by the right-hand side of (5.1). Suppose that m is a positive integer and fix a representation of the form (5.2) such that Q(m) = P

i=1

a

. If t = 1 in (5.2), then |m| = a

= q

, and (5.1) holds trivially. Assume therefore that t ≥ 2. By (5.2), P

i=2

ε

a

is an admissible representation for the number m − ε

a

, and we therefore have Q(m − ε

a

) ≤ P

i=2

a

. It follows that

(5.3) Q(m) = a

+

X

a

≥ a

+ Q(m − ε

a

).

Thus, to obtain the lower bound in (5.1), it suffices to show that ε

= 1 and a

= q or a

= q + 1 whenever m is sufficiently large.

Suppose first that a

≤ p

m/2. Then (5.2) implies

Q(m) = X

i=1

a

≥ 1 a

X

a

≥ 1 a

X

ε

a

   = m

a

≥ √ 2m,

which contradicts the upper bound of (3.6) if m is sufficiently large. If a

>

p m/2 and ε

= −1, then (5.3) and part (i) of the lemma give Q(m) ≥ a

+ p

m + a

≥ p

m/2 + p 3m/2, which again yields a contradiction to the upper bound of (3.6).

Finally, suppose that a

> p

m/2, ε

= 1, but a

6∈ {q, q + 1}. In this case we obtain from (5.3) and part (i) of the lemma the bound

(5.4) Q(m) ≥ a

+ p

|m − a

|.

Now, note that the function x + p

|m − x

| is decreasing for p

m/2 < x

< √

m and increasing for x > √

m. Since q ≤ √

m < q + 1, it follows that over the ranges p

m/2 < a

≤ q − 1 and a

≥ q + 2 the right-hand side of

(5.4) is minimal when a

= q − 1 or a

= q + 2, and in either case is bounded

from below by q − 1 + min( p

m − (q − 1)

, p

(q + 2)

− m)

≥ q − 1 + q

m − ( √

m − 1)

= √ m + √

2m

+ O(1).

Since this bound exceeds the upper bound (3.6) for large enough m, we conclude that for sufficiently large m, a

must be equal to either q or q + 1, as we wanted to show.

To obtain the reverse inequality, it suffices to note that under the con- ditions Q(m − q

) < q and Q(m − (q + 1)

) < q + 1 we obtain admissible representations of m by adding q

to any admissible representation of m−q

or by adding (q + 1)

to any admissible representation of m − (q + 1)

and therefore have Q(m) ≤ min(q + Q(m − q

), q + 1 + Q(m − (q + 1)

)). In view of the inequalities 0 ≤ m − q

≤ 2q and 0 ≤ (q + 1)

− m ≤ 2q + 1 and the bound Q(m)  p

|m|, the two conditions are satisfied provided m is sufficiently large.

R e m a r k. The recurrence formula (5.1) could be used in principle to evaluate Q(m) for any m to within an error term O(1), but it is unlikely that it would lead to a simple explicit expression for Q(m) or provide a simple algorithm for computing Q(m) for any particular value of m without the knowledge of the prior values of the function Q. The reason for this is that it seems hard to decide a priori, which of the two terms on the right of the formula achieves the minimum; in particular, since the function Q(m) is not monotonic, the minimum is not necessarily attained (or even approximately attained) at the term in which the argument of Q (i.e., m−q

or m − (q + 1)

) has smaller absolute value.

P r o o f o f (3.6), l o w e r b o u n d. In view of part (i) of Lemma 5.1 we may assume that m is sufficiently large and positive. Writing θ = θ

= k √

mk and q = b √

mc, we have √

m = q+θ if q

≤ m ≤ q

+q, √

m = q+1−θ if q

+ q + 1 ≤ m < (q + 1)

, and in any case

min(|m − q

|, |m − (q + 1)

|) ≥ 2qθ + O(1).

Applying Lemma 5.1, we therefore obtain Q(m) ≥ q + min( p

|m − q

|, p

|m − (q + 1)

|)

≥ q + p

2qθ + O(1) ≥ √ m + √

2θm

+ O(1), which proves the lower bound of (3.6).

P r o o f o f (3.8). We first note that the recurrence relation m

= m

+ m

implies b √

m

c = b √

m

+ 1c = m

. Thus, if m = m

or m = m

+ 1, then we have, in the notation of Lemma 5.1, q = m

.

Moreover, the numbers m − q

and m − (q + 1)

are equal to m

and

−(m

+ 1), respectively, if m = m

, and to m

+ 1 and −m

if m = m

+ 1. Setting

Q

= min(Q(m

), Q(m

+ 1))

and noting that Q(m) = Q(−m) we therefore obtain from (5.1) the inequal- ity

Q

≥ m

+ Q

for all sufficiently large k, say k ≥ k

. Iterating this inequality, we deduce (5.5) Q(m

) ≥ Q

≥

k−1

X

i=k₀−1

m

+ Q

=

k−1

X

i=1

m

+ O(1) for k ≥ k

.

To estimate the sum on the right of (5.5), we show by induction that for 0 ≤ i ≤ k

(5.6) m

≤ m

≤ m

+ 1 − 2

.

For i = 0, (5.6) holds trivially. Assuming (5.6) holds for some i ≤ k − 1, we deduce

m

≥ m

≥ (m

)

= m

and

m

≤ (m

+ 1 − 2

)

= (m

+ m

+ 1 − 2

)

≤ (m

+ 2(1 − 2

)m

)

< (m

+ 1 − 2

)

, which implies (5.6) for i + 1 and completes the induction.

Applying first (5.6) with i = k − 1 we obtain

2 = m

≤ m

≤ m

+ 1 − 2

= 3 − 2

,

which implies k = L

/ log 2 + O(1) = t

+ O(1). Using this inequality and the upper bound of (5.6) we get

X

m

≥ X

(m

− 1) =

t_mk

X

i=0

√ m

− k + O(1)

= f ( √

m

) − 1

log 2 L

+ O(1),

since by (5.6) the terms m

with i = k +O(1) are of order O(1). Combined with (5.5), this gives the desired estimate.

6. Proof of Theorem 1; lower bounds. Recall that R(s) is the largest

integer r not expressible in the form (3.4) with integers h

and k

satisfying

(3.2) and (3.3). For 0 ≤ r

< 4s let R(s, r

) denote the largest such integer r that lies in the residue class r

modulo 4s. Then clearly

(6.1) R(s) = max

0≤r0<4s

R(s, r

).

We shall obtain the lower bounds of Theorem 1 by considering R(s, r

) for suitable choices of r

.

We begin with a lemma which gives a bound for R(s, r

) in terms of the function Q(m) defined in Theorem 3.

Lemma 6.1. We have

(6.2) R(s, r

) ≥ 2s min {Q(2s − d), Q(2s + d)} + O(s), where |d| ≤ 2s is chosen so that

(6.3) d ≡

 r

mod 4s if r

is odd, 2s − r

mod 4s if r

is even.

P r o o f. It suffices to show that any integer r ≡ r

mod 4s which has a representation of the form

(6.4) r = 2s

X

(h

+ k

) + X

i=1

(k

− h

) = 2sΣ

+ Σ

,

say, with integers h

and k

satisfying (3.2) and (3.3), is bounded from below by the right-hand side of (6.2).

We first observe that, by the upper bound Q(m) ≤ p

|m| + O(|m|

) of Theorem 3, the right-hand side of (6.2) is bounded from above by

2s min( √

2s − d, √

2s + d) + O(s

) ≤ 2s √

2s + O(s

).

Thus, if

(6.5) r ≥ 4s

+ O(s),

then r is bounded from below by the right-hand side of (6.2).

Next, note that under the conditions 0 ≤ h

< s and k

> 0, which are implied by (3.2) and (3.3), the right-hand side of (6.4) is an increasing function of each of the variables h

and k

. Hence, for any λ with 0 < λ ≤ 1, (6.4) implies

r ≥ 2s X

i=1

(λh

+ λk

) + X

((λk

)

− (λh

)

)

≥ 2sλΣ

+ λ

Σ

≥ 2sλ p

|Σ

| + λ

Σ

, since trivially

Σ

≥ X

(h

+ k

) ≥ |Σ

|.

If now |Σ

| ≥ 4s, then choosing λ = p

4s/|Σ

| we obtain r ≥ 4s

− 4s and hence (6.5). Thus, it remains to consider the case when

(6.6) |Σ

| < 4s.

Observe that the sums Σ

and Σ

in (6.4) have the same parity, since x ≡ ±x

mod 2 for any integer x. Hence, if r ≡ r

mod 4s with r

even, (6.4) implies that both sums are even and that r

≡ Σ

mod 4s. If r

is odd, both sums are odd, and in this case (6.4) yields r

≡ Σ

− 2s mod 4s. In either case we have |Σ

| ≡ 2s + d or |Σ

| ≡ 2s − d with d given by (6.3). In view of (6.6), this implies

(6.7) |Σ

| ∈ {2s ± d}.

The conditions (3.2) and (3.3) imply that in the representation Σ

= P

i=1

(k

− h

) the numbers h

are mutually distinct and non-negative and the numbers k

are mutually distinct and positive, although the two sets of numbers are not necessarily disjoint. However, by dropping any pairs (h

, k

) with h

= k

as well as 0 if it occurs among the numbers h

and relabeling the remaining numbers h

and k

we obtain a representation of the form

Σ

=

t₁

X

i=1

k

−

t₂

X

i=1

h

in which the integers h

and k

are mutually distinct and strictly positive.

The latter representation is an admissible representation in the definition of Q(Σ

), and we therefore have

Q(|Σ

|) = Q(Σ

) ≤

t1

X

i=1

k

+

t2

X

i=1

h

≤ Σ

.

Combining this inequality with (6.7) and (6.4) yields the desired lower bound for r.

This completes the proof of the lemma.

P r o o f o f (1.2), l o w e r b o u n d. By (6.1) and Lemma 6.1 we have (6.8) R(s) ≥ 2s max

|d|≤√

2s

min {Q(2s − d), Q(2s + d)} + O(s).

To bound the right-hand side, we use the bound of (3.6) of Theorem 3 together with the estimates

√ 2s ± d = √

2s ± d 2 √

2s + O(s

) (|d| ≤ √ 2s), (6.9)

(2s ± d)

= (2s)

+ O(s

) (|d| ≤ √ 2s).

We thus obtain for |d| ≤ √ 2s,

(6.10) min(Q(2s − d), Q(2s + d)) ≥ √

2s + p

2µ(2s)

+ O(s

),

where

µ = µ(s, d) = min(k √

2s + dk, k √

2s − dk).

By (6.9) we have k √

2s ± dk =

√ 2s ± d 2 √

2s

+ O(s

) and therefore

(6.11) max

|d|≤√ 2s

µ(s, d) = max

|δ|≤1/2

min{kδ + √

2sk, kδ − √

2sk} + O(s

).

It is easy to see that the maximum on δ is attained either at δ = 0 or at δ = 1/2 and thus is equal to max(k √

2sk, k √

2s − 1/2k) = λ

/2 by the definition of λ

. It follows that the left-hand side of (6.11) is equal to λ

/2 + O(s

), which combined with (6.10) and (6.8) proves the lower bound of (1.2).

P r o o f o f (1.4). We set s

= m

/2 for k ≥ 1 with m

defined as in part (iii) of Theorem 3. Clearly s

= 1 and s

= 2s

+ 1 for k ≥ 2, so that s

is an odd integer. Applying the bound of Lemma 6.1 with r

= 2s

(so that d = 0), together with the estimate (3.8) of Theorem 3, we obtain R(s

) ≥ R(s

, 2s

) ≥ 2s

Q(2s

) + O(s

)

≥ 2s

{f ( √

m

) + O(L

)} + O(s

) = 2s

{f ( √

2s

) + O(L

)}, which proves (1.4).

7. Proof of Theorem 1; upper bounds. To obtain the upper bounds (1.2) and (1.3) for R(s), we need to show that if r is greater than the right-hand side of (1.2) or (1.3) then r is expressible in the form (3.4), i.e.,

(7.1) r = 2s

X

(h

+ k

) + X

i=1

(k

− h

),

with integers h

and k

satisfying (3.2) and (3.3). In fact, it will be convenient to also consider such representations with (3.2) and (3.3) replaced by the slightly stronger conditions

1 ≤ h

≤ s − 1, h

distinct, (7.2)

1 ≤ k

≤ s − 1, k

distinct, (7.3)

which have the advantage of being symmetric in h

and k

. We denote by R

(s) the set of integers r expressible in the form (3.2)–(3.4), and by R

(s) the set of integers expressible in the form (7.1)–(7.3). Needless to say, empty sums are to be interpreted as zero, so that R

(s) = R

(s) = {0}. We further set R(s) = S

t≥0

R

(s), R

(s) = S

t≥0

R

(s), and for any residue

class r

mod 4s we put

R

(s, r

) = {r ∈ R

(s) : r ≡ r

mod 4s},

and define R

(s, r

), R(s, r

), and R

(s, r

) analogously. Note that R

(s) ⊂ R(s).

The following three propositions contain the key steps of the proof and will be proved in turn in the remainder of this section. The second and third of these propositions will be used again in Section 9 to obtain an explicit numerical bound.

Proposition 7.1. For any residue class r

mod 4s there exists a non- negative integer r ∈ R

(s, r

) for some t  L

satisfying

r ≤ 2s{ √

2s + λ

(2s)

+ O(s

)}, (7.4)

r ≤ 2s{f ( √

2s) + O(L

)}.

(7.5)

Proposition 7.2. If s ≥ 150 and r ∈ R

(s) for some t ≤ s/25, then r + 4sq ∈ R

(s) for every q satisfying

(7.6) 4t + 3 ≤ q ≤ b(s + 5)/6cs.

Proposition 7.3. Suppose that s ≥ 50 and that R

(s − 1) contains every integer in the interval [(s − 1)

/6, (s − 1)

/2]. Then R(s) contains every integer ≥ 2s

/3.

P r o o f o f T h e o r e m 1; u p p e r b o u n d s. We may clearly assume that s is sufficiently large. The first two propositions imply that R

(s), and hence also R(s), contains every integer r in the ranges

2s{ √

2s + λ

(2s)

+ c

s

} ≤ r ≤ 4b(s + 5)/6cs

, (7.7)

2s{f ( √

2s) + c

L

} ≤ r ≤ 4b(s + 5)/6cs

, (7.8)

provided c

and c

are sufficiently large absolute constants. Since for large s the ranges (7.7) and (7.8) contain the interval [s

/6, 2s

/3], it follows by the third proposition that, if s is sufficiently large, then R(s) also contains every integer ≥ 2s

/3. Therefore, R(s) = max{r : r 6∈ R(s)} is bounded by the left-hand sides of (7.7) and (7.8), and we obtain the upper bounds of (1.2) and (1.3).

P r o o f o f P r o p o s i t i o n 7.1. In the case r

≡ 0 mod 4s, r = 0 be- longs to R

(s, 0) and (7.4) and (7.5) are trivially satisfied. We can therefore assume that r

6≡ 0 mod 4s.

As a first step, we show that for sufficiently large s and every integer m with 0 < m < 4s there exist integers h

and k

(1 ≤ i ≤ t) satisfying (7.2) and (7.3) with

(7.9) t  L

,

such that

(7.10) m =

X

(k

− h

) and

(7.11)

X

(h

+ k

) ≤

 √ m + √

2θ

m

+ O(m

), f ( √

m) + O(L

), where θ

is defined as in Theorem 3.

An application of Theorem 3 yields a representation

(7.12) m =

t1

X

i=1

h

−

t2

X

i=1

k

with distinct positive integers h

, 1 ≤ i ≤ t

, and k

, 1 ≤ i ≤ t

, whose sum is bounded by the right-hand side of (7.11) and such that

(7.13) t

+ t

≤ CL

,

where C is the constant in (3.9). The bound (7.11) implies that the integers h

and k

are bounded by  √

m < √

4s, and hence are ≤ s − 1 if s is sufficiently large. The conditions (7.2) and (7.3) are therefore satisfied for these integers, and if t

= t

then (7.9)–(7.11) follow immediately with t = t

= t

.

If t

6= t

, we will obtain (7.9)–(7.11) by suitably enlarging the sets {h

} and {k

} to two sets having the same cardinality t, while leaving the value of P

i

h

− P

i

k

unchanged. Without loss of generality, assume that t

> t

and set

l = t

− t

, t = t

+ l = t

+ 2l.

By (7.13) we have t ≤ t

+ l ≤ 2t

≤ 2CL

, so that (7.9) is satisfied. We define additional integers h

and k

by setting

(7.14) h

= 5a

, k

= 3a

, k

= 4a

(1 ≤ i ≤ l) with distinct positive integers a

to be chosen later. This definition ensures that

X

(k

− h

) =

t1

X

i=1

k

−

t2

X

i=1

h

,

which in view of (7.12) yields (7.10). Moreover, if we restrict the integers a

to the residue class 1 modulo 3, then the sets {3a

}, {4a

}, and {5a

} are

pairwise disjoint, and the numbers defined in (7.14) are therefore mutually

distinct positive integers. Thus, in order to satisfy the conditions (7.2) and

(7.3), it remains to ensure that these numbers are distinct from the numbers

h

, 1 ≤ i ≤ t

, and k

, 1 ≤ i ≤ t

, and are bounded by s − 1.

We consider the set of positive integers a ≤ 12CL

+ 3, where C is the constant in (7.13). Clearly, at least 4CL

of these integers satisfy the congruence a ≡ 1 mod 3, and at most 3(t

+ t

) integers can be of the form λh

, 1 ≤ i ≤ t

, or λk

, 1 ≤ i ≤ t

, with λ = 1/3, 1/4, or 1/5. Since by (7.13), 4CL

− 3(t

+ t

) ≥ t

+ t

≥ l, there exist l of these integers, say a

, . . . , a

, with a

≡ 1 mod 3, such that none of the integers (7.14) is equal to one of the numbers h

, 1 ≤ i ≤ t

, or k

, 1 ≤ i ≤ t

. Moreover, since l ≤ t

≤ CL

and a

≤ 12CL

+ 3, the integers in (7.14) are bounded by

 L

≤ L

(and thus are ≤ s − 1 for large enough s), and we have X

i=t1+1

h

+ X

k

 X

a

 L

.

Thus, extending the summation in P

i=1

h

and P

i=1

k

to the full range 1 ≤ i ≤ t increases the two sums by at most O(L

), and therefore does not affect the upper bound (7.11). Hence (7.9)–(7.11) hold in any case.

Now, let r

mod 4s be a given non-zero residue class and define |d| < 2s by the congruence

(7.15) d ≡

 r

mod 4s (r

odd), 2s + r

(r

even).

We apply the above construction with m = m

= 2s ± d to obtain integers h

and k

(1 ≤ i ≤ t

) satisfying (7.2), (7.3), and (7.9)–(7.11), and set for ε = ±

(7.16) r

= 2sΣ

+ εm

,

where Σ

= P

i=1

(h

+ k

). We shall show that at least one of the integers r

has the properties claimed in the proposition.

First note that the numbers r

are both non-negative, since 0 < m

< 4s and Σ

≥ 2. Also, both numbers lie in the residue class r

mod 4s, since by (7.10), Σ

≡ m

≡ d mod 2 and therefore

r

≡ 2sd + ε(2s + εd) ≡ 2s(d + 1) + d ≡ r

mod 4s.

Moreover, by (7.16) and (7.10), r

has a representation of the required form (7.1) with t

 L

terms, and interchanging the roles of h

and k

in (7.16) shows that the same is true for r

. Therefore, we have r

∈ R

(s, r

) with t

 L

, and it remains to show that at least one of the integers r

is bounded by the right-hand sides of (7.4) and (7.5).

By (7.11) and (7.16) we have

(7.17) r

≤

 2s{M

(d) + O(s

)}, 2s{f ( √

2s + εd) + O(L

)},

where

M

(d) = √

2s + εd + p

2θ

(2s + εd)

.

The second estimate in (7.17) together with the monotonicity of the function f (x) immediately gives the upper bound

min(r

, r

) ≤ 2s{f ( √

2s) + O(L

)}, and hence the estimate (7.5) for one of the integers r = r

.

The proof of (7.4) is more involved. By (7.17) it suffices to show that for any d with |d| < 2s,

(7.18) min{M

(d), M

(d)} ≤ √

2s + λ

(2s)

+ O(s

).

To prove this estimate, we may clearly assume that d ≥ 0. If d > (2s)

, then we have

√ 2s − d ≤ √

2s − d 2 √

2s ≤ √

2s −

(2s)

, and therefore

M

(d) ≤ √

2s +

(2s)

, which implies (7.18) since λ

≥ 1/ √

2 for all s. Thus it remains to consider the case when 0 ≤ d ≤ (2s)

. Setting

δ = d(2s)

, µ = δ(2s)

= d(2s)

, we have 0 ≤ δ ≤ 1 and hence obtain by Taylor’s formula

√ 2s + εd = √

2s + εd 2 √

2s − d

8(2s)

+ O(s

)

= √

2s + εµ/2 − δ

/8 + O(s

) and

(2s + εd)

= (2s)

+ O(1).

Thus,

M

(d) = √

2s + (2s)

{εδ/2 + q

2k √

2s + εµ/2 − δ

/8k} + O(s

), and to prove (7.18) it suffices to show that the coefficient of (2s)

here is at most λ

for at least one of the choices of ε = ± . This is a consequence of the following lemma.

Lemma 7.4. For ε = ± and real numbers θ, δ, and µ, let λ

(θ, δ, µ) = εδ/2 + p

2kθ + εµ/2 − δ

/8k and put

λ(θ, δ, µ) = min {λ

(θ, δ, µ), λ

(θ, δ, µ)} .