[PPT] - On a problem of S ark ozy and S os for multivariate linear PowerPoint Presentation

SLIDE 1

INTRODUCTION RESULT PROOF REMARKS

On a problem of S ´ ark ¨

zy and S ´
s

for multivariate linear forms

Juanjo Ru´ e Christoph Spiegel

Discrete Mathematics Days

Sevilla, June 2018

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INTRODUCTION RESULT PROOF REMARKS

Some general Motivation: Gauss’ Circle Problem Q: How many integer lattice points are in a circle with radius r centred at the origin?

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INTRODUCTION RESULT PROOF REMARKS

Some general Motivation: Gauss’ Circle Problem Q: How many integer lattice points are in a circle with radius r centred at the origin? A: #{(x, y) ∈ Z2 : x2 + y2 ≤ r2} = π r2 + E(r)

SLIDE 4

INTRODUCTION RESULT PROOF REMARKS

Some general Motivation: Gauss’ Circle Problem Q: How many integer lattice points are in a circle with radius r centred at the origin? A: #{(x, y) ∈ Z2 : x2 + y2 ≤ r2} = π r2 + E(r)

Theorem (Huxley 2003)

We have E(r) = O(r131/208).

SLIDE 5

INTRODUCTION RESULT PROOF REMARKS

Some general Motivation: Gauss’ Circle Problem Q: How many integer lattice points are in a circle with radius r centred at the origin? A: #{(x, y) ∈ Z2 : x2 + y2 ≤ r2} = π r2 + E(r)

Theorem (Huxley 2003)

We have E(r) = O(r131/208).

Theorem (Hardy 1915; Landau 1915)

We cannot have E(r) = o(r1/2 log(r)1/4).

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INTRODUCTION RESULT PROOF REMARKS

Additive representation functions

Definition

For any infinite set A ⊆ N0 and n ∈ N0, let rA(n) = #

(a1, a2) ∈ A2 : a1 + a2 = n
.

(1)

SLIDE 7

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions

Definition

For any infinite set A ⊆ N0 and n ∈ N0, let rA(n) = #

(a1, a2) ∈ A2 : a1 + a2 = n
.

(1)

Remark

Trivially rA(n) is odd if n = 2a for some a ∈ A and even otherwise.

SLIDE 8

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions

Definition

For any infinite set A ⊆ N0 and n ∈ N0, let rA(n) = #

(a1, a2) ∈ A2 : a1 + a2 = n
.

(1)

Remark

Trivially rA(n) is odd if n = 2a for some a ∈ A and even otherwise.

Theorem (Erd˝

s and Fuchs 1956)

For any infinite A ⊆ N and c > 0 we cannot have

N

n=1

rA(n) = cN + o

N1/4 log N−1/2

. (2)

SLIDE 9

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions

Definition

For any infinite set A ⊆ N0 and n ∈ N0, let rA(n) = #

(a1, a2) ∈ A2 : a1 + a2 = n
.

(1)

Remark

Trivially rA(n) is odd if n = 2a for some a ∈ A and even otherwise.

Theorem (Erd˝

s and Fuchs 1956)

For any infinite A ⊆ N and c > 0 we cannot have

N

n=1

rA(n) = cN + o

N1/4 log N−1/2

. (2)

Corollary

Considering the case where A = {m2 : m ∈ N}, c = π/4 and N = r2 − 4r/π, it follows that we cannot have E(r) = o

r1/2 log(r)−1/2

.

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INTRODUCTION RESULT PROOF REMARKS

Additive representation functions S´ ark¨

zy and S´
s ’97: For which k1, . . . , kd ∈ N does there exist an

infinite set A ⊆ N0 and n0 ≥ 0 such that rA(n; k1, . . . , kd) = #

(a1, . . . , ad) ∈ Ad : k1 a1 + · · · + kd ad = n
is constant for n ≥ n0?

SLIDE 11

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions S´ ark¨

zy and S´
s ’97: For which k1, . . . , kd ∈ N does there exist an

infinite set A ⊆ N0 and n0 ≥ 0 such that rA(n; k1, . . . , kd) = #

(a1, . . . , ad) ∈ Ad : k1 a1 + · · · + kd ad = n
is constant for n ≥ n0?

Remark

We already observed that rA(n; 1, 1) cannot become constant.

SLIDE 12

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions S´ ark¨

zy and S´
s ’97: For which k1, . . . , kd ∈ N does there exist an

infinite set A ⊆ N0 and n0 ≥ 0 such that rA(n; k1, . . . , kd) = #

(a1, . . . , ad) ∈ Ad : k1 a1 + · · · + kd ad = n
is constant for n ≥ n0?

Remark

We already observed that rA(n; 1, 1) cannot become constant. We can extend this to rA(n; 1, . . . , 1).

SLIDE 13

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions S´ ark¨

zy and S´
s ’97: For which k1, . . . , kd ∈ N does there exist an

infinite set A ⊆ N0 and n0 ≥ 0 such that rA(n; k1, . . . , kd) = #

(a1, . . . , ad) ∈ Ad : k1 a1 + · · · + kd ad = n
is constant for n ≥ n0?

Remark

We already observed that rA(n; 1, 1) cannot become constant. We can extend this to rA(n; 1, . . . , 1).

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k, k2, . . . , kd−1) = 1.

SLIDE 14

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions S´ ark¨

zy and S´
s ’97: For which k1, . . . , kd ∈ N does there exist an

infinite set A ⊆ N0 and n0 ≥ 0 such that rA(n; k1, . . . , kd) = #

(a1, . . . , ad) ∈ Ad : k1 a1 + · · · + kd ad = n
is constant for n ≥ n0?

Remark

We already observed that rA(n; 1, 1) cannot become constant. We can extend this to rA(n; 1, . . . , 1).

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k, k2, . . . , kd−1) = 1.

Theorem (Ru´ e and Cilleruelo 2009)

For any k1, k2 ≥ 2 and A ⊆ N0, rA(n; k1, k2) cannot become constant.

SLIDE 15

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k, k2, . . . , kd−1) = 1.

Theorem (Ru´ e and Cilleruelo 2009)

For any k1, k2 ≥ 2 and A ⊆ N0, rA(n; k1, k2) cannot become constant.

SLIDE 16

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k, k2, . . . , kd−1) = 1.

Theorem (Ru´ e and Cilleruelo 2009)

For any k1, k2 ≥ 2 and A ⊆ N0, rA(n; k1, k2) cannot become constant.

Theorem (Ru´ e and S. 2018+)

If there are pairwise co-prime integers q1, . . . , qm ≥ 2 such that ki = qb(i,1)

1

· · · qb(i,m)

m

≥ 2 (3) where b(i, j) ∈ {0, 1}, then rA(n; k1, . . . , kd) cannot become constant for any infinite A ⊆ N0.

SLIDE 17

INTRODUCTION RESULT PROOF REMARKS

Additive representation functions

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k, k2, . . . , kd−1) = 1.

Theorem (Ru´ e and Cilleruelo 2009)

For any k1, k2 ≥ 2 and A ⊆ N0, rA(n; k1, k2) cannot become constant.

Theorem (Ru´ e and S. 2018+)

If there are pairwise co-prime integers q1, . . . , qm ≥ 2 such that ki = qb(i,1)

1

· · · qb(i,m)

m

≥ 2 (3) where b(i, j) ∈ {0, 1}, then rA(n; k1, . . . , kd) cannot become constant for any infinite A ⊆ N0. This includes the case of pairwise co-prime k1, . . . , kd ≥ 2.

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INTRODUCTION RESULT PROOF REMARKS

The proof of Moser’s result

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k) = 1 for all n ≥ 0.

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INTRODUCTION RESULT PROOF REMARKS

The proof of Moser’s result

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k) = 1 for all n ≥ 0.

Proof.

The generating function of A is fA(z) =

a∈A za.

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INTRODUCTION RESULT PROOF REMARKS

The proof of Moser’s result

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k) = 1 for all n ≥ 0.

Proof.

The generating function of A is fA(z) =

a∈A za. We have

fA(z) fA(zk) =

(a,a′)∈A2

za+ka′ =

∞

n=0

r(n; 1, k) zn

!

=

∞

n=0

zn = 1 1 − z. (4)

SLIDE 21

INTRODUCTION RESULT PROOF REMARKS

The proof of Moser’s result

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k) = 1 for all n ≥ 0.

Proof.

The generating function of A is fA(z) =

a∈A za. We have

fA(z) fA(zk) =

(a,a′)∈A2

za+ka′ =

∞

n=0

r(n; 1, k) zn

!

=

∞

n=0

zn = 1 1 − z. (4) Writing fA(z) = (1 − z)−1f −1

A (zk) and repeatedly substituting,

SLIDE 22

INTRODUCTION RESULT PROOF REMARKS

The proof of Moser’s result

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k) = 1 for all n ≥ 0.

Proof.

The generating function of A is fA(z) =

a∈A za. We have

fA(z) fA(zk) =

(a,a′)∈A2

za+ka′ =

∞

n=0

r(n; 1, k) zn

!

=

∞

n=0

zn = 1 1 − z. (4) Writing fA(z) = (1 − z)−1f −1

A (zk) and repeatedly substituting, we get

fA(z) =

∞

j=0
1 + z(k2)j

+ z2(k2)j + · · · + +z(k−1)(k2)j .

SLIDE 23

INTRODUCTION RESULT PROOF REMARKS

The proof of Moser’s result

Theorem (Moser 1962)

For any k ≥ 2 there exists A ⊆ N0 such that rA(n; 1, k) = 1 for all n ≥ 0.

Proof.

The generating function of A is fA(z) =

a∈A za. We have

fA(z) fA(zk) =

(a,a′)∈A2

za+ka′ =

∞

n=0

r(n; 1, k) zn

!

=

∞

n=0

zn = 1 1 − z. (4) Writing fA(z) = (1 − z)−1f −1

A (zk) and repeatedly substituting, we get

fA(z) =

∞

j=0
1 + z(k2)j

+ z2(k2)j + · · · + +z(k−1)(k2)j . This is the representation function of the set of all integers whose k2-ary representation has only digits strictly smaller than k.

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INTRODUCTION RESULT PROOF REMARKS

Proof Outline of Main Result

Proof Outline.

From here on we will abbreviate rA(n) = rA(n; k1, . . . , kd).

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INTRODUCTION RESULT PROOF REMARKS

Proof Outline of Main Result

Proof Outline.

From here on we will abbreviate rA(n) = rA(n; k1, . . . , kd). Generalising the previous approach, we have fA(zk1) · · · fA(zkd) =

(a1,...,ad)∈Ad

zk1a1+···+kdad =

∞

n=0

rA(n) zn. (5)

SLIDE 26

INTRODUCTION RESULT PROOF REMARKS

Proof Outline of Main Result

Proof Outline.

From here on we will abbreviate rA(n) = rA(n; k1, . . . , kd). Generalising the previous approach, we have fA(zk1) · · · fA(zkd) =

(a1,...,ad)∈Ad

zk1a1+···+kdad =

∞

n=0

rA(n) zn. (5) If rA(n) becomes constant, that is there exist c > 0 and n0 ≥ 0 such that rA(n) = c for n ≥ n0 ≥ 0,

SLIDE 27

INTRODUCTION RESULT PROOF REMARKS

Proof Outline of Main Result

Proof Outline.

From here on we will abbreviate rA(n) = rA(n; k1, . . . , kd). Generalising the previous approach, we have fA(zk1) · · · fA(zkd) =

(a1,...,ad)∈Ad

zk1a1+···+kdad =

∞

n=0

rA(n) zn. (5) If rA(n) becomes constant, that is there exist c > 0 and n0 ≥ 0 such that rA(n) = c for n ≥ n0 ≥ 0, then

∞

n=0

rA(n) zn =

n0−1

n=0

rA(n) zn +

∞

n=n0

c zn = Q(z) + c zn0 1 − z = P(z) 1 − z (6)

SLIDE 28

INTRODUCTION RESULT PROOF REMARKS

Proof Outline of Main Result

Proof Outline.

From here on we will abbreviate rA(n) = rA(n; k1, . . . , kd). Generalising the previous approach, we have fA(zk1) · · · fA(zkd) =

(a1,...,ad)∈Ad

zk1a1+···+kdad =

∞

n=0

rA(n) zn. (5) If rA(n) becomes constant, that is there exist c > 0 and n0 ≥ 0 such that rA(n) = c for n ≥ n0 ≥ 0, then

∞

n=0

rA(n) zn =

n0−1

n=0

rA(n) zn +

∞

n=n0

c zn = Q(z) + c zn0 1 − z = P(z) 1 − z (6) where Q ∈ N0[z] and P ∈ Z[z] are polynomials and P(1) = 0.

SLIDE 29

INTRODUCTION RESULT PROOF REMARKS

Proof Outline of Main Result

Proof Outline.

From here on we will abbreviate rA(n) = rA(n; k1, . . . , kd). Generalising the previous approach, we have fA(zk1) · · · fA(zkd) =

(a1,...,ad)∈Ad

zk1a1+···+kdad =

∞

n=0

rA(n) zn. (5) If rA(n) becomes constant, that is there exist c > 0 and n0 ≥ 0 such that rA(n) = c for n ≥ n0 ≥ 0, then

∞

n=0

rA(n) zn =

n0−1

n=0

rA(n) zn +

∞

n=n0

c zn = Q(z) + c zn0 1 − z = P(z) 1 − z (6) where Q ∈ N0[z] and P ∈ Z[z] are polynomials and P(1) = 0. We have fA(zk1) · · · fA(zkd) = P(z) 1 − z. (7)

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INTRODUCTION RESULT PROOF REMARKS

Introducing cyclotomic poylinomials The cyclotomic polynomial of order n is given by Φn(z) =

ξ∈φn

(z − ξ) (8) where φn = {ξ ∈ C : ξk = 1 iff k ≡ 0 mod n}.

SLIDE 31

INTRODUCTION RESULT PROOF REMARKS

Introducing cyclotomic poylinomials The cyclotomic polynomial of order n is given by Φn(z) =

ξ∈φn

(z − ξ) (8) where φn = {ξ ∈ C : ξk = 1 iff k ≡ 0 mod n}. We have

SLIDE 32

INTRODUCTION RESULT PROOF REMARKS

Introducing cyclotomic poylinomials The cyclotomic polynomial of order n is given by Φn(z) =

ξ∈φn

(z − ξ) (8) where φn = {ξ ∈ C : ξk = 1 iff k ≡ 0 mod n}. We have (i) Φn ∈ Z[z] for all n

SLIDE 33

INTRODUCTION RESULT PROOF REMARKS

Introducing cyclotomic poylinomials The cyclotomic polynomial of order n is given by Φn(z) =

ξ∈φn

(z − ξ) (8) where φn = {ξ ∈ C : ξk = 1 iff k ≡ 0 mod n}. We have (i) Φn ∈ Z[z] for all n and (ii) Φn is irreducible over Z[z].

SLIDE 34

INTRODUCTION RESULT PROOF REMARKS

Introducing cyclotomic poylinomials The cyclotomic polynomial of order n is given by Φn(z) =

ξ∈φn

(z − ξ) (8) where φn = {ξ ∈ C : ξk = 1 iff k ≡ 0 mod n}. We have (i) Φn ∈ Z[z] for all n and (ii) Φn is irreducible over Z[z]. Recall that fA(zk1) · · · fA(zkd) = P(z)/(1 − z) for some P(z) ∈ Z[z] satisfy- ing P(1) = 0.

SLIDE 35

INTRODUCTION RESULT PROOF REMARKS

Introducing cyclotomic poylinomials The cyclotomic polynomial of order n is given by Φn(z) =

ξ∈φn

(z − ξ) (8) where φn = {ξ ∈ C : ξk = 1 iff k ≡ 0 mod n}. We have (i) Φn ∈ Z[z] for all n and (ii) Φn is irreducible over Z[z]. Recall that fA(zk1) · · · fA(zkd) = P(z)/(1 − z) for some P(z) ∈ Z[z] satisfy- ing P(1) = 0. Now, for any n there exists a unique sn ∈ N0 s.t. Pn(z) = P(z) Φ−sn

n

(z) (9) satisfies Pn(z) ∈ Z[z] as well as Pn(ξ) = 0 for any ξ ∈ φn.

SLIDE 36

INTRODUCTION RESULT PROOF REMARKS

Factoring out the generating function If rA(n) becomes constant for some A ⊆ N0, then fA(zk1) · · · fA(zkd) = P(z) 1 − z where P(1) = 0 (7)

SLIDE 37

INTRODUCTION RESULT PROOF REMARKS

Factoring out the generating function If rA(n) becomes constant for some A ⊆ N0, then fA(zk1) · · · fA(zkd) = P(z) 1 − z where P(1) = 0 (7) and there exist sn ∈ N0 such that Pn(z) = P(z) Φ−sn

n

(z) satisfies Pn(ξ) = 0 for ξ ∈ φn. (9)

SLIDE 38

INTRODUCTION RESULT PROOF REMARKS

Factoring out the generating function If rA(n) becomes constant for some A ⊆ N0, then fA(zk1) · · · fA(zkd) = P(z) 1 − z where P(1) = 0 (7) and there exist sn ∈ N0 such that Pn(z) = P(z) Φ−sn

n

(z) satisfies Pn(ξ) = 0 for ξ ∈ φn. (9)

Proposition

For any (j1, . . . , jd) ∈ Nd

0 there exist rj satisfying

lim

ω→1 f(ωξ) · Φ −rj k

j1 1 ···k jd d

(ωξ) / ∈ {0, ±∞} (10) for any ξ ∈ φk

j1 1 ···k jd d .

SLIDE 39

INTRODUCTION RESULT PROOF REMARKS

Factoring out the generating function If rA(n) becomes constant for some A ⊆ N0, then fA(zk1) · · · fA(zkd) = P(z) 1 − z where P(1) = 0 (7) and there exist sn ∈ N0 such that Pn(z) = P(z) Φ−sn

n

(z) satisfies Pn(ξ) = 0 for ξ ∈ φn. (9)

Proposition

For any (j1, . . . , jd) ∈ Nd

0 there exist rj satisfying

lim

ω→1 f(ωξ) · Φ −rj k

j1 1 ···k jd d

(ωξ) / ∈ {0, ±∞} (10) for any ξ ∈ φk

j1 1 ···k jd d . These exponents satisfy r0 = −1/d

SLIDE 40

INTRODUCTION RESULT PROOF REMARKS

Factoring out the generating function If rA(n) becomes constant for some A ⊆ N0, then fA(zk1) · · · fA(zkd) = P(z) 1 − z where P(1) = 0 (7) and there exist sn ∈ N0 such that Pn(z) = P(z) Φ−sn

n

(z) satisfies Pn(ξ) = 0 for ξ ∈ φn. (9)

Proposition

For any (j1, . . . , jd) ∈ Nd

0 there exist rj satisfying

lim

ω→1 f(ωξ) · Φ −rj k

j1 1 ···k jd d

(ωξ) / ∈ {0, ±∞} (10) for any ξ ∈ φk

j1 1 ···k jd d . These exponents satisfy r0 = −1/d and

r(j1⊖b(1,1), ..., jd⊖b(d,1)) + · · · + r(j1⊖b(1,m), ..., jd⊖b(d,m)) = dsj (11) for all j ∈ Nm

0 \ {0} where a ⊖ b = max(a − b, 0) and sj = sk

j1 1 ···k jd d .

SLIDE 41

INTRODUCTION RESULT PROOF REMARKS

Finding a contradiction in the exponents Consider the case of Ru´ e and Cilleruelo, that is we have d = 2.

SLIDE 42

INTRODUCTION RESULT PROOF REMARKS

Finding a contradiction in the exponents Consider the case of Ru´ e and Cilleruelo, that is we have d = 2. The proposition gives the existence of {rj : j ∈ N2

0} satisfying

(i) r(0,0) = −1/2, (ii) r(j+1,0) = s(j+1,0) − r(j,0), (iii) r(0,j+1) = s(0,j+1) − r(0,j) and (iv) r(j1+1,j2) + r(j1,j2+1) = s(j1+1,j2+1).

SLIDE 43

INTRODUCTION RESULT PROOF REMARKS

Finding a contradiction in the exponents Consider the case of Ru´ e and Cilleruelo, that is we have d = 2. The proposition gives the existence of {rj : j ∈ N2

0} satisfying

(i) r(0,0) = −1/2, (ii) r(j+1,0) = s(j+1,0) − r(j,0), (iii) r(0,j+1) = s(0,j+1) − r(0,j) and (iv) r(j1+1,j2) + r(j1,j2+1) = s(j1+1,j2+1). Inductively, as s⋆ ∈ N0, we have r⋆ / ∈ Z and therefore r⋆ = 0 due to (i).

SLIDE 44

INTRODUCTION RESULT PROOF REMARKS

Finding a contradiction in the exponents Consider the case of Ru´ e and Cilleruelo, that is we have d = 2. The proposition gives the existence of {rj : j ∈ N2

0} satisfying

(i) r(0,0) = −1/2, (ii) r(j+1,0) = s(j+1,0) − r(j,0), (iii) r(0,j+1) = s(0,j+1) − r(0,j) and (iv) r(j1+1,j2) + r(j1,j2+1) = s(j1+1,j2+1). Inductively, as s⋆ ∈ N0, we have r⋆ / ∈ Z and therefore r⋆ = 0 due to (i). As P is a polynomial there exists ℓ0 such that sj1,j2 = 0 if j1 + j2 ≥ ℓ0.

SLIDE 45

INTRODUCTION RESULT PROOF REMARKS

Finding a contradiction in the exponents Consider the case of Ru´ e and Cilleruelo, that is we have d = 2. The proposition gives the existence of {rj : j ∈ N2

0} satisfying

(i) r(0,0) = −1/2, (ii) r(j+1,0) = s(j+1,0) − r(j,0), (iii) r(0,j+1) = s(0,j+1) − r(0,j) and (iv) r(j1+1,j2) + r(j1,j2+1) = s(j1+1,j2+1). Inductively, as s⋆ ∈ N0, we have r⋆ / ∈ Z and therefore r⋆ = 0 due to (i). As P is a polynomial there exists ℓ0 such that sj1,j2 = 0 if j1 + j2 ≥ ℓ0. Assume w.l.o.g. that ℓ0 is odd.

SLIDE 46

INTRODUCTION RESULT PROOF REMARKS

Finding a contradiction in the exponents Consider the case of Ru´ e and Cilleruelo, that is we have d = 2. The proposition gives the existence of {rj : j ∈ N2

0} satisfying

(i) r(0,0) = −1/2, (ii) r(j+1,0) = s(j+1,0) − r(j,0), (iii) r(0,j+1) = s(0,j+1) − r(0,j) and (iv) r(j1+1,j2) + r(j1,j2+1) = s(j1+1,j2+1). Inductively, as s⋆ ∈ N0, we have r⋆ / ∈ Z and therefore r⋆ = 0 due to (i). As P is a polynomial there exists ℓ0 such that sj1,j2 = 0 if j1 + j2 ≥ ℓ0. Assume w.l.o.g. that ℓ0 is odd. Now

◮ r(ℓ0+1,0) = −r(ℓ0,0) due to (ii),

SLIDE 47

INTRODUCTION RESULT PROOF REMARKS

Finding a contradiction in the exponents Consider the case of Ru´ e and Cilleruelo, that is we have d = 2. The proposition gives the existence of {rj : j ∈ N2

0} satisfying

(i) r(0,0) = −1/2, (ii) r(j+1,0) = s(j+1,0) − r(j,0), (iii) r(0,j+1) = s(0,j+1) − r(0,j) and (iv) r(j1+1,j2) + r(j1,j2+1) = s(j1+1,j2+1). Inductively, as s⋆ ∈ N0, we have r⋆ / ∈ Z and therefore r⋆ = 0 due to (i). As P is a polynomial there exists ℓ0 such that sj1,j2 = 0 if j1 + j2 ≥ ℓ0. Assume w.l.o.g. that ℓ0 is odd. Now

◮ r(ℓ0+1,0) = −r(ℓ0,0) due to (ii), ◮ r(0,ℓ0+1) = −r(0,ℓ0) due to (iii),

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INTRODUCTION RESULT PROOF REMARKS

Finding a contradiction in the exponents Consider the case of Ru´ e and Cilleruelo, that is we have d = 2. The proposition gives the existence of {rj : j ∈ N2

0} satisfying

(i) r(0,0) = −1/2, (ii) r(j+1,0) = s(j+1,0) − r(j,0), (iii) r(0,j+1) = s(0,j+1) − r(0,j) and (iv) r(j1+1,j2) + r(j1,j2+1) = s(j1+1,j2+1). Inductively, as s⋆ ∈ N0, we have r⋆ / ∈ Z and therefore r⋆ = 0 due to (i). As P is a polynomial there exists ℓ0 such that sj1,j2 = 0 if j1 + j2 ≥ ℓ0. Assume w.l.o.g. that ℓ0 is odd. Now

◮ r(ℓ0+1,0) = −r(ℓ0,0) due to (ii), ◮ r(0,ℓ0+1) = −r(0,ℓ0) due to (iii), ◮ r(ℓ0,0) = r(0,ℓ0) and r(ℓ0+1,0) = −r(0,ℓ0+1) due to (iv)

SLIDE 49

INTRODUCTION RESULT PROOF REMARKS

Finding a contradiction in the exponents Consider the case of Ru´ e and Cilleruelo, that is we have d = 2. The proposition gives the existence of {rj : j ∈ N2

0} satisfying

(i) r(0,0) = −1/2, (ii) r(j+1,0) = s(j+1,0) − r(j,0), (iii) r(0,j+1) = s(0,j+1) − r(0,j) and (iv) r(j1+1,j2) + r(j1,j2+1) = s(j1+1,j2+1). Inductively, as s⋆ ∈ N0, we have r⋆ / ∈ Z and therefore r⋆ = 0 due to (i). As P is a polynomial there exists ℓ0 such that sj1,j2 = 0 if j1 + j2 ≥ ℓ0. Assume w.l.o.g. that ℓ0 is odd. Now

◮ r(ℓ0+1,0) = −r(ℓ0,0) due to (ii), ◮ r(0,ℓ0+1) = −r(0,ℓ0) due to (iii), ◮ r(ℓ0,0) = r(0,ℓ0) and r(ℓ0+1,0) = −r(0,ℓ0+1) due to (iv)

implying the contradiction r(ℓ0,0) = r(0,ℓ0) = r(ℓ0+1,0) = r(0,ℓ0+1) = 0.

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INTRODUCTION RESULT PROOF REMARKS

Remarks and Open Problems

Conjecture

The cases covered by Moser, that is 1, k, k2, . . . , kd−1, are the only ones for which rA(n) can become constant.

1. What about cases not covered by our result, e.g. rA(n; 2, 3, 4) or

rA(1, 2, 6)?

2. What about the unordered variant

RA(n; k1, . . . , kd) = #

{a1, . . . , ad} ∈ 2A : k1 a1 + · · · + kd ad = n
?
3. What about an Erd˝
s-Fuchs-type result for k1 = 2 and k2 = 3?

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INTRODUCTION RESULT PROOF REMARKS