Modified Erd\H{o}s-Ginzburg-Ziv Constants for $(\mathbb{Z}/n\mathbb{Z})^2$

TL;DR

This paper investigates bounds for the modified Erdős-Ginzburg-Ziv constants in specific finite abelian groups, extending known results and exploring the behavior of zero-sum subsequences of various lengths.

Contribution

It provides new bounds for the modified Erdős-Ginzburg-Ziv constants in groups like  and , and analyzes the constants for subsequences of different lengths.

Findings

Bounds for s'_{t}(G) in ()^2 and () groups.

Bounds for s'_{t}(G) in ()^d with specified subsequence lengths.

Analysis of the Erds-Ginzburg-Ziv constant for ()^2 with subsequences of length tn.

Abstract

For an abelian group $G$ and an integer $t > 0$, the modified Erd\H{o}s-Ginzburg-Ziv constant $s'_t(G)$ is the smallest integer $\ell$ such that any zero-sum sequence of length at least $\ell$ with elements in $G$ contains a zero-sum subsequence (not necessarily consecutive) of length $t$. We compute bounds for $s'_{t}(G)$ for $G = \left(\mathbb{Z}/n\mathbb{Z}\right)^2$ and $G = \left(\mathbb{Z}/n_1\mathbb{Z} \times \mathbb{Z}/n_2\mathbb{Z}\right)$. We also compute bounds for $G = \left(\mathbb{Z}/p\mathbb{Z}\right)^d$ where the subsequence can be any length in $\{p, \dots, (d-1)p\}$. Lastly, we investigate the Erd\H{o}s-Ginzburg-Ziv constant for $G = \left(\mathbb{Z}/n\mathbb{Z}\right)^2$ and subsequences of length $tn$.

Equations

$$s_{pt}^{\prime }\left((\mathbb{Z}/p\mathbb{Z}{)}^2\right)=(t+2)p-2.$$

$$s_{3t}^{\prime }\left((\mathbb{Z}/3\mathbb{Z}{)}^2\right)=3(t+1).$$

$$(t+2)n-k+1\le s_{nt}^{\prime }\left((\mathbb{Z}/n\mathbb{Z}{)}^2\right)\le (t+2)n+m-3,$$

$$s_{n_2}\left(\mathbb{Z}/n_1\mathbb{Z}\times \mathbb{Z}/n_2\mathbb{Z}\right)=2n_1+2n_2-3.$$

$$2n_2-\ell \le s_{n_2}^{\prime }\left(\mathbb{Z}/n_1\mathbb{Z}\times \mathbb{Z}/n_2\mathbb{Z}\right)\le 2n_1+2n_2-\ell +1,$$

$$s_{n_{2}t}^{\prime }\left(\mathbb{Z}/n_1\mathbb{Z}\times \mathbb{Z}/n_2\mathbb{Z}\right)\le 2n_1+(t+1)n_2-\ell +1.$$

$$s_{pt}\left((\mathbb{Z}/p\mathbb{Z}{)}^2\right)=(t+2)p-2.$$

$$(t+2)n-2\le s_{nt}\left((\mathbb{Z}/n\mathbb{Z}{)}^2\right)\le (t+2)n+m-3.$$

$$(0,0)a$$

$$(1,0)b$$

$$(0,1)c$$

$$(1,1)d$$

$$(tn-1+n-k+2)r+(n-k+2+k-3)(r+1)$$

$$(tn-1+n-k+2)s+(n-k+2+k-3)(s+1)$$

$$k(-r)$$

$$k(-s)$$

$$\displaystyle(0,0)\$$

$$\displaystyle(1,0)\$$

$$\displaystyle(0,1)\$$

$$\displaystyle(1,1)\$$

$$3((t+1)+1)\ge 4(3)-3,$$

$$2m_1+2m_2-3\ge 2m_1+2m_2-\ell +1,$$

$$\displaystyle(0,0)\ \$$

$$\displaystyle(1,1)\ \$$

$$r(-\ell )$$

$$s(-\ell )$$

$$(0,0)a$$

$$(1,0)b$$

$$(0,1)c$$

$$i\equiv 0(\mathrm{mod}n)\text{ and }j\equiv 0(\mathrm{mod}n).$$

$$1-(p\mid J)+(2p\mid J)\equiv 0(\mathrm{mod}p).$$

$$(p\mid J)\equiv 1(\mathrm{mod}p).$$

$$(t+2)p-2\ge 4p-3.$$

$$(t+2)n-2+(m-1)\ge 4n-3.$$

$$1-(p-1\mid J)-(p\mid J)+(2p-1\mid J)+(2p\mid J)-(3p-1\mid J)-(3p\mid J)\equiv 0\mathrm{mod}p.$$

Full text

1. Introduction

In 1961, Erdős, Ginzburg, and Ziv proved the following theorem.

Theorem 1.1** (Erdős-Ginzburg-Ziv [4]).**

Any sequence of length $2n-1$ in $\mathbb{Z}/n\mathbb{Z}$ contains a zero-sum subsequence of length $n$.

Many different proofs of this theorem have been given since the original in 1961. Perhaps the simplest proof makes use of the Chevalley-Warning theorem. Here, we don’t require the subsequence to be consecutive, and a sequence is zero-sum if its elements sum to zero. This theorem has inspired many follow-up questions on zero-sum sequences.

For the general case we consider the following problem: Let $G$ be an abelian group and let $\mathcal{L}\subseteq\mathbb{N}$. Then $s_{\mathcal{L}}(G)$ is defined to be the minimal $\ell$ such that any sequence of length $\ell$ with elements in $G$ contains a zero-sum subsequence whose length is in $\mathcal{L}$. When $\mathcal{L}=\mathrm{exp}(G)$, this is the Erdős-Ginzburg-Ziv constant.

In this paper we will also study the modified Erdős-Ginzburg-Ziv constant $s^{\prime}_{\mathcal{L}}(G)$ defined as the smallest $\ell$ such that any zero-sum sequence of length at least $\ell$ with elements in $G$ contains a zero-sum subsequence whose length is in $\mathcal{L}$. When $\mathcal{L}=\{t\}$ is a singleton set, we ignore the bracket notation. Note that one may also study the problem for subsets $G_{0}\subseteq G$. However, in this paper we will always consider the modified or unmodified constant of the entire group $G$. In 2019, Berger and Wang determined modified EGZ constants in the finite cyclic case and some extensions. In particular they prove:

Theorem 1.2** ([3], Theorem 1.3).**

The modified EGZ constant of $\mathbb{Z}/n\mathbb{Z}$ is given by $s^{\prime}_{nt}(\mathbb{Z}/n\mathbb{Z})=(t+1)n-\ell+1$, where $\ell$ is the smallest integer such that $\ell\mathchar 13613\relax n$.

Theorem 1.3** ([3], Theorem 1.4).**

We have $s^{\prime}_{n}\left((\mathbb{Z}/n\mathbb{Z})^{2}\right)=4n-\ell+1$ where $\ell$ is the smallest integer such that $\ell\geq 4$ and $\ell\mathchar 13613\relax n$.

They also state the following problem:

Problem 1.1** ([3], Problem 4.3).**

Compute $s^{\prime}_{nt}\left((\mathbb{Z}/n\mathbb{Z})^{2}\right)$ for $t>1$.

Our first two results provide partial answers for Problem 1.1.

Theorem 1.4**.**

If $p\neq 3$ is prime and $t\geq 2$ then the modified EGZ constant of $\left(\mathbb{Z}/p\mathbb{Z}\right)^{2}$ is given by

[TABLE]

If $p=3$, we have

[TABLE]

Theorem 1.5**.**

Let $t\geq 2$ and write $n=pm$. Then the modified EGZ constant of $\left(\mathbb{Z}/n\mathbb{Z}\right)^{2}$ satisfies the bounds

[TABLE]

where $k$ is the smallest integer such that $k\geq 3$ and $\gcd(n,k)=1$. Note that when $n=p$ is prime, the upper and lower bounds match and we obtain Theorem 1.4.

In 2006, Halter-Koch and Geroldinger obtained the following result.

Theorem 1.6** ([5], Theorem 5.8.3).**

The EGZ constant of $\left(\mathbb{Z}/n_{1}\mathbb{Z}\times\mathbb{Z}/n_{2}\mathbb{Z}\right)$, where $n_{1}\mid n_{2}$, is given by

[TABLE]

We investigate the problem of computing the modified constant for this group first posed in [3].

Problem 1.2** ([3], Problem 4.4).**

Compute $s^{\prime}_{n_{2}t}\left(\mathbb{Z}/n_{1}\mathbb{Z}\times\mathbb{Z}/n_{2}\mathbb{Z}\right)$ for $t\geq 1$ and $n_{1}\mid n_{2}$.

We give bounds for this case. We split it up into two theorems. In Theorem 1.7, we provide upper and lower bounds when $t=1$. For $t>1$, we are able to prove the following upper bound for the modified EGZ constant.

Theorem 1.7**.**

The modified EGZ constant of $\left(\mathbb{Z}/n_{1}\mathbb{Z}\times\mathbb{Z}/n_{2}\mathbb{Z}\right)$, where $n_{1}\mid n_{2}$, satisfies the bounds

[TABLE]

where $\ell$ is the smallest integer such that $\ell\geq 4$ and $\ell\mathchar 13613\relax n_{2}$.

Theorem 1.8**.**

Let $\ell$ be the smallest integer such that $\ell\geq 4$ and $\ell\mathchar 13613\relax n_{1},n_{2}$. Let $t\geq 1$ and $n_{1}\mid n_{2}$. Then

[TABLE]

Lastly, we investigate the (unmodified) EGZ constant of $\left(\mathbb{Z}/n\mathbb{Z}\right)^{2}$. In 1983, Kemnitz [7] conjectured that $s_{n}\left((\mathbb{Z}/n\mathbb{Z})\right)^{2}=4n-3$. In 1993, Alon and Dubiner [1] proved that $s_{n}\left((\mathbb{Z}/n\mathbb{Z}\right)^{2}\leq 6n-5$ and showed $s_{p}\left((\mathbb{Z}/p\mathbb{Z})^{2}\right)\leq 5p-2$ for sufficiently large primes $p$. In 2000, Róyai [10] proved $s_{p}\left((\mathbb{Z}/p\mathbb{Z})^{2}\right)\leq 4p-2$. Finally, in 2007, Reiher [8] resolved Kemnitz’s Conjecture.

Theorem 1.9** ([8], Theorem 3.2).**

If $J$ is a sequence of length $4n-3$ in $\left(\mathbb{Z}/n\mathbb{Z}\right)^{2}$, then $(n\mid J)>0$.

We consider the EGZ constant when $\mathcal{L}=\{nt\}$, $t\geq 1$, i.e., the minimal $\ell$ such that any sequence of length $\ell$ contains a zero-sum subsequence of length $2n$ or $3n$, etc.

Theorem 1.10**.**

If $t\geq 2$ and $n=p$ is prime, then we have

[TABLE]

Corollary 1.11**.**

Let $t\geq 2$ and write $n=pm$. We have

[TABLE]

Note if $n=p\neq 3$ is prime, then $m=1$ and we recover Theorem 1.10.

2. Proofs of Theorems 1.5 and 1.4

In this section we give the proof of Theorem 1.5. As in [3], if $J$ is a sequence of elements of $\left(\mathbb{Z}/n\mathbb{Z}\right)^{2}$, we use $(k\mid J)$ to denote the number of zero-sum subsequences of $J$ of size $k$. The proof closely follows the ideas presented in [3].

Proposition 2.1**.**

Let $3\leq k\leq n-1$ be the least integer such that $\gcd(n,k)=1$. Then there exists a zero-sum sequence in $(\mathbb{Z}/n\mathbb{Z})^{2}$ of length $(t+2)n-k$ which contains no zero-sum subsequences of length $nt$.

Proof.

Consider a sequence of the form

[TABLE]

where $a$ denotes the number of $(0,0)$’s, etc. It suffices to show that there is no zero-sum subsequence of any length among the nonzero elements, otherwise we could add copies of $(0,0)$ until we have a zero-sum subsequence of length $nt$. Indeed the sum of the nonzero elements is $(n-1,n-1)$ and there are at most $n-1$ nonzero elements with value $1$ being summed in each coordinate, so there is no zero-sum subsequence modulo $n$. Note also that since $k\geq 3$, $b,c\leq n-1$ and $d\leq n-4$, we cannot form a zero-sum subsequence using copies of only one basis element. We claim there exists $(r,s)\in(\mathbb{Z}/n\mathbb{Z})^{2}$ such that adding $(r,s)$ to each term of the above sequence will result in a zero-sum sequence. Note that adding $(r,s)$ to each term does not change the fact that there is no zero-sum subsequence of length $nt$. Indeed, we only need to satisfy the divisibilty relations

[TABLE]

which reduce to

[TABLE]

We can solve for $(r,s)$ since $\gcd(n,k)=1$. ∎

Proposition 2.2**.**

There exists a zero-sum sequence in $(\mathbb{Z}/3\mathbb{Z})^{2}$ of length $(3t+2)$ which contains no zero-sum subsequence of length $3t$.

Proof.

Consider a sequence of the form

[TABLE]

There is clearly no zero-sum subsequence of length $3t$. We claim there exists $(r,s)\in(\mathbb{Z}/n\mathbb{Z})^{2}$ such that adding $(r,s)$ to each term of the above sequence will result in a zero-sum sequence. It is easy to check that $(2,2)$ works. ∎

Proposition 2.2 provides the lower bound for the $p=3$ case of Theorem 1.4. Proposition 2.1 provides the lower bound for both Theorem 1.4 and Theorem 1.5, by noting that when $n=p$ is prime, $k=3$.

Proposition 2.3**.**

If $J$ is a zero-sum sequence in $(\mathbb{Z}/3\mathbb{Z})^{2}$ of length $3(t+1)$ then $(3t\mid J)>0$.

Proof.

We induct on $t$. Note that when $t=2$, we have $3(t+1)=9=4(3)-3$. By Theorem 1.9, we can remove a zero-sum subsequence of length $3$, leaving us with $6$ elements. Since the original sequence of length $9$ was zero-sum, the remaining $6$ elements are zero-sum, so we have found our zero-sum subsequence of length $3t$. Now suppose the statement is true for all positive integers at most $t\geq 2$. Consider a zero-sum sequence of length $3((t+1)+1)$. We have

[TABLE]

since $t\geq 1$. So we remove a zero-sum subsequence of length $3$. This leaves a zero-sum sequence with $3(t+1)$ elements. By the induction hypothesis, this has a zero-sum subsequence of length $3t$. Combining this subsequence with the zero-sum subsequence of length $3$ we removed yields a zero-sum subsequence of length $3(t+1)$, as desired. ∎

Note that in general the modified EGZ constant is bounded above by the EGZ constant. If any sequence of some length has a zero-sum subsequence, then surely any zero-sum sequence of that same length will have a zero-sum subsequence. Theorem 1.10 and Corollary 1.11 provide the upper bounds to finish the proofs of Theorems 1.4 and 1.5. Note in the case $p=3$, the value of the upper bound provided by Theorem 1.10 is exactly one more than the length in Proposition 2.3.

Now we prove an analogue of a key lemma from [3].

Lemma 2.4** ([3], Lemma 3.4).**

If $J$ is a zero-sum sequence of length $3n$ in $(\mathbb{Z}/n\mathbb{Z})^{2}$, then $(n\mid J)$.

Proposition 2.5**.**

If $J$ is a zero-sum sequence of length $3n$ in $(\mathbb{Z}/n\mathbb{Z})^{2}$ then $(2n\mid J)>0$.

Proof.

By Lemma 2.4, $(n\mid J)>0$. If $(n\mid J)=1$, then the complement sequence of length $2n$ is zero-sum since $J$ is zero-sum. Otherwise $(n\mid J)\geq 2$, in which case we can pick $2$ of the zero-sum subsequences of length $n$ and combine them to obtain a zero-sum subsequence of length $2n$. ∎

Now we generalize Proposition 2.5 for all $t\geq 2$.

Corollary 2.6**.**

If $J$ is a zero-sum sequence of length $(t+1)n$ in $(\mathbb{Z}/n\mathbb{Z})^{2}$ and $t\geq 2$, then $(tn\mid J)>0$.

Proof.

The $t=2$ case is Proposition 2.5. Assume $t\geq 3$. Then $(t+1)n>4n-3$. By Theorem 1.9, we can remove zero-sum subsequences of length $n$ until there are exactly $3n$ remaining. This gives us $t-2$ zero-sum subsequences of length $n$. Since $J$ is zero-sum, the $3n$ remaining elements are zero-sum. Hence by Proposition 2.5, there is a zero-sum subsequence of length $2n$. Combining this with the $t-2$ zero-sum subsequences of length $n$ gives a zero-sum subsequence of length $nt$. ∎

3. Proof of Theorems 1.7 and 1.8

Proposition 3.1**.**

Let $\ell$ be the smallest positive integer greater than or equal to $4$ such that $\ell\mathchar 13613\relax n_{1}$. If $J$ is a zero-sum sequence in $G=\left(\mathbb{Z}/n_{1}\mathbb{Z}\times\mathbb{Z}/n_{2}\mathbb{Z}\right)$ with $n_{1}\mid n_{2}$ and $J$ has length at least $2n_{1}+2n_{2}-\ell+1$, then $(n_{2}\mid J)>0$.

Proof.

Assume $n_{1}\neq n_{2}$, otherwise this is just the $(\mathbb{Z}/n\mathbb{Z})^{2}$ case. We proceed by strong induction on the exponent of the group. Note that $\exp(G)=n_{2}$ in this case. Let $d$ be a divisor of $n_{1}$ such that $d\mid n_{1}$, $d<n_{2}$ and write $n_{1}=dm_{1}$ and $n_{2}=dm_{2}$. Note that $H=\left(\mathbb{Z}/m_{1}\mathbb{Z}\times\mathbb{Z}/m_{2}\mathbb{Z}\right)$ is a subgroup of $G$. When $\exp(G)=2$, the claim is clearly true. Suppose the claim is true for all $\exp(G)<n_{2}$. First consider a zero-sum sequence of length $2n_{1}+2n_{2}-d$. Note that $2n_{1}+2n_{2}-d\geq 4d\geq 4d-3$, so by Theorem 1.3 we can remove subsequences of length $d$ with sum $0\pmod{d}$ until there are exactly $3d$ remaining. Then by Lemma 2.4, we can break off another $d$ elements to obtain $2m_{1}+2m_{2}-3$ blocks of size $d$, with sums $dx_{1},\dots,dx_{2m_{1}+2m_{2}-3}$, for some $x_{i}$. By the induction hypothesis, since

[TABLE]

some $m_{2}$ of the $x_{i}$ must sum to [math] in $(\mathbb{Z}/m_{1}\mathbb{Z}\times\mathbb{Z}/m_{2}\mathbb{Z})$. Combining the corresponding blocks gives a subsequence of length $n_{2}$ whose sum is zero in $\left(\mathbb{Z}/n_{1}\mathbb{Z}\times\mathbb{Z}/n_{2}\mathbb{Z}\right)$. Now note that since $\ell$ is the least integer such that $\ell\mathchar 13613\relax n_{1}$, we have $\ell-1\mid n_{1}$. Since $n_{1}\mid n_{2}$, we also have $\ell-1\mid n_{2}$. Letting $d=\ell-1$ finishes the proof. ∎

Now we will show that if $|J|$ were any smaller, there couldn’t be a zero-sum subsequence of length $n_{2}$.

Proposition 3.2**.**

Suppose $4\leq\ell\mathchar 13613\relax n_{2}$. There exists a zero-sum sequence in $\left(\mathbb{Z}/n_{1}\mathbb{Z}\times\mathbb{Z}/n_{2}\mathbb{Z}\right)$ of length $2n_{2}-\ell$ which contains no zero-sum subsequences of length $n_{2}$.

Proof.

Let $g:=\gcd(\ell,n_{2})$. Consider a sequence of the form

[TABLE]

It is easy to verify that this does not contain a zero-sum subsequence of length $n_{2}$. We claim there exists $(r,s)\in\left(\mathbb{Z}/n_{1}\mathbb{Z}\times\mathbb{Z}/n_{2}\mathbb{Z}\right)$ such that adding $(r,s)$ to each term will result in a zero-sum sequence. Note again that adding $(r,s)$ to each element won’t change the fact that there is no zero-sum subsequence of length $n_{2}$. Since $\ell\mathchar 13613\relax n_{2}$, $g\leq\ell/2$. Therefore $a\leq n_{2}-\ell/2\leq n_{2}-2$, and $g\geq 1$, so $b\leq n_{2}-1$. To find $(r,s)$ we need only to satisfy the divisibility relations

[TABLE]

By the definition of $g$, we can find solutions $(r,s)$ to make the sequence zero-sum. ∎

Proposition 3.1 and 3.2 together imply Theorem 1.7. For the proof of Theorem 1.8, we begin with the following corollary.

Corollary 3.3**.**

Let $\ell$ be the smallest integer such that $\ell\geq 4$ and $\ell\mathchar 13613\relax n_{2}$. Let $t\geq 1$ and $n_{1}\mid n_{2}$. If $J$ is a zero-sum sequence in $\left(\mathbb{Z}/n_{1}\mathbb{Z}\times\mathbb{Z}/n_{2}\mathbb{Z}\right)$ of length at least $2n_{1}+(t+1)n_{2}-\ell+1$, then $(n_{2}t\mid J)>0$.

Proof.

We proceed by induction on $t$. Note the base case $t=1$ is given by Proposition 3.1. Now suppose the statement is true for positive integers less than $t>1$. Then $J$ contains a zero-sum sequence of length $(t-1)n_{2}$. Remove this sequence from $J$. Then $J$ has $2n_{1}+2n_{2}-\ell+1$ elements remaining, which sum to zero since $J$ was zero-sum. This reduces to the base case, so $J$ contains a zero-sum subsequence of length $n_{2}$. Combining this with the $(t-1)n_{2}$ length sequence gives a zero-sum subsequence of length $n_{2}t$. ∎

Corollary 3.3 proves Theorem 1.8.

4. Proofs of Theorem 1.10 and Corollary 1.11

Proposition 4.1**.**

Let $t\geq 1$ and $n\geq 2$. There exists a sequence in $\left(\mathbb{Z}/n\mathbb{Z}\right)^{2}$ of length $(t+2)n-3$ which contains no zero-sum subsequence of length $nt$.

Proof.

Consider the following sequence:

[TABLE]

We clearly cannot make a sequence of $tn$ $(0,0)$’s. It suffices to verify that there does not exist a zero-sum subsequence of any length among the nonzero elements. Otherwise, we could just add enough $(0,0)$’s to get a zero-sum subsequence of length $tn$. Suppose we use $i$ $(1,0)$’s, and $j$ $(0,1)$’s, where $0\leq i,j\leq n-1$. In order for the subsequence to be zero-sum, necessarily we would need

[TABLE]

Since $0\leq i,j\leq n-1$, the only solution is $i=j=0$. Hence there is no zero-sum subsequence. ∎

This gives the lower bound in both Theorem 1.10 and Corollary 1.11.

To prove Theorem 1.10, we will need the following preliminary lemma.

Lemma 4.2** ([8], Corollary 2.3).**

Let $p$ be a prime, and let $J$ be a sequence of elements in $\left(\mathbb{Z}/p\mathbb{Z}\right)^{2}$. If $|J|=3p-2$ or $|J|=3p-1$, then

[TABLE]

Proposition 4.3**.**

If $J$ is a sequence in $\left(\mathbb{Z}/p\mathbb{Z}\right)^{2}$ of length $4p-2$, then $(2p\mid J)>0$.

Proof.

Note that $4p-2>4p-3$. By Theorem 1.9, $J$ contains a zero-sum subsequence of length $p$. Removing the sequence from $J$, we are left with $3p-2$ elements. By Lemma 4.2 we have

[TABLE]

If $(2p\mid J)>0$, we’re done and have found our zero-sum subsequence of length $2p$. Otherwise, $(2p\mid J)=0$ which implies

[TABLE]

Therefore, $(p\mid J)>0$, so there is another zero-sum subsequence of length $p$. Combining this with the first one gives a zero-sum subsequence of length $2p$. ∎

Corollary 4.4**.**

Let $t\geq 2$. If $J$ is a sequence in $\left(\mathbb{Z}/p\mathbb{Z}\right)^{2}$ of length $(t+2)p-2$, then $(tp\mid J)>0$.

Proof.

We proceed by induction on $t$. The case $t=2$ follows from Proposition 4.3. Suppose the statement is true for positive integers less than $t>2$. Since $t\geq 2$, we have

[TABLE]

By Theorem 1.9 $J$ has a zero-sum subsequence of length $p$. Now remove the sequence so that $J$ has $((t-1)+2)p-2$ elements remaining. By the induction hypothesis, $J$ has a zero-sum subsequence of length $(t-1)p$. Combining this with the zero-sum subsequence of length $p$ yields a zero-sum subsequence of length $tp$. ∎

Proposition 4.1 and Corollary 4.4 imply Theorem 1.10.

Now we prove a version of Proposition 4.3 for non-prime $n$.

Proposition 4.5**.**

Write $n=pm$. If $J$ is a sequence in $\left(\mathbb{Z}/n\mathbb{Z}\right)^{2}$ of length $4n-2+(m-1)$, then $(2n\mid J)>0$.

Proof.

Note that $4n-2+(m-1)>4m-3$, so we can find some $m$ elements whose sum is $0\pmod{m}$. Denote their sum by $mx_{1}$ and remove the elements from $J$. We can continue doing this until there are exactly $3m-3$ elements remaining. This gives us $4p-2$ blocks of size $m$ whose sums are $mx_{1},\dots,mx_{4p-2}$ for some $x_{i}$’s. By Proposition 4.3, there is some $2p$ of the $x_{i}$’s summing to $0\pmod{p}$. Combining the blocks gives us $2n$ elements whose sum is $0\pmod{n}$. ∎

Corollary 4.6**.**

Write $n=pm$ and let $t\geq 2$. If $J$ is a sequence in $\left(\mathbb{Z}/n\mathbb{Z}\right)^{2}$ of length $(t+2)n-2+(m-1)$, then $(tn\mid J)>0$.

Proof.

We induct on $t$. The $t=2$ case is Proposition 4.5. Suppose the statement is true for positive integers less than $t>2$. Since $t\geq 2$, we have

[TABLE]

By Theorem 1.9, $J$ has a zero-sum subsequence of length $n$. Removing it leaves us with $((t-1)+2)n-2+(m-1)$ elements. By the induction hypothesis, we can remove a zero-sum subsequence of length $(t-1)n$. Combining these elements with the zero-sum subsequence of length $n$ yields a zero-sum subsequence of length $tn$, as desired. ∎

Corollary 4.6 and Proposition 4.1 imply Corollary 1.11.

5. Bounds for Modified EGZ Constants in $(\mathbb{Z}/p\mathbb{Z})^{d}$

Proposition 5.1**.**

https://www.overleaf.com/project/5cf1d1637138370431c623d4 Let $p>3$ be prime, and let $J$ be a sequence of elements in $\left(\mathbb{Z}/p\mathbb{Z}\right)^{3}$. Then if $|J|=4p-4$, then

[TABLE]

To prove the proposition, we use the following classical theorem.

Theorem 5.2** (Chevalley-Warning).**

Let $n,d_{1},\dots,d_{r}$ be positive integers such that $d_{1}+\dots+d_{r}<n$. For each $1\leq i\leq r$ let $P_{i}(t_{1},\dots,t_{n})\in\mathbb{F}_{q}[t_{1},\dots,t_{n}]$ be a polynomial of degree $d_{i}$ with zero constant term. Then there exists $0\neq x=(x_{1},\dots,x_{n})\in\mathbb{F}_{q}^{n}$ such that $P_{i}(x)=0$ for all $1\leq i\leq r$. Furthermore, let

[TABLE]

Then $Z\equiv 0\mod p$.

Now we will prove Proposition 5.1.

Proof.

Let $J=\{(a_{n},b_{n},c_{n}):1\leq n\leq 4p-4$}. Consider the following polynomials over $\mathbb{F}_{p}[t_{1},\dots,t_{4p-3}]$:

[TABLE]

Since $4p-3>4p-4$, by Theorem 5.2, there exists $0\neq x=(x_{1},\dots,x_{4p-3})$ such that $P_{1}(x)=\dots=P_{4}(x)=0.$ We partition the solutions according to $(x_{1},\dots,x_{4p-4},0)$ and $(x_{1},\dots,x_{4p-4},\text{ nonzero})$.

First we consider solutions of the form $(x_{1},\dots,x_{4p-4},0)$. Let $I=\{1\leq i\leq 4p-4\ :\ x_{i}\neq 0\}$. Note that $x^{p-1}=1$ if $x$ is nonzero and $x^{p-1}=0$ if $x=0$. Then since $P_{1}(x)=\dots P_{4}(x)=0$, we have

[TABLE]

Therefore, $|I|\equiv 0\mod p$ and, since $0<|I|\leq 4p-4$, we have $|I|=p,2p$, or $3p$. Note that this set of solutions contains the zero solution, so the total number of solutions where $x_{4p-3}=0$ is

[TABLE]

Now we consider the set of solutions of the form $(x_{1},\dots,x_{4p-4},\text{ nonzero})$. In this case, define $I$ the same way and since $P_{1}(x)=\dots=P_{4}(x)=0$, we have

[TABLE]

Therefore, $|I|\equiv-1\mod p$, and since $0<|I|\leq 4p-4$, we have $|I|=p-1,2p-1$, or $3p-1$. Thus the number of solutions is

[TABLE]

Reducing modulo $p$ and combining these with the other set of solutions yields the result. ∎

This proof leads us to the following corollary.

Corollary 5.3**.**

If $|J|=4p-3,4p-2$, or $4p-1$ then

[TABLE]

Corollary 5.4**.**

Suppose $J$ is a zero-sum sequence in $(\mathbb{Z}/p\mathbb{Z})^{3}$ and $|J|=4p$. Then $(p\mid J)>0$ or $(2p\mid J)>0$.

Proof.

Let $x\in J$ be arbitrary. Suppose towards a contradiction that $(p\mid J)=0$ and $(2p\mid J)=0$. Then we must also have $(p\mid J-\{x\})=0$ and $(2p\mid J-\{x\})=0$. Since $|J-\{x\}|=4p-1$, by Corollary 5.3, $(3p\mid J-\{x\})\equiv-1\mod p$. So $(3p\mid J-\{x\})>0$. Since $J$ is zero sum, note that if there was a zero-sum subsequence of length $p$, its complement sequence of length $3p$ must also be zero-sum. In other words,

[TABLE]

contradicting $(p\mid J)=0$. ∎

Note that the preceding few results are amenable to the exact same methods for higher dimensions. In general, for $(\mathbb{Z}/p\mathbb{Z})^{d}$, one would construct $(d+1)$ polynomials using the Chevalley-Warning method. This would yield the following:

If $|J|=(d+1)(p-1)$, then

[TABLE]

Furthermore, if $|J|=(d+1)p-m$ for some $1\leq m\leq d$, then

[TABLE]

Lastly, this would imply that if $J$ is zero sum and $|J|=(d+1)p$, then at least one of $(p\mid J),\dots,\left((d-1)p\mid J\right)$ is greater than zero. This leads us to the following corollary.

Corollary 5.5**.**

Let $p$ be prime, $G=\left(\mathbb{Z}/p\mathbb{Z}\right)^{d}$, and $\mathcal{L}=\{p,2p,\dots,(d-1)p\}$. Then

[TABLE]

6. Open Problems

In 1973, Harborth [6] considered the problem of computing $s_{n}\left((\mathbb{Z}/n\mathbb{Z})^{d}\right)$ for higher dimensions. In particular, he proved the following bounds.

Theorem 6.1** (Harborth, [6]).**

We have

[TABLE]

In general the lower bound is not tight, but Harborth showed we have equality for $n=2^{k}$.

In 2019, this was improved by Naslund resulting in the following bounds.

Theorem 6.2** (Naslund, [9]).**

[TABLE]

where $J(p)$ is a constant satisfying $0.8414<J(p)<0.91837$.

In 2019, Berger and Wang made the following conjecture.

Conjecture 6.3** (Conjecture 4.2, [3]).**

If $n=2^{k}$ and $d\geq 1$, we have

[TABLE]

where $\ell$ is the smallest integer such that $\ell\geq 2^{d}$ and $\ell\mathchar 13613\relax n$.

We make the following conjecture.

Conjecture 6.4**.**

Let $n,t,d\geq 1$ be positive integers. We have

[TABLE]

where $\ell$ is the smallest integer such that $\ell\geq 2^{d}$ and $\ell\mathchar 13613\relax n$.

We also have not determined the EGZ constant $s_{nt}\left((\mathbb{Z}/n\mathbb{Z})^{2}\right)$ for non-prime $n$.

Problem 6.5**.**

Compute $s_{nt}\left((\mathbb{Z}/n\mathbb{Z})^{2}\right)$ for non-prime $n$ and $t\geq 2$.

Acknowledgements

This research was conducted at the University of Minnesota Duluth REU and was supported by NSF / DMS grant 1659047 and NSA grant H98230-18-1-0010. I would like to thank Joe Gallian for running the program and Aaron Berger for helpful conversations and encouragement.