# On linear combinations of units with bounded coefficients and double-base digit expansions

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## Abstract

Let \(\mathfrak o \) be the maximal order of a number field. Belcher showed in the 1970s that every algebraic integer in \(\mathfrak o \) is the sum of pairwise distinct units, if the unit equation \(u+v=2\) has a non-trivial solution \(u,v\in \mathfrak o ^*\). We generalize this result and give applications to signed double-base digit expansions.

### Keywords

Unit sum number Additive unit structure Digit expansions### Mathematics Subject Classification (2010)

11R16 11R11 11A63 11R67## 1 Introduction

In the 1960s Jacobson [7] asked, whether the number fields \(\mathbb Q (\sqrt{2})\) and \(\mathbb Q (\sqrt{5})\) are the only quadratic number fields such that each algebraic integer is the sum of distinct units. Śliwa [11] solved this problem for quadratic number fields and showed that even no pure cubic number field has this property. These results were extended to cubic and quartic fields by Belcher [2, 3]. In particular, Belcher solved the case of imaginary cubic number fields completely by applying the following criterion, which now bears his name, cf. [3].

**Belcher’s criterion**

The problem of characterizing all number fields in which every algebraic integer is a sum of distinct units is still unsolved. Let us note that this problem is contained in Narkiewicz’ list of open problems in his famous book [9, see page 539, Problem 18].

Recently the interest in the representation of algebraic integers as sums of units arose due to the contribution of Jarden and Narkiewicz [8]. They showed that in a given number field there does not exist an integer \(k\), such that every algebraic integer can be written as the sum of at most \(k\) (not necessarily distinct) units. For an overview on this topic we recommend the survey paper due to Barroero, Frei, and Tichy [1]. Recently Thuswaldner and Ziegler [13] considered the following related problem. Let an order \(\mathfrak o \) of a number field and a positive integer \(k\) be given. Does each element \(\alpha \in \mathfrak o \) admit a representation as a linear combination \(\alpha = c_1 \varepsilon _1 + \cdots + c_\ell \varepsilon _\ell \) of units \(\varepsilon _1,\ldots ,\varepsilon _\ell \in \mathfrak o ^*\) with coefficients \(c_i\in \{1,\ldots , k\}\) ? This problem was attacked by using dynamical methods from the theory of digit expansions. In the present paper we address this problem again. In particular, we wish to generalize Belcher’s criterion in a way to make it applicable to this problem.

In order to get the most general form, we refine the definition of the unit sum height given in [13].

**Definition 1.1**

*unit sum height of*\(\alpha \). In addition we define \(\omega _{R,\varGamma }(0):= 0\) and \(\omega _{R,\varGamma }(\alpha ):= \infty \) if \(\alpha \) admits no representation as a finite linear-combination of elements contained in \(\varGamma \cap R\). Moreover, we define

Let us note that for a number field \(F\) with the group of units \(\varGamma \) of an order \(\mathfrak o \) of \(F\) we have \(\omega _\varGamma (\mathfrak o ) = \omega (\mathfrak o )\), where \(\omega (\mathfrak o )\) is the unit sum height defined in [13].

With those notations our main result is the following.

**Theorem 1.2**

The following section, Sect. 2, is devoted to the proof of Theorem 1.2. In the third section we apply our main theorem, Theorem 1.2, to some special orders of Shanks’ simplest cubic fields. A special case of that theorem yields applications to double-base expansions. There we choose \(F=\mathbb Q \), \(R=\mathbb Z \) and \(\varGamma =\langle -1,p,q\rangle \), where \(p\) and \(q\) are coprime integers. We discuss that in Sect. 4.

## 2 Proof of Theorem 1.2

We start this section by giving a short plan of the proof.

*Plan of Proof*

Let \(\alpha \in R\) be arbitrary. Our goal is to find a representation of \(\alpha \) of the form (1.1) in which the coefficients \(a_1,\ldots , a_\ell \) are all bounded by \(n-1\). We first show that \(\alpha \) can be represented as a linear combination of the form (1.1) with \(\nu _1,\ldots ,\nu _\ell \) chosen in a particular way. The idea of the proof is rather simple and is based on induction over the total weight of this representation (this is the sum of all of its coefficients, see Definition 2.2). Start with a representation of \(\alpha \) as above and choose a coefficient which is greater than or equal to \(n\) (if such a coefficient does not exist, we are finished). Now apply (1.2). This leads to a new representation of \(\alpha \) of the form (1.1) whose total weight does not increase (and actually remains the same after excluding some trivial cases). This process is now repeated until we either have a representation in which all coefficients are bounded by \(n-1\), or the support of the representation contains big gaps. In the first case we are finished. In the second case we can split the representation into two parts which are separated by a large gap. The total weight of each part is less than the total weight of the original representation of \(\alpha \). We thus use the induction hypothesis on both of them, so we get a new representation of each part with coefficients bounded by \(n-1\). Now, since the gap between the supports of these two parts is large, they do not overlap after we apply (1.2) to them in the appropriate way and we can put them together to find a representation as desired also in this case.

\(\square \)

*R*has at least one representation of the form (1.1). The coefficients of that representation are integers, but not necessarily smaller than \(n\).

**A**

*Proof of A*

Note that \(\zeta ^k \eta _\ell \varvec{\varepsilon }^\mathbf{x}\in \varGamma \cap R\). Furthermore, we can choose the coefficients \(a_{k,\ell ,\mathbf{x}}\) to be non-negative, since, by assumption, we have \(-1\in \varGamma \), which allows us to choose the “signs” in our representation. \(\square \)

From now on we suppose that \(\zeta \), \(\eta _1,\ldots ,\eta _L\), and \(\varvec{\varepsilon }\) are fixed and given as in **A**. We use the following convention on representations.

**Convention 2.1**

An important quantity is the weight of a representation. It is defined as follows.

**Definition 2.2**

**A**, i.e.,

*total weight of*\(\alpha \) and write \(w_\alpha \) for it.

As mentioned in the plan of the proof of Theorem 1.2, we apply Eq. (1.2) to an existing representation to get another one. In the following paragraph, we define that replacement step, which will then always be denoted by \(*\).

The following statements **B** and **C** deal with two special cases.

**B**

If \(\alpha \in R\) with \(w_\alpha < I\), then Theorem 1.2 holds.

We use that statement as the basis of our induction on the total weight \(w\).

*Proof of B*

Since \(I \le n\) we have \(w_\alpha < n\). So the sum of all (non-negative) coefficients is smaller than \(n\). Therefore all coefficients themselves are in \([\![ {0},{n-1}]\!]\), which proves the theorem in that special case.\(\square \)

**C**

If \(I<n\), then Theorem 1.2 holds.

*Proof of C*

Assume that there is a coefficient \(a_{k,\ell ,\mathbf{x}} \ge n\) in the representation of \(\alpha \). We apply \(*\) to obtain a new representation. But because \(I<n\), the new one has smaller total weight, which is a contradiction to the fact that \(w\) was chosen minimal. \(\square \)

Because of **B** and **C** we suppose from now that \(w \ge I\) and \(I = n\). As indicated above, we prove Theorem 1.2 by induction on the total weight \(w\) of \(\alpha \). More precisely we want to prove the following claim by induction.

**Claim 2.3**

In order to prove Theorem 1.2 it is sufficient to prove Claim 2.3. As already mentioned, we use induction on the total weight \(w\) of \(\alpha \). Note that the induction basis has been shown above in **B**.

Let us start by looking what happens if one applies \(*\).

**D**

Repeatedly applying \(*\) yields pairwise “essentially different” representations of \(\alpha \).

*Proof of D*

**A**.

Now we look what happens after sufficiently many applications of \(*\).

**E**

- 1.Each coefficient satisfies \(b_{k,\ell ,\mathbf{x}} \in [\![ {0},{n-1}]\!]\) andholds for all \(m \in [\![ {1},{M}]\!]\).$$\begin{aligned} \overline{B}_m - \underline{B}_m \le w + 2(w-1)f(w-1) \end{aligned}$$
- 2.There exists an index \(m\) such thatholds.$$\begin{aligned} \overline{B}_m - \underline{B}_m > w + 2(w-1)f(w-1) \end{aligned}$$

*Proof of E*

**D**, and there are at most \(T(w)\) possibilities to distribute our new coefficients in an interval \([\![ {0},{K-1}]\!]\times [\![ {1},{L}]\!]\times B\) with

**E**. \(\square \)

**F**

With the setup and notations of **E**, a possible “translation of the indices” stays small.

*Proof of F*

The quantity \(r\) is the maximum of all exponents in the representation of the \(u_i\) as powers of the \({\varepsilon }_1,\ldots ,{\varepsilon }_M\). Thus, an application of \(*\) can change the exponents, and therefore the upper and lower bounds, respectively, by at most \(r\). We have at most \(T(w)\) applications of \(*\), so the statement follows. \(\square \)

Now we look at the two different cases of **E**. The first one leads to a result directly, whereas in the second one we have to use the induction hypothesis to get a representation as desired.

**G**

If we are in case (1) of **E**, then we are “finished”.

*Proof of G*

**H**

If we are in case (2) of **E**, then we can split the representation into two parts and between them there is a “large gap”.

*Proof of H*

**E**we have an index \(m \in [\![ {1},{M}]\!]\) with

**I**

If we have the splitting described in **H**, then Claim 2.3 follows for weight \(w\).

*Proof of I*

## 3 The case of simplest cubic fields

Using our main theorem we are able to prove the following result.

**Theorem 3.1**

We have \(\omega (\mathbb Z [\alpha ])\le 2\) for all \(a\in \mathbb Z \).

*Proof*

First let us note some important facts on \(\mathbb Q (\alpha )\) and \(\mathbb Z [\alpha ]\), see for example Shanks’ original paper [10]. We know that \(\mathbb Q (\alpha )\) is Galois over \(\mathbb Q \) with Galois group \(G=\{\text{ i}d, \sigma ,\sigma ^2\}\) and with \(\alpha _2:=\sigma (\alpha )=-1-\frac{1}{\alpha }\). If we set \(\alpha _1:= \alpha \), then \(\alpha _1\) and \(\alpha _2\) are a fundamental system of units. Now we know enough about the structure of \(\mathbb Z [\alpha ]\) to apply Theorem 1.2.

## 4 Application to signed double-base expansions

We start with the definition of a signed double-base expansion of an integer.

**Definition 4.1**

*Signed double-base expansion*) Let \(p\) and \(q\) be different integers. Let \(n\) be an integer with

*signed*\(p\)-\(q\)-

*double*-

*base expansion of*\(n\). The pair \((p,q)\) is called

*base pair*.

A natural first question is, whether each integer has a signed double-base expansion for a fixed base pair.

If one of the bases \(p\) and \(q\) is either \(2\) or \(3\), then existence follows since every integer has a *binary representation* (base \(2\) with digit set \({\left\{ {0,1}\right\} }\)) and a *balanced ternary representation* (base \(3\) with digit set \({\left\{ {-1,0,1}\right\} }\)), respectively. To get the existence results for general base pairs, we use the following theorem, cf. [5]

**Theorem 4.2**

(Birch) Let \(p\) and \(q\) be coprime integers. Then there is a positive integer \(N(p,q)\) such that every integer larger than \(N(p,q)\) may be expressed as a sum of distinct numbers of the form \(p^iq^j\) all with non-negative integers \(i\) and \(j\).

**Corollary 4.3**

Let \(p\) and \(q\) be coprime integers. Then each integer has a signed \(p\)-\(q\)-double-base expansion.

Next we want to give an efficient algorithm that allows to calculate a signed double base expansion of a given integer. Birch’s theorem, or more precisely the proof in [5], does not provide an efficient way to do that. However, using our main result, there is a way to compute such expansions efficiently at least for certain base pairs.

**Corollary 4.4**

*Proof*

We start to prove the first part of the corollary and therefore apply Theorem 1.2 with \(\mathbb F =\mathbb Q \), \(R=\mathbb Z \) and \(\varGamma \) is the multiplicative group generated by \(-1,p\) and \(q\). Since by assumption \(2=\pm (p^x-q^y)\) we have a solution to (1.2) and Theorem 1.2 yields that \(p\)-\(q\)-double-base expansions exist.

**Claim 4.5**

*Proof*

We prove the claim by induction on \(w\). If \(w\le 1\) the statement of the claim is obvious. Further, if all the \(a_i\) are in \({\left\{ {-1,0,1}\right\} }\) we are done. Therefore we assume that there is at least one index \(i\) with \( \left|{a_i}\right|>1\).

Now we want to give some examples for base pairs, where the corollary can be used.

*Example 4.6*

*twin prime pair*, i.e., we have \(q=p+2\) and both \(p\) and \(q\) are primes. Then clearly

*Example 4.7*

^{1}was done in Sage [12].

One can find pairs \((p,q)\) where Corollary 4.4 does not work. The following remark discusses some of those pairs.

*Remark 4.8*

So in the cases given in the remark above, as well as in a lot of other cases, we cannot use the corollary to compute a signed double-base expansion efficiently. This leads to the following question.

*Question 4.9*

Is there an efficient (polynomial time) algorithm for each base pair \((p,q)\) to compute a signed \(p\)-\(q\)-double-base expansion for all integers?

There is also another way to use Theorem 1.2. For some combinations of \(p\) and \(q\) we can get a weaker result. First, we define an extension of the signed double-base expansion: we allow negative exponents in the \(p^i q^j\), too.

**Definition 4.10**

*Extended signed double-base expansion*) Let \(p\) and \(q\) be different integers (usually coprime). Let \(z\in \mathbb Q \). If we have

*extended signed*\(p\)-\(q\)

*-double-base expansion of*\(z\).

With that definition, we can prove the following corollary to Theorem 1.2.

**Corollary 4.11**

*Remark 4.12*

If we have a solution to the equation in Corollary 4.4, then Corollary 4.11 works, too. But more can be said about the existence and efficient computability of extended double-base expansions for the elements of \(\mathbb Z [1/p,1/q]\). If each integer has an efficient computable signed \(p\)-\(q\)-double-base expansion, then each element of \(\mathbb Z [1/p,1/q]\) has an extended signed \(p\)-\(q\)-double-base expansion which can be computed efficiently. This result is not difficult to prove.

Now we prove the corollary.

*Proof of Corollary 4.11*

The proof of this corollary runs along the same lines as the proof of Corollary 4.4.

We apply Theorem 1.2 with \(\mathbb F =\mathbb Q \), \(R=\mathbb Z [1/p,1/q]\) and \(\varGamma \) is the multiplicative group generated by \(-1\), \(p\) and \(q\). Since, by assumption, \(2=\pm (p^aq^b-p^cq^d)\) we have a solution to (1.2), Theorem 1.2 yields that \(p\)-\(q\)-double-base expansions exist.

We can use the corollary proved above to get the following examples.

*Example 4.13*

*Sophie Germain prime*and \(q=2p+1\). We obtain

The case when \(p\) is a prime and \(q=2p-1\) is a prime works analogously.

The end of this section is dedicated to a short discussion. All the results on efficient computability in this section needed a special representation of 2. We have given some pairs \((p,q)\) where the methods given here do not work.

Further, one could ask, whether the representations we get have a special structure. Of particular interest would be an algorithm to get expansions with a small number of summands (small number of non-zero digits). For a given base pair \((p,q)\) this leads to the following question

*Question 4.14*

How to compute a signed \(p\)-\(q\)-double-base expansion with minimal weight for a given integer?

A greedy approach for solving this question can be found in Berthé and Imbert [4], some further results can be found in Dimitrov and Howe [6].

## Footnotes

- 1.
The source code can be found on http://www.danielkrenn.at/belcher/. Further a full list of expansions of the natural numbers up to \(10,000\) can be found there.

## Notes

### Open Access

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