Heights and transcendence of p-adic continued fractions

Abstract

Special kinds of continued fractions have been proved to converge to transcendental real numbers by means of the celebrated Subspace Theorem. In this paper we study the analogous p–adic problem. More specifically, we deal with Browkin p–adic continued fractions. First we give some new remarks about the Browkin algorithm in terms of a p–adic Euclidean algorithm. Then, we focus on the heights of some p–adic numbers having a periodic p–adic continued fraction expansion and we obtain some upper bounds. Finally, we exploit these results, together with p–adic Roth-like results, in order to prove the transcendence of three families of p–adic continued fractions.

1 Introduction

Continued fractions are one of the most successful methods to construct transcendental numbers. The first studies in this direction are due to Liouville [23] who dealt with unbounded partial quotients. Later on, Maillet [25] and Baker [5] exhibited continued fractions with bounded partial quotients converging to transcendental numbers. Furthermore, Baker’s results have been recently improved by Adamczewski and Bugeaud [1]. The continued fractions studied in these works are quasi-periodic, in the sense that they have the form

$$\begin{aligned} a_0 + \frac{1}{a_1+ \frac{1}{a_2+ \frac{1}{\ddots }}} = [a_0, a_1, a_2, \ldots ] \end{aligned}$$

where $a_i = a_{i+1}$ holds for infinitely many i. In [3], the authors proved that the real number $[0, a_1, a_2, \ldots ]$ with $a_i = 1 + (\lfloor i \theta \rfloor \bmod k)$, where $0< \theta < 1$ is irrational and $k \ge 2$ an integer, is transcendental for any $i\ge 1$. This improves a result of Davison [15]. Adamczewski and Bugeaud [2] showed that if a continued fraction begins with an arbitrarily long palindromic string, then it converges to a transcendental number. Further results on this topic can be found in [4, 11, 16, 18, 19]. The main tool used in the above works is the celebrated Roth’s theorem and subsequent developments due to Schmidt and later Evertse. This strategy goes as follows: describe a real number $\alpha$ by its continued fraction expansion and use the latter to obtain infinitely many sufficiently good rational approximations so that, by Roth’s theorem, $\alpha$ cannot be algebraic.

Over the years, many people have worked to translate these studies into the field of p–adic numbers. The first idea for a continued fraction algorithm over $\mathbb {Q}_p$ is essentially due to Mahler [24], whose definition was later improved in particular by Ruban [27] and Browkin [8, 9], leading to different algorithms. A significant difference between them is that Ruban’s approach leads to a periodic expansion for some rational numbers, while Browkin’s continued fractions are always finite for $\alpha \in \mathbb {Q}$, precisely like in the classical archimedean version. We also remark that, despite the existence of several definitions of continued fractions in $\mathbb {Q}_p$, a completely satisfactory p–adic counterpart is still missing. The main problem is that it is not known if there exists a p–adic continued fraction algorithm which eventually becomes periodic if the input is a quadratic irrational. In other words, there is no analogue of Lagrange’s theorem. Actually, Lagrange’s theorem fails with respect to Ruban’s algorithm [26] (see also [14] for an effective criterion of periodicity), while for Browkin’s algorithm it remains an open question and only some partial results have been found: see for instance [6] and [13]. In Sect. 2.1 we will discuss some more reasons to think of Browkin’s approach as the “right” one, in the spirit of Ostrowski’s theorem.

The main goal of this paper is the construction of transcendental p–adic continued fractions by means of Browkin’s algorithm. In recent works there has been an increasing interest towards many families of Ruban’s p–adic continued fractions converging to transcendental numbers. Ooto [26] obtained results inspired by Baker’s, and the authors of [7] proved that some quasi–periodic and palindromic p–adic continued fractions converge to transcendental numbers.

Our main results are summarized in the following theorem.

Theorem A

Let

$$\begin{aligned}\alpha = [0, b_1, b_2, b_3, \ldots ]\end{aligned}$$

be a non–periodic Browkin p–adic continued fraction. Assume that one of the following holds:

• (a) the sequence $(|b_i |_p)_{i \ge 1}$ is bounded and there are infinitely many subsequences

  $$\begin{aligned}b_{n_i} = \ldots = b_{n_i + \lambda _i k_i -1} = p^{-1}\end{aligned}$$

  (with further technical conditions on $n_i$’s, $\lambda _i$’s and $k_i$’s, to be specified in Theorem 3 and Hypothesis 1);

• (b) the sequence $(b_i)_{i\ge 1}$ begins with arbitrarily long palindromes, $|b_n|_p \ge p^{4}$ for all $n \gg 0$ and $\alpha$ is not a quadratic irrational.

Then $\alpha$ is transcendental. Moreover, if the hypothesis of (a) is weakened to $b_{h+k_i} = b_{h}$, for $n_i \le h \le n_i + (\lambda _i-1)k_i -1$, for every i (with small changes in the technical conditions, see Theorem 4), then $\alpha$ is either transcendental or quadratic irrational.

Part (a) corresponds to Baker [5] and Ooto [26], part (b) to Adamczewski-Bugeaud [2]. The proofs are given in Sect. 4 (Theorems 3, 4 and 5). We also note that the above constructions yield uncountable many transcendental numbers.

In Sect. 3 we study the height of $periodic$ continued fractions and provide some upper bounds for it. These results, stronger than their analogues in [26] and of general interest in the context of bounding the height of algebraic numbers, play a crucial role in the proofs of our main theorems. The other tools we are going to use are the p–adic versions of Roth’s theorem and the Subspace Theorem.

2 Preliminaries

Let p be an odd prime. We denote by $v_p(\cdot )$, $|\cdot |_p$ and $|\cdot |_\infty$ the p–adic valuation, the p–adic absolute value and the Euclidean absolute value, respectively.

Given $\alpha _0 \in \mathbb {Q}_p$, a p–adic continued fraction expansion $[b_0, b_1, \ldots ]$ is obtained choosing a function $s: \mathbb {Q}_p \rightarrow \mathbb {Q}$ and then iterating the following steps

$$\begin{aligned} {\left\{ \begin{array}{ll} b_i = s(\alpha _i) \\ \alpha _{i+1} = \frac{1}{\alpha _i - b_i} \end{array}\right. } \end{aligned}$$

(1)

for all $i \ge 0$ and $\alpha _i \not = b_i$; if $\alpha _i=b_i$, the process stops and the continued fraction expansion is finite. The function s is defined so to have $|x-s(x)|_p<1$ for every $x\in \mathbb {Q}_p$ and plays the same role of the floor function in the classical case. In Browkin’s algorithm the idea is to take the integers between $-(p-1)/2$ and $(p-1)/2$ as a set of representatives for $\mathbb {Z}$ modulo p. Then every $\alpha \in \mathbb {Q}_p$ is represented uniquely as a power series with coefficients in this set and s is defined as

$$\begin{aligned} s:&\mathbb {Q}_p \longrightarrow \mathbb {Q} \nonumber \\&\alpha = \sum _{i=r}^\infty a_ip^i \mapsto \sum _{i=r}^0 a_ip^i \;\;\text { with }r \in \mathbb {Z}\text { and }a_i\in \left\{ -\frac{p-1}{2}, \ldots , \frac{p-1}{2}\right\} . \end{aligned}$$

(2)

In particular, we have $s(\alpha )=0$ if $\alpha$ is in the maximal ideal of the p-adic integers and $|s(\alpha )|_p=p^k$ for some $k\in \mathbb {N}$ otherwise.

The $b_i$’s are called partial quotients and the $\alpha _i$’s complete quotients. We define the sequences $(A_i)_{i \ge -2}$ and $(B_i)_{i \ge -2}$ as follows:

$$\begin{aligned} {\left\{ \begin{array}{ll} A_{-2} = 0, A_{-1} = 1 \\ A_i = b_iA_{i-1} + A_{i-2}, \forall i \ge 0 \end{array}\right. }, \quad {\left\{ \begin{array}{ll} B_{-2} = 1, B_{-1} = 0 \\ B_i = b_iB_{i-1} + B_{i-2}, \forall i \ge 0 \end{array}\right. } \end{aligned}$$

(3)

so that their ratios give the convergents of the p–adic continued fraction, i.e., $\frac{A_i}{B_i} = [b_0, \ldots , b_i]$, for all $i \ge 0$. The classical identities

$$\begin{aligned} \alpha _0 = [b_0, \ldots , b_i, \alpha _{i+1}] = \frac{\alpha _{i+1}A_i+A_{i-1}}{\alpha _{i+1}B_i+B_{i-1}}, \quad A_{i-1}B_i - B_iA_{i-1} = (-1)^i \end{aligned}$$

(4)

still hold for p–adic continued fractions, since they are obtained in a formal way.

In the following proposition, we summarize some well-known facts about Browkin p–adic continued fractions.

Proposition 1

Given $\alpha _0 = [b_0, b_1, \ldots ]$, we have

• (i) $v_p(b_{n}) = v_p(\alpha _n)$, i.e., $|b_n |_p = |\alpha _n |_p$, for all $n \ge 1$.

• (ii) $v_p(b_n) < 0$, i.e., $|b_n |_p \ge p$, for all $n \ge 1$.

• (iii) $v_p(A_n) = v_p(b_0) + \ldots + v_p(b_n)$, $|A_n |_p = |b_0 \cdots b_n |_p$, for all $n \ge 0$, if $b_0 \not =0$.

• (iv) $v_p(A_n) = v_p(b_2) + \ldots + v_p(b_n)$, $|A_n |_p = |b_2 \cdots b_n |_p$, for all $n \ge 2$, if $b_0=0$.

• (v) $v_p(B_n) = v_p(b_1) + \ldots + v_p(b_n)$, $|B_n |_p = |b_1 \cdots b_n |_p$, for all $n \ge 1$.

• (vi) $v_p\left( \alpha _0 - \frac{A_n}{B_n} \right) = -v_p(B_n B_{n+1})$, for all $n \ge 0$.

Proof

The proofs of claims (i), (ii), (v) and (vi) can be found in [8, section 2]. As for (iii) and (iv), they are easily checked for $n=0$, $n=2$ respectively and if they hold up to n then

$$\begin{aligned} |A_{n+1}|_p=|b_{n+1}A_n+A_{n-1}|_p=|b_{n+1}A_n|_p \end{aligned}$$

since $|b_{n+1}A_n|_p>|A_n|_p\ge |A_{n-1}|_p$ by point (ii), so one can conclude by induction. $\square$

Remark 1

The $A_i$’s and $B_i$’s are rational numbers whose denominators are powers of p depending on the valuations of the partial quotients.

2.1 Some remarks on Browkin’s approach to p-adic continued fractions

For any prime $\ell$, let $|\cdot |_\ell$ and $\bar{B}_\ell (0,1)=\{x\in \mathbb {Q}:|x|_\ell \le 1\}$ denote respectively the $\ell$-adic absolute value and the closed unit ball in $\mathbb {Q}$ with respect to $|\cdot |_\ell$. Then we have

$$\begin{aligned} \mathbb {Z}=\bigcap _{\ell \text { prime}}\bar{B}_\ell (0,1)\;\; \text{ and } \;\;\mathbb {Z}[p^{-1}]=\bigcap _{\ell \ne p}\bar{B}_\ell (0,1) \end{aligned}$$

(the sets of S-integers of $\mathbb {Q}$, with S being respectively $\{\infty \}$ and $\{\infty ,p\}$). Moreover, putting $B_\infty (0,r)=\{x\in \mathbb {Q}:|x|_\infty <r\}$ for any $r>0$, a simple computation (as performed in [8, page 69]) shows

$$\begin{aligned} s(\mathbb {Q}_p)\subseteq \left\{ x\in \mathbb {Z}[p^{-1}]:|x|_\infty <\frac{p}{2}\right\} =B_\infty \!\left( 0,\frac{p}{2}\right) \cap \bigcap _{\ell \ne p}\bar{B}_\ell (0,1)\,. \end{aligned}$$

(5)

The value p/2 for the radius of the archimedean ball is optimal (see again [8, page 69]).

Thus, both the classical floor function and Browkin’s definition of s can be seen as instances of a function $s_v:\mathbb {Q}_v\rightarrow \mathbb {Q}$ (with v a place of $\mathbb {Q}$ and $\mathbb {Q}_v$ the completion with respect to v) such that

• (a) $|x-s_v(x)|_v<1$ for every $x\in \mathbb {Q}_v$;

• (b) The function $s_v$ takes values in the closed unit balls at all places outside $S=\{\infty ,v\}$ (i.e., $s_v(\mathbb {Q}_v)$ is a subset of the ring of S-integers);

• (c) If $v\ne \infty$, then $s_v(\mathbb {Q}_v)$ is contained in a ball $B_\infty (0,r)$ with small radius (as small as possible, compatibly with the previous conditions).

Condition (c) is what differentiates Browkin’s definition from alternative approaches, like Ruban’s (which admits greater values for $|s(x)|_\infty$), and in our opinion is the key reason for a closer analogy with the real case. In Ruban’s algorithm, the map s is still defined as in (2), but with the coefficients $a_i$ chosen in the set $\{0, \ldots , p-1\}$. One consequence is that Ruban’s continued fraction expansion of a rational number $\alpha$ need not be finite, even when $\alpha$ is an integer (e.g., take $\alpha =-p$; see [22, page 303] for details). With Browkin’s choice, instead, a p-adic number is rational if and only if it has a finite continued fraction expansion, like it happens with the reals. Equivalently, Browkin’s continued fractions correspond to a p-adic version of the Euclidean algorithm with the property of terminating on $\mathbb {Q}$ (analogous to the classical property on $\mathbb {Z}$), as we are going to illustrate.

Proposition 2

Given any x and y in $\mathbb {Q}_p$, with $y \not =0$, there exist unique $q \in \mathbb {Z}\left[ \frac{1}{p}\right]$ and $r \in \mathbb {Q}_p$ with $|q |_\infty <\frac{p}{2}$ and $|r |_p < |y |_p$ such that $x = q y + r$.

Proof

See [21, Theorem 1].

Recall that for $\alpha \in \mathbb {Q}_p$ Browkin’s s yields $s(\alpha ) \in \mathbb {Z}[p^{-1}]$, $|\alpha -s(\alpha )|_p<1$ and $|s(\alpha )|_\infty < \frac{p}{2}$. Thus comparing the two sides of

$$\begin{aligned} s\left( \frac{x}{y} \right) +\left( \frac{x}{y}-s\left( \frac{x}{y} \right) \right) = \frac{x}{y} = q + \frac{r}{y} \end{aligned}$$

we see that the quotient q defined in Proposition 2 coincides with s(x/y)

As described in [21, Definition 2], the p-adic Euclidean algorithm is obtained by iteration of the division-with-remainder process described in Proposition 2. If we start with inputs x, y both in $\mathbb {Q}$, then the algorithm stops (i.e., the remainder becomes 0) after a finite number of steps [21, Theorem 2]. In Browkin’s algorithm, the continued fraction of x/y results by iterating the same steps. Indeed, putting $x=x_0, y=x_1$, with $|x_0 |_p \ge |x_1 |_p \not = 0$, we can write $x_0 = b_0 x_1 + x_2$ where $b_0 = s(x_0 / x_1)$ and $|x_2 |_p < |x_1 |_p$. Repeating this, we get

$$\begin{aligned} x_i = b_i x_{i+1} + x_{i+2}, \end{aligned}$$

for $i = 0, 1, \ldots$, where $b_i = s(x_i / x_{i+1})$. Thus, setting $\alpha _i = \frac{x_i}{x_{i+1}}$, we obtain the continued fraction expansion of $\alpha _0$, since we have

$$\begin{aligned} \alpha _i = b_i + \frac{1}{\alpha _{i+1}} \end{aligned}$$

and the $b_i$’s and $\alpha _i$’s are obtained as in (2).

Finally, we note that Browkin’s proof that the continued fraction of $\alpha \in \mathbb {Q}$ is finite ([8, Theorem 3]) and Lager’s proof that the p-adic Euclidean algorithm stops if one starts with $x,y\in \mathbb {Q}$ ([21, Theorem 2]) are both based on the same computation.

Remark 2

Let K be a number field, $S_\infty$ the set of its non-archimedean places and $\mathfrak P$ a non-archimedean place of K. The question of finiteness in a $\mathfrak P$-adic continued fraction expansion for elements of K is discussed at length in [12]. Here we want to mention that a sufficient condition for ensuring such finiteness arises exactly from adapting condition (c) above: that is, choosing a “floor function” $s_{\mathfrak P}$ so that its values are S-integers (with $S=S_\infty \cup \{\mathfrak P\}$) which are “small” at the archimedean places. See [12, Theorems 4.5 and 4.6] for precise statements. (One should also compare our conditions (a) and (b) with the more precise axiomatization of $\mathfrak P$-adic floor function in [12, Definition 3.1].)

2.1.1 On the axiomatization of s

Let K be a field complete with respect to a discrete valuation $\nu$ and let $\mathfrak {m}$ be the maximal ideal in its ring of integers. In [8, section 1] one can find an axiomatic definition of a function $s:K\rightarrow K$ which should play the role of the floor function for continued fractions in K: namely, s is required to satisfy the three properties

• (S1) $s(0)=0$ ;

• (S2) $\nu \circ s=\nu$ ;

• (S3) $(a-b)\in \mathfrak {m}\Longrightarrow s(a)=s(b)$.

However, there are some problems with this definition. To begin with, (S2) and (S3) are not compatible (since $\nu$ takes infinitely many distinct values in $\mathfrak {m}$). Moreover, these conditions do not imply $x-s(x)\in \mathfrak {m}$ (i.e., the analogue of our condition (a) above): indeed, the function $s:\mathbb {Q}_p\rightarrow \mathbb {Q}_p$ defined by

$$\begin{aligned}s(a)={\left\{ \begin{array}{ll}0\;\;\;\;\;\;\;\;\text { if }a\in p\mathbb {Z}_p\\ p^{v_p(a)}\;\text { otherwise}\end{array}\right. }\end{aligned}$$

satisfies (S1), (S3) and a modified version of (S2), but $2p^{-1}-s(2p^{-1})$ is not in $p\mathbb {Z}_p$. It is clear from the rest of the paper that what the author had really in mind when stating (S2) and (S3) was the condition $s=s_1\circ \pi$, where $\pi :K\rightarrow K/\mathfrak {m}$ is the quotient map and $s_1$ a section of $\pi$.

3 Auxiliary results

We collect some auxiliary results needed for proving the main theorems in the next section. In particular, we give some bounds for the (naive) height of algebraic numbers with a specific expansion in Browkin p–adic continued fractions. These results may be also of general interest independently of their use. Moreover, we state the p–adic version of the Subspace Theorem, which is an essential tool for proving the transcendence of numbers also in the real case. We also give two weaker versions of it (Corollaries 1 and 3), which are eventually used in our proofs.

3.1 Properties of partial quotients, convergents and height bounds

In the following, unless otherwise stated, all continued fractions developments are based on Browkin’s choice of s, as in (2), and the sequences $(A_i),(B_i)$ are always defined by (3).

Lemma 1

If $\alpha , \beta \in \mathbb {Q}_p$ have the same $n + 1$ first partial quotients in the p–adic continued fraction expansion, then $|\alpha - \beta |_p < \frac{1}{|B_n |_p^2}$.

Proof

By hypothesis, the n–th and $(n+1)$–th convergents of the p-adic continued fraction of $\alpha$ and $\beta$ are the same, namely $\frac{A_n}{B_n}$ and $\frac{A_{n+1}}{B_{n+1}}$. By Proposition 1, we have

$$\begin{aligned}\left|\alpha - \frac{A_n}{B_n} \right|_p = \frac{1}{|B_{n+1} |_p |B_n |_p} = \frac{1}{|b_{n+1} |_p |B_n |_p^2} \end{aligned}$$

since $v_p(B_{n+1}) = v_p(b_{n+1}B_n + B_{n-1}) = v_p(b_{n+1}) + v_p(B_n).$ The same holds for $\beta$ and thus we conclude. $\square$

Lemma 2

Given $\alpha _0 = [b_0, b_1, \ldots ]$, we have $|A_n |_\infty < |A_n |_p$ and $|B_n |_\infty < |B_n |_p$, for all $n\gg 0$.

Proof

Let $x_0,x_1\in \mathbb {Q}$ be arbitrary and put

$$\begin{aligned}x_n:=b_nx_{n-1}+x_{n-2}.\end{aligned}$$

Choose M so that $|x_0|_\infty <M$ and $\displaystyle |x_1|_\infty <M\left( \frac{p}{2}+1\right)$ both hold. Then

$$\begin{aligned} |x_k|_\infty <M\left( \frac{p}{2}+1\right) ^{\!k} \end{aligned}$$

(6)

is true for every $k\in \mathbb {N}$. Indeed, assuming by induction that (6) is satisfied up to $n-1$, one finds

$$\begin{aligned} |x_n|_\infty\le & {} |b_n|_\infty \,|x_{n-1}|_\infty +|x_{n-2}|_\infty \\\le & {} \frac{p}{2}|x_{n-1}|_\infty +|x_{n-2}|_\infty \hspace{70pt}\text { by }(5) \\< & {} \frac{p}{2} M\left( \frac{p}{2}+1\right) ^{\!n-1}+M\left( \frac{p}{2}+1\right) ^{\!n-2}=M\left( \frac{p}{2}+1\right) ^{\!n-2}\left( \frac{p^2}{4}+\frac{p}{2}+1\right) \\< & {} M\left( \frac{p}{2}+1\right) ^{\!n}. \end{aligned}$$

In particular, one can choose M such that that (6) applies to both sequences $(A_n)$ and $(B_n)$. Now it is enough to recall Proposition 1 and note that

$$\begin{aligned}M\left( \frac{p}{2}+1\right) ^{\!n}<p^{n-2}\le \min \big \{|A_n|_p,|B_n|_p\big \}\end{aligned}$$

is true for $n\gg 0$. $\square$

Remark 3

With slightly more effort, one can check that $|B_n |_\infty \le |B_n |_p$ holds for all $n\ge -2$. If $|b_0|_p\ne 1$, one also has $|A_n |_\infty \le |A_n |_p$ for all $n \ge -2$.

We recall the following classical definition. Let $\alpha$ be an algebraic number of degree d: then $\alpha$ is a root of a non-zero polynomial with integer coefficients $c_0,...,c_d$. The naive height of $\alpha$ is defined as

$$\begin{aligned}h(\alpha ) = \frac{\max \{ |c_i|_\infty : i = 0, \ldots , d \}}{\gcd (c_0, \ldots , c_d)}.\end{aligned}$$

Lemma 3

Assume $\alpha \in \mathbb {Q}_p$ has a periodic Browkin continued fraction

$$\begin{aligned}\alpha = [0, b_1, \ldots , b_{k},\overline{b_{k+1}, \ldots , b_{k+t+1}}].\end{aligned}$$

Then

$$\begin{aligned}h(\alpha ) \le \frac{8}{p^2}|B_{k+t+1} |_p^2 |B_k |_p^2.\end{aligned}$$

There is no loss of generality in assuming $k \ge 1$, since one can always start the period after the first few terms.

Proof

Let us consider $\beta = [\overline{b_{k+1}, \ldots , b_{k+t+1}}]$ and attach to it sequences $(\tilde{A}_i),(\tilde{B}_i)$ by (3). Using (4), we get

$$\begin{aligned} \beta = \frac{\beta \tilde{A}_t + \tilde{A}_{t-1}}{\beta \tilde{B}_t + \tilde{B}_{t-1}}, \quad \alpha = \frac{\beta A_k + A_{k-1}}{\beta B_k + B_{k-1}} \end{aligned}$$

from which

$$\begin{aligned} C_0 \alpha ^2 + C_1 \alpha + C_2 = 0 \end{aligned}$$

(7)

where

$$\begin{aligned} {\left\{ \begin{array}{ll} C_0 = - \tilde{A}_{t-1} B_k^2 + \tilde{A}_t B_k B_{k-1} - \tilde{B}_{t-1} B_k B_{k-1} + \tilde{B}_t B_{k-1}^2 \\ C_1 \!=\! 2 \tilde{A}_{t-1} A_k B_k \!-\! \tilde{A}_{t} A_{k-1} B_{k} \!+\! \tilde{B}_{t-1} A_{k-1} B_k \!-\! \tilde{A}_t A_k B_{k-1} \!+\! \tilde{B}_{t-1}A_k B_{k-1}\! -\! 2 \tilde{B}_t A_{k-1} B_{k-1} \\ C_2 = -\tilde{A}_{t-1} A_k^2 + \tilde{A}_t A_k A_{k-1} -\tilde{B}_{t-1} A_k A_{k-1} + \tilde{B}_t A_{k-1}^2 \end{array}\right. }\end{aligned}$$

By construction we have

$$\begin{aligned} C_i = \sum _{j=1}^{h_i} C_{ij} \in \mathbb {Z}\left[ p^{-1}\right] \end{aligned}$$

where $h_0=h_2=4$ or $h_1=8$ and the summands $C_{ij}$ are given by the formulae above. In order to bound $h(\alpha )$ by (7), we need to multiply by a power of p, so to have integer coefficients.

Putting

$$\begin{aligned} |\tilde{A}_i |_p = p^{\tilde{e}_i}, \quad |\tilde{B}_i |_p = p^{\tilde{f}_i}, \quad |A_i |_p = p^{e_i}, \quad |B_i |_p = p^{f_i}, \end{aligned}$$

we obtain four increasing sequences of integers $(\tilde{e}_i), (\tilde{f}_i), (e_i)$ and $(f_i)$. By Proposition 1, we have $\tilde{e}_i> \tilde{f}_i>0$ for $i \ge 0$, because $|b_{k+1}|_p\ge p$, and $f_i> e_i>0$ for $i \ge 2$, since $b_0 = 0$; moreover $\tilde{e}_i + f_k=f_{k+i+1}$ holds for $i\ge 0$. By these observations, one can check that

$$\begin{aligned}|C_{ij}|_p \le p^{\tilde{e}_t + 2f_k -1} = \frac{1}{p} |B_{k+t+1} |_p |B_{k} |_p\end{aligned}$$

for all i, j. Therefore multiplying (7) by $p^{\tilde{e}_t + 2f_k-1}$ we obtain a polynomial with integer coefficients satisfied by $\alpha$. This implies

$$\begin{aligned} h(\alpha ) \le \frac{1}{p} |B_{k+t+1} |_p |B_{k} |_p \max \{ |C_0 |_\infty , |C_1 |_\infty , |C_2 |_\infty \}. \end{aligned}$$

To conclude note that by Lemma 2 (and Remark 3, using $b_0 = 0$ and $|b_{k+1}| \ge p$) we have

$$\begin{aligned}|C_i |_\infty \le \sum _{j=1}^{h_i} |C_{ij}|_p \le \frac{8}{p} |B_{k+t+1} |_p |B_{k} |_p.\end{aligned}$$

$\square$

Lemma 4

For every $k \ge 1$, given $b_i = \frac{\hat{b}_i}{p^{a_i}}$ such that $a_i, \hat{b}_i \in \mathbb {Z}$, $a_i > 0$, $|\hat{b}_i|_\infty < \sqrt{\frac{3}{14}} \cdot \frac{p^a}{F_{k+1}}$ (for the $b_i \not = p^{-1}$), where $a = \min \{ a_i: 1 \le i \le k, a_i \not = 1 \}$ and $(F_i)_{i \ge 0} = (0, 1, 1, 2, 3, 5, \ldots )$ is the Fibonacci sequence, then $\alpha = [0, b_1, b_2, \ldots , b_{k}, \overline{p^{-1}}]$ satisifies

$$\begin{aligned}h(\alpha ) \le |B_{k} |_p^2.\end{aligned}$$

Proof

We can write

$$\begin{aligned}\quad A_i = \frac{\hat{A}_i}{p^{e_i}}, \quad B_i = \frac{\hat{B}_i}{p^{f_i}} \end{aligned}$$

where $\hat{A}_i, \hat{B}_i \in \mathbb {Z}$, $v_p(\hat{b}_i) = v_p(\hat{A}_i) = v_p(\hat{B}_i) = 0$, $e_i = a_2 + \ldots + a_i$, $f_i = a_1 + \ldots + a_i$, for any $1 \le i \le k$. Note that we have

$$\begin{aligned} = \frac{A_i}{B_i} = \frac{p^{a_1} \hat{A}_i}{\hat{B}_i}\end{aligned}$$

for any $1 \le i \le k$. We have that $\hat{B}_{i}$ is the sum of $F_{i+1}$ elements of the kind $\hat{b}_{i_1}\cdots \hat{b}_{i_h}p^{a_{j_1}}\cdots p^{a_{j_l}}$ such that $\{i_1, \ldots , i_h, j_1, \ldots j_l\} = \{1, \ldots , i\}$ and $h+l = i$, for any $1 \le i \le k$. We prove this by induction. For $i=1$, we have

$$\begin{aligned}\frac{A_1}{B_1} = \frac{p^{a_1} \hat{A}_1}{\hat{B}_1} = [0, b_1] = \frac{1}{b_1} = \frac{p^{a_1}}{\hat{b}_1}, \quad \hat{B}_1 = \hat{b}_1\end{aligned}$$

For $i = 2$, we have

$$\begin{aligned} \frac{A_2}{B_2} = \frac{p^{a_1}\hat{A}_2}{\hat{B}_2} = [0, b_1, b_2] = \frac{p^{a_1} \hat{b}_2}{\hat{b}_1 \hat{b}_2 + p^{a_1}p^{a_2}}, \quad \hat{B}_2 = \hat{b}_1 \hat{b}_2 + p^{a_1}p^{a_2}. \end{aligned}$$

For any $3 \le i \le k$, we have

$$\begin{aligned} \frac{A_i}{B_i} = \frac{b_{i} A_{i-1} + A_{i-2}}{b_{i} B_{i-1} + B_{i-2}} = \frac{\frac{\hat{b}_{i}}{p^{a_{i}}} \cdot \frac{\hat{A}_{i-1}}{p^{e_{i-1}}} + \frac{\hat{A}_{i-2}}{p^{e_{i-2}}}}{\frac{\hat{b}_{i}}{p^{a_{i}}} \cdot \frac{\hat{B}_{i-1}}{p^{f_{i-1}}} + \frac{\hat{B}_{i-2}}{p^{f_{i-2}}}} = \frac{p^{a_1}(\hat{b}_{i} \hat{A}_{i-1} + p^{a_{i-1}}p^{a_i} \hat{A}_{i-2})}{\hat{b}_i \hat{B}_{i-1} + p^{a_{i-1}}p^{a_i}\hat{B}_{i-2}}. \end{aligned}$$

By inductive hypothesis, $\hat{B}_{i-1}$ has $F_{i}$ addends of the kind $\hat{b}_{i_1}\cdots \hat{b}_{i_h}p^{a_{j_1}}\cdots p^{a_{j_l}}$ with

$\{i_1, \ldots , i_h, j_1, \ldots j_l\}$ $= \{1, \ldots , i-1\}$ and $h+l = i-1$. Thus, $\hat{b}_{i} \hat{B}_{i-1}$ has $F_{i+1}$ addends of the kind $\hat{b}_{i_1}\cdots \hat{b}_{i_h}p^{a_{j_1}}\cdots p^{a_{j_l}}$ with $\{i_1, \ldots , i_h, j_1, \ldots j_l\}$ $= \{1, \ldots , i\}$ and $h+l = i$.

Similarly, $p^{a_{i-1}}p^{a_i} \hat{B}_{i-2}$ has $F_{i-1}$ addends of the kind $\hat{b}_{i_1}\cdots \hat{b}_{i_h}p^{a_{j_1}}\cdots p^{a_{j_l}}$ with $\{i_1, \ldots , i_h, j_1, \ldots j_l\} = \{1, \ldots , i\}$ and $h+l = i$.

Finally, we have that $\hat{B}_i$ has $F_{i}+ F_{i-1} = F_{i+1}$ addends of the kind $\hat{b}_{i_1}\cdots \hat{b}_{i_h}p^{a_{j_1}}\cdots p^{a_{j_l}}$ with $\{i_1, \ldots , i_h, j_1, \ldots j_l\} = \{1, \ldots , i\}$ and $h+l = i$.

By hypothesis we have $|\hat{b}_i |_\infty < \sqrt{\frac{3}{14}}\cdot \frac{p^a}{F_{k+1}}$, where $a = \min \{ a_i: 1 \le i \le k, a_i\not = 1 \}$ or $\hat{b}_i = 1 < p$ (if $b_i = p^{-1}$) for any $1 \le i \le k$, then $|B_{k} |_\infty < \sqrt{\frac{p}{4p + 2}}$. Indeed, $B_{k} = \frac{\hat{B}_{k}}{p^{f_{k}}}$ and we have seen that $\hat{B}_{k}$ is a sum of $F_{k+1}$ elements of the kind $\hat{b}_{i_1}\cdots \hat{b}_{i_h}p^{a_{j_1}}\cdots p^{a_{j_l}}$ with $\{i_1, \ldots , i_h, j_1, \ldots j_l\} = \{1, \ldots , i\}$ and $h+l = k$, that is

$$\begin{aligned} B_k = \sum \frac{\hat{b}_{i_1} \cdots \hat{b}_{i_h}}{p^{a_1} \cdots p^{a_{h}}} \end{aligned}$$

where there are $F_{k+1}$ addends. Thus, by hypothesis we have that any addend satisfies

$$\begin{aligned} \hat{b}_{i_1}\cdots \hat{b}_{i_h} < \sqrt{\frac{3}{14}} \cdot \frac{p^{a_{i_1}+\ldots +a_{i_h}}}{F_{k+1}}, \end{aligned}$$

from which we obtain

$$\begin{aligned} |B_k |_\infty < \sqrt{\frac{3}{14}} \le \sqrt{\frac{p}{4p+2}} \end{aligned}$$

for every odd prime p. Similar arguments can be also used to show that $|A_{k} |_\infty < \sqrt{\frac{p}{4p + 2}}$. Let us observe that, fixed an integer k, we can take the partial quotients $b_1, \ldots , b_k$ such that $|\hat{b}_i |_\infty < \sqrt{\frac{3}{14}}\cdot \frac{p^a}{F_{k+1}}$ in infinitely many different ways. Indeed, it is sufficient to take the minimum of the $a_i$’s sufficiently large and observe that the integers $a_i$’s can be chosen in infinitely different ways. Now, consider $\beta = [\overline{p^{-1}}]$, we have that $\beta$ is a root of $p \beta ^2 - \beta - p = 0$. Moreover, we have

$$\begin{aligned} \alpha = \frac{\beta A_k + A_{k-1}}{\beta B_k + B_{k-1}}, \quad \beta = \frac{A_{k-1} - \alpha B_{k-1}}{B_k \alpha - A_k}. \end{aligned}$$

Thus, we obtain the minimal polynomial of $\alpha$ as

$$\begin{aligned}&(B_kB_{k-1}-B_k^2p+B_{k-1}^2p)\alpha ^2 + \\ (2A_kB_k-2A_{k-1}B_{k-1}p-&A_kB_{k-1}-A_{k-1}B_k)\alpha +(A_kA_{k-1}-A_k^2p+A_{k-1}^2p) = 0, \end{aligned}$$

which can be also written as

$$\begin{aligned}&(\hat{B}_k \hat{B}_{k-1}p^{a_k-1}-\hat{B}_k^2+\hat{B}_{k-1}^2p^{2a_k})\alpha ^2 + \\ (2\hat{A}_k \hat{B}_k&p^{a_1}- 2 \hat{A}_{k-1} \hat{B}_{k-1} p^{a_1+2a_k} - \hat{A}_k \hat{B}_{k-1}p^{a_1+a_k-1} - \hat{A}_{k-1}\hat{B}_k p^{a_1+a_k-1})\alpha + \\&(\hat{A}_k \hat{A}_{k-1} p^{2a_1+a_k-1}-\hat{A}_k^2 p^{2a_1}+ \hat{A}_{k-1}^2 p^{2a_1+2a_k}) = 0. \end{aligned}$$

Under the condition $|A_{k} |_{\infty }, |B_{k} |_{\infty } < \sqrt{\frac{p}{4p + 2}}$, we get

• $|\hat{B}_k \hat{B}_{k-1}p^{a_k-1}-\hat{B}_k^2+\hat{B}_{k-1}^2p^{2a_k} |_\infty \le p^{2(a_1 + \ldots + a_k)}$, since $|B_kB_{k-1} |_{\infty } \cdot \frac{1}{p} + B_{k-1}^2 + B_k^2 \le 1$

• $|2\hat{A}_k \hat{B}_k p^{a_1}- 2 \hat{A}_{k-1} \hat{B}_{k-1} p^{a_1+2a_k} - \hat{A}_k \hat{B}_{k-1}p^{a_1+a_k-1} - \hat{A}_{k-1}\hat{B}_k p^{a_1+a_k-1} |_{\infty } \le p^{2(a_1 + \ldots + a_k)}$, since $2|A_kB_k |_{\infty } + 2|A_{k-1}B_{k-1} |_{\infty } + |A_kB_{k-1}|_{\infty }\cdot \frac{1}{p}+|A_{k-1}B_k|_{\infty } \frac{1}{p} \le 1$

• $|\hat{A}_k \hat{A}_{k-1} p^{2a_1+a_k-1}-\hat{A}_k^2 p^{2a_1}+ \hat{A}_{k-1}^2 p^{2a_1+2a_k}|_{\infty } \le p^{2(a_1 + \ldots + a_k)}$, since $|A_kA_{k-1} |_\infty \cdot \frac{1}{p} + A_k^2 + A_{k-1}^2 \le 1$.

Thus, we have $h(\alpha ) \le p^{2(a_1 + \ldots + a_k)} = |B_{k} |_p^2$. $\square$

Example 1

With the notation of Lemma 4, for $p = 5$, $k = 2$ consider the continued fraction

$$\begin{aligned} \alpha = \left[ 0, \frac{4}{5^2}, -\frac{3}{5^3}, \overline{\frac{1}{5}}\;\right] . \end{aligned}$$

We observe that $|\hat{b}_1 |_\infty = 4$ and $|\hat{b}_2 |_\infty = 3$ are smaller than $\sqrt{\frac{3}{14}}\cdot \frac{5^2}{2} = 5.78\ldots$ . The minimal polynomial of $\alpha$ is $9.129.469 x^2 + 5.530.075 x - 9.713.125$ and $|B_2 |_p^2 = p^{10} = 9.765.625$. Clearly, we can construct infinitely many continued fractions of this kind with $k = 2$. Indeed, we can take $b_1$ and $b_2$ with a denominator consisting of any possible power of p (and large values of this powers allow also more possible choices for $\hat{b}_1$ and $\hat{b}_2$).

3.2 p-adic Subspace theorem

We begin restating a p-adic version of the celebrated Roth’s theorem and its generalizations due to Schmidt, Evertse, Schlickewei; see [17]. This is currently one of the main technical tools of diophantine approximation applied to archimedean continued fractions, see, for instance, [2].

Let us recall the definition of the Weil (sometimes called absolute) height function H of an algebraic number x. The absolute value $|\cdot |_v$ is the unique absolute value on $\mathbb {Q}(x)$ which restricts to the usual v-adic absolute value over $\mathbb {Q}$. Let also n and $n_v$ be respectively the global and local degrees of x. Then the Weil height is defined as

$$\begin{aligned}H(x):= \prod _v\max \{1,|x|^{n_v}_v\}^{1/n},\end{aligned}$$

for v running over all places of $\mathbb {Q}(x)$.

Remark 4

The relation between the naive height h and the absolute height H is well known. For $\alpha \in \mathbb {Q}$, the two heights coincide, on the other hand, for $\alpha \in \overline{\mathbb {Q}}$ of degree D, we have

$$\begin{aligned} H(\alpha )\le (D+1)^{1/2}h(\alpha ),\;\; h(\alpha )\le 2^D H(\alpha ). \end{aligned}$$

For the proof and much more, see [20, Part B, p.177].

We recall the general formulation of Roth’s theorem, as stated in [20].

Theorem 1

Let K be a number field, let S be a finite set of absolute values on K with each absolute value extended in some way to $\overline{K}$. Let $\alpha \in \overline{ K}$ and $\varepsilon > 0$ be given. Suppose that

$$\begin{aligned}\xi : S \rightarrow [0, 1] \text { is a function satisfying } \sum _{v \in S} \xi (v) = 1.\end{aligned}$$

Then, there are only finitely many $\beta \in K$ with the property that

$$\begin{aligned} |\alpha - \beta |_v \le \frac{1}{H(\beta )^{(2+\varepsilon )\xi (v)}} \end{aligned}$$

for all $v \in S$.

Proof

See [20, Theorem D.2.2]. $\square$

In our first main Theorem, we need the following version of Roth’s theorem as described in the next corollary.

Corollary 1

Consider K a quadratic field, $S =\{ \infty , p\}$ and $\xi (\infty ) = 0$, $\xi (p) = 1$. Let $\varepsilon , C > 0$, then there are finitely many $\beta \in K$ such that

$$\begin{aligned} |\alpha - \beta |_p\le C H(\beta )^{-2-\epsilon } \end{aligned}$$

holds.

We now state a p–adic version of the Subspace Theorem.

Theorem 2

Let $n \ge 2$ and $S=\{\infty , p\}$ contain two places of $\mathbb {Q}$. Consider $L_{1,\infty },...,L_{n,\infty }$ independent linear forms in n variables with real algebraic coefficients and $L_{1,p},...,L_{n,p}$ independent linear forms in n variables with p-adic algebraic coefficients. Then, for every $\epsilon > 0$ there are a finite number of proper rational subspaces $T_1, \ldots , T_r$ such that every $x \in \mathbb {Z}[p^{-1}]^n - \{0\}$ satisfying

$$\begin{aligned} \prod _{v\in S} \prod _{i\le n}|L_{i,v}(x)|_v < \max \{|x_1|_\infty ,...,|x_n|_\infty \}^{-\epsilon }\max \{|x_1|_p,...,|x_n|_p\}^{-\epsilon } \end{aligned}$$

lies in one of these subspaces.

Proof

See [10, Corollary 7.2.5] $\square$

We exploit Theorem 2 to prove the following results which will be applied for proving part (b) of Theorem A.

Lemma 5

Let $\alpha , \beta \in \mathbb {Q}_p$ be algebraic and $1, \alpha , \beta$ linearly independent over $\mathbb {Q}$. Then for every $\epsilon > 0$ there are only finitely many $(x, y, z) \in \mathbb {Z}[p^{-1}]^3$, with $z \not =0$ and $|z|_p = \max \{ |x|_p, |y|_p, |z|_p \}$, such that

$$\begin{aligned}|(z\alpha - x)(z\beta - y)|_p |z|_p^{2\epsilon + 7/4} < 1\end{aligned}$$

and $|x|_\infty < |x|_p^{1/4}$, $|y|_\infty < |y|_p^{1/4}$, $|z|_\infty < |z|_p^{1/4}$.

Proof

We introduce the following linearly independent forms in three variables

$$\begin{aligned} L_{1,p}(x,y,z)=\alpha z - x \quad L_{2,p}(x,y,z)=\beta z - y, \quad L_{3,p}(x,y,z)= z \end{aligned}$$

and

$$\begin{aligned} L_{1,\infty }(x,y,z) = x, \quad L_{2,\infty }(x,y,z) = y, \quad L_{3, \infty }(x,y,z) = z. \end{aligned}$$

In what follows, we always consider triples $(x, y, z) \in \mathbb {Z}[p^{-1}]^3$ such that $|x|_\infty < |x|_p^{1/4}$, $|y|_\infty < |y|_p^{1/4}$, $|z|_\infty < |z|_p^{1/4}$. By Theorem 2, the solutions $(x, y, z) \in \mathbb {Z}[p^{-1}]^3$ of

$$\begin{aligned} |(z\alpha - x)(z\beta - y)z|_p|xyz|_\infty \max (|x|_\infty , |y|_\infty , |z|_\infty )^\epsilon \max (|x|_p, |y|_p, |z|_p)^\epsilon < 1 \end{aligned}$$

(8)

lie in a finite number of rational subspaces. Hence for every solution (x, y, z) in a fixed subspace $x = Ay + Bz$ (with $A, B \in \mathbb {Q}$), formula (8) becomes

$$\begin{aligned}{} & {} |(z(\alpha -B) - Ay)(z\beta -y) z|_p|(Ay+Bz)yz|\max (|Ay+Bz|, |y|, \nonumber \\{} & {} |z|)^\epsilon \max (|Ay+Bz|_p, |y|_p, |z|_p)^\epsilon < 1, \end{aligned}$$

(9)

where we denote by $|\cdot |$ what before was denoted by $|\cdot |_\infty$.

Now observe that the solutions $(x,y,z) \in \mathbb {Z}[p^{-1}]^3$ of

$$\begin{aligned} |(z(\alpha -B) - Ay)(z\beta - y)|_p|(Ay+Bz)y|_\infty \max (|y|_\infty , |z|_\infty )^\epsilon \max (|y|_p, |z|_p)^\epsilon < 1 \end{aligned}$$

(10)

lie, by Theorem 2, in a finite number of rational subspaces so that we can consider $z = Cy$, for $C \in \mathbb {Q}$. Thus (10) becomes

$$\begin{aligned}|y|_p^{\epsilon + 2} |y|_\infty ^{\epsilon +2} K < 1,\end{aligned}$$

for some constant $K \in \mathbb {Q}$. This inequality has a finite number of solutions in $\mathbb {Z}[p^{-1}]$ (given that $|y|_\infty < |y|_p^{1/4}$). Hence also (10) has a finite number of solutions and thus the same holds for (9) as well. Indeed, every solution of (9) is a solution of (10) (caveat, the converse may fail), since $|z|_\infty |z|_p\ge 1$. Consequently there are finitely many solutions of (8).

Now we use the hypothesis $|z|_p = \max (|x|_p, |y|_p, |z|_p)$. The left side of (8) becomes

$$\begin{aligned}F(x,y,z):= |(z \alpha - x)(z \beta - y)|_p|z|_p^{\epsilon + 1} |xyz|_\infty \max (|x|_\infty , |y|_\infty , |z|_\infty )^{\epsilon }\end{aligned}$$

and we have

$$\begin{aligned}|(z\alpha - x)(z \beta - y)|_p|z|_p^{2\epsilon + 7/4} > F(x,y,z)\end{aligned}$$

for every $(x,y,z) \in \mathbb {Z}[p^{-1}]^3$. This completes the proof. $\square$

Corollary 2

Let $\alpha , \beta \in \mathbb {Q}_p$ be algebraic and $1, \alpha , \beta$ linearly independent over $\mathbb {Q}$. Then for every $\epsilon >0$ there are only finitely many $(x,y,z) \in \mathbb {Z}[p^{-1}]^3$, with $z \not =0$, such that

$$\begin{aligned} \left| \alpha - \frac{x}{z} \right| _p< |z|_p^{-\epsilon - 15/8}, \quad \left| \beta - \frac{y}{z}\right| _p < |z|_p^{-\epsilon - 15/8}\end{aligned}$$

and $|x|_\infty < |x|_p^{1/4}$, $|y|_\infty < |y|_p^{1/4}$, $|z|_\infty < |z|_p^{1/4}$, with $|z|_p = \max \{ |x|_p, |y|_p, |z|_p \}$.

Proof

Every triple $(x,y,z) \in \mathbb {Z}[p^{-1}]^3$ satisfying the hypotheses of the Corollary satisfies also the hypotheses of Lemma 5. Therefore there are finitely many such triples. $\square$

Corollary 3

Let $\alpha \in \mathbb {Q}_p$ be non rational and non–quadratic. If there exists a real number $\delta > 15/8$ and infinitely many $(x,y,z)\in \mathbb {Z}[p^{-1}]^3$, with $z \not =0$ and $|z|_p = \max \{ |x|_p, |y|_p, |z|_p \}$, such that

$$\begin{aligned} \max \{|\alpha -x/z|_p, |\alpha ^2-y/z|_p \}<|z |_p^{-\delta }, \end{aligned}$$

and $|x|_\infty < |x|_p^{1/4}$, $|y|_\infty < |y|_p^{1/4}$, $|z|_\infty < |z|_p^{1/4}$ then $\alpha$ is transcendental.

Proof

Let us suppose $\alpha$ algebraic, non–rational and non–quadratic. Then $1, \alpha , \alpha ^2$ are linearly independent over $\mathbb {Q}$ and the thesis follows by Corollary 2. $\square$

4 Main results

In this section, we prove three results on the transcendence of Browkin p–adic continued fractions. We state the following hypothesis which contains the conditions that must be satisfied by the partial quotients in order that Lemma 4 holds.

Hypothesis 1

Let $(b_1, \ldots , b_k)$ be a finite sequence such that $b_i = \frac{\hat{b}_i}{p^{a_i}}$, with $a_i, \hat{b}_i \in \mathbb {Z}$, $a_i > 0$, $|\hat{b}_i|_\infty < \sqrt{\frac{3}{14}} \cdot \frac{p^a}{F_{k+1}}$ (for the $b_i \not = p^{-1}$), where $a = \min \{ a_i: 1 \le i \le k, a_i\not =1 \}$ and $(F_i)_{i \ge 0}$ the Fibonacci sequence.

Theorem 3

Let $\alpha = [0, b_1, b_2, b_3, \ldots ]$ be a non–periodic Browkin p–adic continued fraction such that $(b_i)$ is bounded in $\mathbb {Q}_p$, with $D=\max _i\{|b_i|_p\}$ and let $(n_i)_{i \ge 0}$, $(k_i)_{i \ge 0}$, $(\lambda _i)_{i \ge 0}$ be sequences of positive integers such that

• the $k_i$’s are bounded;

• $n_{i+1} \ge n_i + \lambda _i k_i$ and there exists $C > 2\frac{\log D}{\log p} - 1$ such that $\lambda _i > C n_i$ for all i sufficiently large;

• $b_{n_i} = \ldots = b_{n_i + \lambda _i k_i -1} = p^{-1}$ for every i;

• the finite sequence $(b_1, \ldots , b_{n_{i-1}})$ satisfies Hypothesis 1 for every $i\ge 1$.

Then $\alpha$ is transcendental.

Proof

Let us suppose that $\alpha$ is an algebraic number, we define infinitely many irrational numbers of the kind

$$\begin{aligned} \beta ^{(i)} = [0, b_1, \ldots , b_{n_i-1}, \overline{p^{-1}}], \end{aligned}$$

such that $h(\beta ^{(i)}) \le |B_{n_i-1} |_p^2$. The existence of infinitely many irrational numbers of this kind is ensured by Lemma 4 (taking $k = n_{i-1}$). By construction, the $\beta ^{(i)}$ have the first $n_i+k_i\lambda _i$ partial quotients equal to the ones of $\alpha$, thus, by Lemma 1, we have

$$\begin{aligned} |\alpha - \beta ^{(i)} |_p < |B_{n_i+k_i\lambda _i - 1} |_p^{-2}. \end{aligned}$$

(11)

Moreover, we have

$$\begin{aligned} |\alpha - \beta ^{(i)} |_p > |B_{n_i - 1} |_p^{-2\omega } \end{aligned}$$

(12)

with $\omega > 2$. Indeed, if $|\alpha - \beta ^{(i)} |_p \le |B_{n_i - 1} |_p^{-2\omega }$, then by Remark 4, we have

$$\begin{aligned} |\alpha - \beta ^{(i)} |_p \le h(\beta ^{(i)})^{-\omega } \le C H(\beta ^{(i)})^{-\omega } \end{aligned}$$

with $C > 0$, in contradiction with Corollary 1. Thus, using (11) and (12), we obtain

$$\begin{aligned} |B_{n_i+k_i\lambda _i - 1} |_p < |B_{n_i-1} |_p^{\omega }. \end{aligned}$$

(13)

Since $|b_i|_p$ is bounded by D, we have $p \le |b_i|_p \le D$ for $i \ge 1$, from which we get $p^i \le |B_i |_p \le D^i$. Hence, considering $k_i$ bounded and remembering point (v) of Proposition 1, in order that (13) holds, we must have $\frac{\lambda _i}{n_i} < w \frac{\log D}{\log p} - 1.$ On the other hand, by hypothesis we have $\frac{\lambda _i}{n_i} > C$ and taking

$$\begin{aligned}\delta = C - \frac{2 \log D}{\log p} + 1\end{aligned}$$

we should have

$$\begin{aligned} 2 + \frac{\log p}{\log D}\delta < \omega \end{aligned}$$

for any $\omega > 2$, where $\delta >0$, which is not possible. $\square$

Theorem 4

Let $\alpha = [0, b_1, b_2, b_3, \ldots ]$ be a non–periodic Browkin p–adic continued fraction such that $(b_i)$ is bounded in $\mathbb {Q}_p$, with $D=\max _i\{|b_i|_p\}$ and let $(n_i)_{i \ge 0}$, $(k_i)_{i \ge 0}$, $(\lambda _i)_{i \ge 0}$ be sequences of positive integers such that

• the $k_i$’s are bounded;

• $n_{i+1} \ge n_i + \lambda _i k_i$ and there exists $C > 4\frac{\log D}{\log p} - 1$ such that $\lambda _i > C n_i$ for all i sufficiently large;

• $b_{h+k_i} = b_{h}$, for $n_i \le h \le n_i + (\lambda _i-1)k_i - 1$, for every i;

• the finite sequence $(b_1, \ldots b_{n_i})$ satisfies Hypothesis 1 for every i.

Then $\alpha$ is transcendental or a quadratic irrational.

Proof

Let us suppose $\alpha$ an algebraic number of degree $> 2$. Since $(k_i)_{i \ge 0}$ is a bounded sequence of positive integers, there exist infinitely many j such that

$$\begin{aligned}k_j = k, \quad b_{n_j} = \hat{b}_{1}, \quad \ldots , \quad b_{n_j + k - 1} = \hat{b}_{k}\end{aligned}$$

for fixed $k \in \mathbb {Z}$ and $\hat{b}_1, \ldots \hat{b}_k \in \mathbb {Z}[p^{-1}]$. We define infinitely many irrational numbers of the kind

$$\begin{aligned} \beta ^{(j)} = [0, b_1, \ldots , b_{n_j-1}, \overline{\hat{b}_1, \ldots , \hat{b}_k}]. \end{aligned}$$

By construction, the $\beta ^{(j)}$’s have the first $n_j + k_j\lambda _j$ partial quotients equal to the ones of $\alpha$, thus, by Lemma 1, it follows

$$\begin{aligned} | \alpha - \beta ^{(j)} |_p < |B_{n_j+k_j\lambda _j-1}|_p^{-2}. \end{aligned}$$

(14)

Moreover, we have

$$\begin{aligned} | \alpha - \beta ^{(j)} |_p > |B_{n_j+k-1}|_p^{-4\omega } \end{aligned}$$

(15)

with $\omega > 2$. Indeed, if $| \alpha - \beta ^{(j)} |_p \le |B_{n_j+k-1}|_p^{-4\omega }$, then by Lemma 3 and Remark 4, we have

$$\begin{aligned} | \alpha - \beta ^{(j)} |_p \le (2|B_{n_j+k-1}|_p^4)^{\varepsilon - \omega } \le C H(\beta ^{(i)})^{\varepsilon - \omega }, \end{aligned}$$

with $0< \varepsilon < \omega - 2$ and $C > 0$, in contradiction with Corollary 1. Thus, using (14) and (15), we obtain

$$\begin{aligned} |B_{n_j+k_j\lambda _j -1}|_p < |B_{n_j-k-1}|_p^{2\omega }. \end{aligned}$$

(16)

Since $|b_i|_p$ is bounded by D, we have $p \le |b_i|_p \le D$, for $i \ge 1$, from which we get $p^i \le |B_i|_p \le D^i$. Hence, in order that (16) holds, there exists $a > 0$ such that

$$\begin{aligned} \lambda _j < a + \left( \frac{1}{2}(C+1)\omega -1 \right) n_j. \end{aligned}$$

Considering that $Cn_j < \lambda _j$ and $C > 4 \frac{\log D}{\log p} - 1$, we get

$$\begin{aligned} \left( 1 - \frac{\omega }{2} \right) (C+1) + \delta < \frac{a}{n_j} \end{aligned}$$

where $\delta = 4 \frac{\log D}{\log p} - 1$, which can not be satisfied for each $\omega > 2$ and j sufficiently large. $\square$

Theorem 5

Let $(b_i)_{i\ge 1}$ be a sequence beginning with arbitrarily long palindromes and $|b_n|_p \ge p^{4}$, for all $n \gg 0$. Consider the Browkin continued fraction $\alpha = [0, b_1, b_2, \ldots ]$, then $\alpha$ is either transcendental or quadratic irrational.

Proof

Let n be a fixed natural number. Let $A_n/B_n$ be the nth convergent of $\alpha$, with $(A_i)$ and $(B_i)$ as in (3). By classical results on continued fractions, we have the following unique decomposition

$$\begin{aligned} M_n=\begin{pmatrix} B_n &{} B_{n-1}\\ A_n &{} A_{n-1}\\ \end{pmatrix} = \begin{pmatrix} b_1 &{} 1\\ 1 &{} 0\\ \end{pmatrix}\begin{pmatrix} b_2 &{} 1\\ 1 &{} 0\\ \end{pmatrix}\cdots \begin{pmatrix} b_n &{} 1\\ 1 &{} 0\\ \end{pmatrix}. \end{aligned}$$

Indeed, by definition, since $A_n=b_nA_{n-1}+A_{n-2}$ and $B_n=b_nB_{n-1}+B_{n-2}$, we have

$$\begin{aligned} M_n=M_{n-1} \begin{pmatrix} b_n &{} 1\\ 1 &{} 0\\ \end{pmatrix} \end{aligned}$$

and by a straightforward induction we obtain the required decomposition. Thus the matrix $M_n$ is symmetrical if and only if $(b_1,...,b_n)$ is palindromic. Assume this is the case, we immediately see that $A_n=B_{n-1}$. Then, by Lemma 1, we have

$$\begin{aligned}\left|\alpha -\frac{A_n}{B_n}\right|_p < \frac{1}{|B_n|_p^2}.\end{aligned}$$

Moreover, recalling that $A_n=B_{n-1}$ and $A_nB_{n+1}-A_{n-1}B_n=(-1)^n$, using Lemma 1, we obtain

$$\begin{aligned} \left|\alpha ^2 - \frac{A_{n-1}}{B_n}\right|_p&= \left|\alpha ^2 -\frac{A_{n-1}}{B_{n-1}}\frac{A_{n}}{B_n}\right|_p=\left|\left( \alpha + \frac{A_{n-1}}{B_{n-1}}\right) \left( \alpha -\frac{A_n}{B_n}\right) +\frac{(-1)^n\alpha }{B_nB_{n-1}} \right|_p \\ {}&\le \max \left\{ \left|\alpha + \frac{A_{n-1}}{B_{n-1}}\right|_p \left|\alpha -\frac{A_n}{B_n} \right|_p ; \frac{|\alpha |_p}{|B_n B_{n+1}|_p} \right\} \\&\le \max \left\{ \left|\frac{\alpha }{B_n^2}\right|_p ; \frac{|\alpha |_p}{|B_nB_{n-1}|_p} \right\} \end{aligned}$$

where the first equality comes from $A_n=B_{n-1}$ and the last one holds for n big enough. Considering that $|b_n|_p \ge p^{4}$ for $n \gg 0$, by the proof of Lemma 2 we know that $|A_n|_\infty < |A_n|_p^{1/4}$, $|A_{n-1}|_\infty < |A_{n-1}|_p^{1/4}$ and $|B_n|_\infty < |B_n|_p^{1/4}$, for $n \gg 0$. Therefore, since $|B_n|_p = \max (|A_n|_p, |A_{n-1}|_p, |B_n|_p)$, we conclude by Corollary 3. $\square$

Remark 5

The continued fraction expansion of a quadratic irrational can easily contain arbitrarily long palindromes, for example if it has a period of length two. On the other hand, recall that it is not known whether quadratic irrationals always have a periodic expansion (contrarily to the classical case). Hence, non-periodicity of $(b_i)_{i\ge 1}$ is not enough to ensure transcendence.

Example 2

Similarly to the archimedean case (see [4]), we introduce p-adic $Sturmian$ continued fractions of slope $\theta$ as an instance of p-adic continued fractions beginning with arbitrarily large palindromes. Consider a real $\theta \in (0,1)$ and two distinct $a,b\in \mathbb {Z}[1/p]$. Then we define

$$\begin{aligned} \sigma _\theta =[0,c_1,...] \end{aligned}$$

where

$$\begin{aligned}c_n= {\left\{ \begin{array}{ll} a,&{} \text { if } \lfloor (n+1)\theta \rfloor -\lfloor n\theta \rfloor =0 \\ b,&{} \text { if } \lfloor (n+1)\theta \rfloor -\lfloor n\theta \rfloor =1 \end{array}\right. }\end{aligned}$$

We also consider the p-adic $Thue$-$Morse$ continued fraction. A Thue-Morse sequence $(c_n)_n$ with values in a, b is defined by $c_n=a$ if the binary expansion of n has an even number of digits 1, and $c_n=b$ otherwise. For instance, the word $c_0...c_{4^n-1}$ is clearly a palindrome. Consider the p-adic number $\theta =[0,c_0,...,c_n]$ where $c_i\in \{a,b\}$. If the sequence of partial quotients $(c_n)_n$ is a Thue-Morse word, then we call $\theta$ a p-adic Thue-Morse continued fraction. By Theorem 5 we immediately obtain the transcendence of both the p-adic Sturmian continued fractions and the p-adic Thue-Morse continued fractions beginning with arbitrarily long palindromes.