Every Salem number is a difference of two Pisot numbers

Abstract

In this note, we prove that every Salem number is expressible as a difference of two Pisot numbers. More precisely, we show that for each Salem number α of degree d, there are infinitely many positive integers n for which $\alpha^{2n-1}-\alpha^n+\alpha$ and $\alpha^{2n-1}-\alpha^n$ are both Pisot numbers of degree d and that the smallest such n is at most $6^{d/2-1}+1$. We also prove that every real positive algebraic number can be expressed as a quotient of two Pisot numbers. Earlier, Salem himself had proved that every Salem number can be written in this way.

1. Introduction

Recall that a Salem number is a real algebraic integer α > 1 whose conjugates over ${\mathbb Q}$ except for α itself all lie in the disc $|z| \leq 1$ with at least one conjugate lying on the boundary $|z|=1$. The Salem number α is reciprocal, so it has even degree $d \geq 4$ over ${\mathbb Q}$, the conjugate α ⁻¹ and d − 2 unimodular conjugates of the form ${\rm e}^{\pm {\rm i}\phi_j}$, $j=1,\dots,d/2-1$, where $0 \lt \phi_1 \lt \cdots \lt \phi_{d/2-1} \lt \pi$. A Pisot number is a real algebraic integer greater than 1 whose other conjugates over ${\mathbb Q}$ (if any) all lie in the open disc $|z| \lt 1$.

Various properties of Salem numbers have been investigated in [6–8, 12, 13, 15, 17] (see also a survey [16]), while their relations with Pisot numbers have been explored in, for example, [1, 2, 5, 9, 10, 18, 19]. For example, an old result of Salem [12] asserts that every Pisot number is a limit point of the set of Salem numbers. In [14], Siegel showed that the smallest Pisot number is the root $\theta=1.3247\dots$ of $x^3-x-1=0$, while the smallest Salem number is not known, and it is not even known whether the set of Salem numbers is bounded away from 1.

In [5], the author investigated various sumsets and difference sets involving Salem and Pisot numbers. In this note, we will prove the following new result in this direction.

Theorem 1. Every Salem number is expressible as a difference of two Pisot numbers.

More explicitly, we will show the following:

Theorem 2. For each Salem number α of degree $d \geq 4$, there exist infinitely many $n \in {\mathbb N}$ for which $\alpha^{2n-1}-\alpha^n+\alpha$ and $\alpha^{2n-1}-\alpha^n$ are both Pisot numbers of degree d. The smallest such n is at most $6^{d/2-1}+1$.

In [12, p. 69] (see also [13, p. 35]), Salem himself proved that every Salem number is expressible as a quotient of two Pisot numbers. On the other hand, the author showed that every positive algebraic number is a quotient of two Mahler measures [4, Theorem 1]. Recall that the Mahler measure $M(\alpha)$ of a non-zero algebraic number α is the modulus of the product of its conjugates lying outside the unit circle and the leading coefficient of its minimal polynomial in ${\mathbb Z}[x]$. Thus, for a real algebraic number α > 1, we have $M(\alpha) \geq \alpha$ with equality if and only if α is a Salem or a Pisot number. Therefore, the following theorem generalizes both these results.

Theorem 3. Every real positive algebraic number α of degree d is expressible as a quotient of two Pisot numbers of degree d from the field ${\mathbb Q}(\alpha)$.

In the next section, we will recall a few simple results, which will be used in the proofs. Then, in § 3, we will prove Theorems 2 and 3. Evidently, Theorem 2 implies Theorem 1.

2. Auxiliary results

In the proof of Theorem 2, we will use the next version of Dirichlet’s approximation theorem [4] (see, e.g., [11, p. 423]).

Lemma 4. Let $\lambda_1, \lambda_2, \ldots, \lambda_N$ be real numbers. Then, for each Q > 1, there is a positive integer $q \leq Q$ such that

\begin{equation*} \|\lambda_j q\| \lt Q^{-1/N} \end{equation*}

for $j=1,2,\ldots,N$.

Throughout, $\|y\|$ stands for the distance between $y \in {\mathbb R}$ and the nearest integer.

Let α be a Salem number of degree $d \geq 4$ with conjugates α ⁻¹ and ${\rm e}^{\pm {\rm i} \phi_j}$, $j=1,\ldots,N$, over ${\mathbb Q}$, where $0 \lt \phi_1 \lt \cdots \lt \phi_N \lt \pi$ and $d=2N+2$. In [13, p. 32], Salem showed that the numbers $\pi,\phi_1,\ldots,\phi_N$ are linearly independent over ${\mathbb Q}$ (the argument is attributed to Pisot). In particular, Salem’s result implies that

Lemma 5. The numbers $\phi_j/\pi$, $j=1,\ldots,N$, are all irrational.

Note that in case $\phi_j/\pi \in {\mathbb Q}$, the conjugate ${\rm e}^{{\rm i}\phi_j}$ of a Salem number must be a root of unity, which is impossible, because all the conjugates of a root of unity over ${\mathbb Q}$ must be roots of unity themselves, but Salem number is not a root of unity. This also implies Lemma 5.

Next, we record the following observation:

Lemma 6. Let α be a real algebraic number of degree $d \geq 2$ with conjugates $\alpha_1=\alpha,\alpha_2,\ldots,\alpha_d$ over ${\mathbb Q}$, and let f be a non-constant polynomial with rational coefficients such that $f(\alpha) \gt 0$ and $|f(\alpha_j)| \lt 1$ for $j=2,\ldots,d$. If $f(\alpha) \in {\mathbb Q}(\alpha)$ is an algebraic integer, then it is a Pisot number of degree d.

Proof. Note that

\begin{equation*}f(\alpha),f(\alpha_2),\ldots,f(\alpha_d)\end{equation*}

is the list of conjugates of an algebraic integer $f(\alpha)$ over ${\mathbb Q}$, possibly repeated several times. In particular, this implies that $f(\alpha_j) \ne 0$ for $j=2,\ldots,d$. Furthermore, $f(\alpha) \geq 1$, since otherwise $0 \lt f(\alpha) \lt 1$, and hence there is a non-zero algebraic integer $f(\alpha)$ with all conjugates in $|z| \lt 1$, including $f(\alpha)$. But then the modulus of the product of the conjugates of $f(\alpha)$ must be smaller than 1, which is impossible. Also, if $f(\alpha) = 1$, then its conjugates $f(\alpha_j)$, $j=2,\dots,d$, are all equal to 1, which is not the case. Consequently, $f(\alpha) \gt 1$. Since $f(\alpha)$ is the only conjugate of $f(\alpha)$ outside the unit circle, all $f(\alpha_j)$, $j=2,\dots,d$, lying in $|z| \lt 1$ must be distinct, whence the result.

3. Proofs of Theorems 2 and 3

Proof of Theorem 2.

Let α be a Salem number of degree $d \geq 4$ with conjugates $\alpha_2=\alpha^{-1}$ and $\{\alpha_3,\ldots,\alpha_d\}=\{{\rm e}^{\pm {\rm i} \phi_1},\ldots,{\rm e}^{\pm {\rm i} \phi_N}\}$, where $N=d/2-1$. Applying Lemma 4 to the N irrational numbers $\lambda_1=\phi_1/(2\pi), \ldots, \lambda_N=\phi_N/(2\pi)$ (see Lemma 5), we derive that for any Q > 1, there is an integer q in the range $1 \leq q \leq Q$ for which

(1)\begin{equation} 0 \lt \|q\phi_j/(2\pi)\| \lt Q^{-1/N}=Q^{-2/(d-2)}. \end{equation}

Put $n=q+1$ and consider the numbers

(2)\begin{equation}\beta=\alpha^{2n-1}-\alpha^n+\alpha \quad \text{and} \quad \gamma=\alpha^{2n-1}-\alpha^n.\end{equation}

We will show that β and γ are both Pisot numbers of degree d in the field ${\mathbb Q}(\alpha)$, provided that

(3)\begin{equation} Q^{-2/(d-2)} \leq \frac{1}{6}, \end{equation}

that is, $Q \geq 6^{d/2-1}$. Of course, by letting $Q \to \infty$ in Equation (1), we will produce infinitely many q satisfying Equation (1), and so infinitely many $n \in {\mathbb N}$ for which $\beta, \gamma \in {\mathbb Q}(\alpha)$ defined in Equation (2) are both Pisot numbers of degree d.

We begin with the number $\gamma=f(\alpha)$, where $f(x)=x^{2n-1}-x^n$ due to Equation (2). First, $\gamma=f(\alpha) \gt 0$ is an algebraic integer lying in the field ${\mathbb Q}(\alpha)$. In order to apply Lemma 6, we need to show that $|f(\alpha_j)| \lt 1$ for $j=2,\dots,d$.

Observe that, by Equation (2),

\begin{equation*}f(\alpha_2)=f(\alpha^{-1})=\alpha^{-2n+1}-\alpha^{-n}.\end{equation*}

It is clear that $-1 \lt \alpha^{-2n+1}-\alpha^{-n} \lt 0$ because α > 1. So $f(\alpha_2)$ lies in $|z| \lt 1$. Next, fix a conjugate $\alpha^{\prime}={\rm e}^{\pm {\rm i} \phi_j}$ of α. It remains to check that for any choice of the sign ± the number

\begin{equation*}f(\alpha^{\prime})=(\alpha^{\prime})^{2n-1}-(\alpha^{\prime})^{n}={\rm e}^{\pm {\rm i} \phi_j n}({\rm e}^{\pm {\rm i} \phi_j (n-1)}-1)={\rm e}^{\pm {\rm i} \phi_j (q+1)}({\rm e}^{\pm {\rm i} \phi_j q}-1)\end{equation*}

lies in $|z| \lt 1$. In view of $|f(\alpha^{\prime})|=2|\,\sin(q\phi_j/2)|$, this is equivalent to $|\sin(q \phi_j/2)| \lt 1/2$. This happens if and only if

\begin{equation*}|q\phi_j/2-\pi k| \lt \pi/6\end{equation*}

for some $k \in {\mathbb Z}$ or, equivalently, $\|q \phi_j/(2\pi)\| \lt 1/6$, which is indeed the case by Equations (1) and (3). This completes our verification. Therefore, $\gamma=f(\alpha) \in {\mathbb Q}(\alpha)$ is a Pisot number of degree d by Lemma 6.

Now, let us consider the number $\beta=f(\alpha)$ defined in Equation (2), where $f(x)=x^{2n-1}-x^n+x$. It is clear that $f(\alpha) \gt \alpha \gt 1$ is an algebraic integer. This time, we find that

\begin{equation*}f(\alpha_2)=f(\alpha^{-1})=\alpha^{-2n+1}-\alpha^{-n}+\alpha^{-1}.\end{equation*}

In view of α > 1 and $n \geq 2$, we obtain $0 \lt \alpha^{-2n+1}-\alpha^{-n}+\alpha^{-1} \lt 1$, so $f(\alpha_2)$ is in $|z| \lt 1$. Next, as above, fix a conjugate $\alpha^{\prime}={\rm e}^{\pm {\rm i} \phi_j}$ of α. This time, we need to show that for any choice of the sign ± the number

\begin{align*}f(\alpha^{\prime}) &=(\alpha^{\prime})^{2n-1}-(\alpha^{\prime})^{n}+\alpha^{\prime}={\rm e}^{\pm {\rm i} \phi_j n}\left({\rm e}^{\pm {\rm i} \phi_j (n-1)}-1+{\rm e}^{\mp {\rm i} \phi_j (n-1)}\right)\\ &= e^{\pm i \phi_j (q+1)}(2\cos(q\phi_j)-1) \end{align*}

lies in the open disc $|z| \lt 1$. This is true if and only if $0 \lt \cos(q\phi_j) \lt 1$. The latter inequalities hold whenever

\begin{equation*}0 \lt |q\phi_j-2\pi k| \lt \pi/2\end{equation*}

for some $k \in {\mathbb Z}$ or, equivalently, $0 \lt \|q \phi_j/(2\pi)\| \lt 1/4$. This is true by Equations (1), (3) and $1/6 \lt 1/4$. As before, by Lemma 6, we conclude that $\beta=f(\alpha) \gt 1$ is a Pisot number of degree d.

Finally, selecting $Q=6^{d/2-1}$, by Equations (1) and (3), we see that the smallest $q \in {\mathbb N}$ for which Equation (1) is true satisfies $1 \leq q \leq 6^{d/2-1}$. This completes the proof of the last assertion of the theorem because the integer $n=q+1$ is in the range $2 \leq n \leq 6^{d/2-1}+1$.

Proof of Theorem 3

Let α be a positive algebraic number of degree d over ${\mathbb Q}$ with conjugates $\alpha_1=\alpha,\alpha_2,\ldots,\alpha_d$. The claim is trivial for d = 1, since every integer $k \geq 2$ is a Pisot number and every positive rational number is a quotient of two such numbers. Assume that $d \geq 2$, and let m be a positive integer for which $m\alpha$ is an algebraic integer.

Fix a positive number u < 1 satisfying

(4)\begin{equation} m u \max(1,|\alpha_2|,\dots,|\alpha_d|) \lt 1, \end{equation}

and a positive number v > 1 satisfying

(5)\begin{equation} m v\alpha \gt 1. \end{equation}

Select a Pisot number $\beta \in {\mathbb Q}(\alpha)$ of degree d (see Theorem 2 in [13, p. 3]). A natural power of β is also a Pisot number of degree d, so by replacing β by its large power if necessary, we can assume that $\beta \gt v$ and that the other d − 1 conjugates of β over ${\mathbb Q}$ are all in $|z| \lt u$.

Write this β in the form $\beta=f(\alpha)$, where f is a non-constant polynomial of degree at most d − 1 with rational coefficients. Then, the numbers $\beta_j=f(\alpha_j)$, $j=1,\ldots,d$, are the conjugates of $\beta=\beta_1$ over ${\mathbb Q}$. Recall that, by the choice of β, we have

\begin{equation*}\beta=f(\alpha) \gt v \quad \text{and} \quad |\beta_j|=|f(\alpha_j)| \lt u \quad \text{for} \ j=2,\dots,d.\end{equation*}

We claim that under assumption on the constants $u \in (0,1)$ as in Equation (4) and v > 1 as in Equation (5), the numbers $m \alpha\beta \in {\mathbb Q}(\alpha)$ and $m\beta \in {\mathbb Q}(\alpha)$ are both Pisot numbers of degree d. This will complete our proof, since their quotient is α.

First, $m\beta$ is a Pisot number, since it is an algebraic integer greater than m > 1, whose other conjugates $m\beta_j$, $j=2,\ldots,d$, all lie in $|z| \lt 1$ by $|\beta_j| \lt u$ and Equation (4). Of course, $m\beta \in {\mathbb Q}(\alpha)$ is of degree d over ${\mathbb Q}$, since so is β.

Second, the number $m\alpha\beta=m\alpha f(\alpha) \in {\mathbb Q}(\alpha)$ is a positive algebraic integer, since so are $m\alpha$ and β. It is greater than 1 by $\beta \gt v$ and Equation (5). Its other conjugates are $m\alpha_j f(\alpha_j)=m \alpha_j \beta_j$, $j=2,\dots,d$. They are all in $|z| \lt 1$ due to $|\beta_j| \lt u$ and Equation (4). Hence, $m \alpha f(\alpha) \in {\mathbb Q}(\alpha)$ is a Pisot number of degree d over ${\mathbb Q}$ by Lemma 6 applied to the polynomial $mxf(x) \in {\mathbb Q}[x]$.

Therefore, $m \alpha\beta \in {\mathbb Q}(\alpha)$ and $m\beta \in {\mathbb Q}(\alpha)$ indeed are both Pisot numbers of degree d, which finishes the proof.