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.

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 Reference Dirichlet4 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 [Reference Salem13, 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.