2. Proof of Proposition 1

To show that $u \geq \ell$ , we will first investigate the case when $\ell =1$ . For $d \geq 2$ , we consider the polynomial

\begin{equation*}f(x)=x^d-H(x^{d-1}+\cdots +1) \in {\mathbb Z}[x].\end{equation*}

We claim that $f$ is irreducible in ${\mathbb Z}[x]$ for each $H \in {\mathbb N}$ .

Indeed, since $f(H)\lt 0$ and $f(H+1)\gt 0$ , $f$ has a root $\beta$ in the interval $(H,H+1)$ . It remains to show that its other $d-1$ roots all lie in $|z|\lt 1$ . (If so, then $f$ cannot be reducible, since the product of its several roots excluding $\beta$ is of modulus less than $1$ .) Consider the polynomial

\begin{equation*}g(x)=f(x)(x-1)=x^{d+1}-(H+1)x^d+H.\end{equation*}

Assume that $f$ has a root $\gamma \ne \beta$ satisfying $|\gamma |\gt 1$ . Select a small positive $\varepsilon \lt \min (|\gamma |-1,\beta -1)$ for which $(H+1)(1+\varepsilon )^d\gt H+(1+\varepsilon )^{d+1}$ , which is possible due to $(H+1)d\gt d+1$ . Then, by Rouché’s theorem applied to the disc with contour $|z|=1+\varepsilon$ , $g$ has the same number of roots in $|z|\lt 1+\varepsilon$ as $(H+1)x^d$ does. Thus, $g$ has $d$ roots in $|z|\lt 1+\varepsilon$ , which is not true, since the roots $\gamma , \beta$ of $g$ of degree $d+1$ have moduli larger than $1+\varepsilon$ . Consequently, all roots of $f$ other than $\beta$ must lie in the disc $|z| \leq 1$ . Assume that some root $\gamma$ of $f$ has modulus $1$ . From $(H+1)\gamma ^d=H+\gamma ^{d+1}$ , by the moduli consideration, we get $H+1=|H+\gamma ^{d+1}|$ , which yields $\gamma ^{d+1}=1$ . This implies $\gamma ^d=(H+\gamma ^{d+1})/{(H+1)=1}$ , and hence, $\gamma =\gamma ^{d+1}/\gamma ^d=1$ . However, $\gamma =1$ is not a root of $f$ , because $f(1)\lt 0$ , a contradiction. Therefore, $f$ has $d-1$ roots in $|z|\lt 1$ as claimed.

From the irreducibility of $f$ , it follows that $\alpha =\beta ^{-1}$ is an algebraic number of degree $d$ and height $H$ . From $\beta \gt H$ , we get $0\lt \alpha \lt 1/H$ . Selecting $r=0$ , we find that $|\alpha -r|=\alpha \lt 1/H$ . This shows that (4) cannot hold if $u\lt 1$ , since we can select $H$ arbitrarily large. Hence, $u \geq 1$ for each $(u,v)$ satisfying (4).

Assume now that $\ell \geq 2$ . The next lemma is a part of [Reference Bugeaud4, Theorem 2.11]:

Lemma 5. Let $d\gt \ell \geq 2$ be integers, and let $\beta$ be a real algebraic number of degree $\ell$ . Then there is a constant $c_3=c_3(\beta ,d)$ and infinitely many real algebraic integers $\alpha$ of degree $d$ such that

\begin{equation*}|\alpha -\beta |\lt \frac {c_3}{H(\alpha )^{\ell }}.\end{equation*}

As above, this shows that (4) cannot hold if $u\lt \ell$ , since we can select $H(\alpha )$ arbitrarily large. Hence, $u \geq \ell$ for each $(u,v)$ satisfying (4). This completes the proof of the first inequality in (5).

Now, we will prove the second inequality in (5), namely, $v \geq \ell +1$ . For this purpose, we can use the following result of Bugeaud [Reference Bugeaud5]:

Lemma 6. Let $n\gt \ell \geq 1$ be integers, and let $\alpha$ be a real algebraic number of degree $n$ . Then there is a positive constant $c=c(\alpha )$ and infinitely many algebraic numbers $\beta$ of degree $\ell$ such that

\begin{equation*}|\alpha -\beta |\lt H(\beta )^{-\ell -1+c(\log \log (3H(\beta ))^{-1/(2\ell +6)}}.\end{equation*}

Assume that the pair $(u,v)$ is suitable with respect to $(d,\ell )$ . Then, by (4), for any fixed real algebraic number $\alpha$ of degree $d$ and infinitely many $\beta$ of degree $\ell$ as indicated in Lemma 6, we deduce

(15) \begin{equation} \frac {c_1}{H(\alpha )^u} \leq H(\beta )^v|\alpha -\beta |\lt H(\beta )^{v-\ell -1+c(\log \log (3H(\beta ))^{-1/(2\ell +6)}}. \end{equation}

Note that $c(\log \log (3H(\beta ))^{-1/(2\ell +6)}$ tends to zero as $H(\beta ) \to \infty$ . So, if $v\lt \ell +1$ , then the exponent on the right-hand side of (15) for $H(\beta )$ tends to the negative number $v-\ell -1$ as $H(\beta ) \to \infty$ . Consequently, the right-hand side of (15) tends to zero as $H(\beta ) \to \infty$ , which is not the case. Hence $v \geq \ell +1$ .

3. Proof of Theorem4

Fix an integer $k\gt 1$ that is not a perfect square and consider the polynomial

(16) \begin{equation} P(x,y)=x^3-y(x^2-x)-k. \end{equation}

In particular, for $y=\sqrt {k}$ , the only real root of the polynomial $P(x,\sqrt {k})$ is $\sqrt {k}$ , while its two complex roots are $\pm i \sqrt [4]{k}$ . Since the roots of $P(x,y)$ as a polynomial in $x$ can be found in terms of the Cardano formulas in parameters $y,k$ , the real root $\alpha (\varepsilon )$ of $P(x,\sqrt {k}-\varepsilon )$ for real $\varepsilon$ of sufficiently small modulus can be expanded as

(17) \begin{equation} \alpha (\varepsilon )=\alpha =\sqrt {k}+B_1\varepsilon +B_2\varepsilon ^2+B_3\varepsilon ^3+\ldots , \end{equation}

with some real numbers $B_1,B_2,B_3,\ldots$ depending on $k$ .

By squaring this expression, we get

\begin{equation*}\alpha ^2=k+2\sqrt {k}B_1 \varepsilon + (2\sqrt {k}B_2+B_1^2)\varepsilon ^2+\ldots .\end{equation*}

Multiplying $\alpha$ and $\alpha ^2$ , we further deduce

\begin{equation*}\alpha ^3=k\sqrt {k}+3kB_1 \varepsilon +3(\sqrt {k}B_1^2+kB_2)\varepsilon ^2+\ldots .\end{equation*}

Since $P(\alpha ,\sqrt {k}-\varepsilon )=\alpha ^3-(\sqrt {k}-\varepsilon )(\alpha ^2-\alpha )-k=0$ , collecting the terms for $\varepsilon$ and $\varepsilon ^2$ in equality $\alpha ^3-k=(\sqrt {k}-\varepsilon )(\alpha ^2-\alpha )$ , we obtain

\begin{equation*}3kB_1=-(k-\sqrt {k})+B_1(2k-\sqrt {k})\end{equation*}

and

\begin{equation*}3(\sqrt {k}B_1^2+kB_2)=-(2\sqrt {k}-1)B_1+(2k-\sqrt {k})B_2+\sqrt {k}B_1^2.\end{equation*}

The first equality of the two above gives

(18) \begin{equation} B_1=-\frac {\sqrt {k}-1}{\sqrt {k}+1}, \end{equation}

while from the second, we get

(19) \begin{equation} (k+\sqrt {k})B_2=-(2\sqrt {k}-1)B_1-2\sqrt {k}B_1^2. \end{equation}

Inserting $B_1$ from (18) into (19) multiplied by $(\sqrt {k}+1)^2$ , we obtain

(20) \begin{equation} B_2= \frac {3k-4\sqrt {k}+1}{\sqrt {k}(\sqrt {k}+1)^3}, \end{equation}

which is equal to $\xi _k$ by (14). Consequently, for real $\varepsilon$ of sufficiently small modulus and for $\delta \gt 0$ , by (17), (18), (20), we derive that

\begin{equation*}\alpha (\varepsilon )=\sqrt {k}-\frac {\sqrt {k}-1}{\sqrt {k}+1} \varepsilon +B\varepsilon ^2,\end{equation*}

where $B=B(k,\varepsilon ) \in ((1-\delta )\xi _k,(1+\delta )\xi _k)$ . Using this formula for $\varepsilon =\varepsilon _n$ as in (11) and for the root $\alpha _n=\alpha (\varepsilon _n)$ of $P_n$ , we deduce (13) for $n$ large enough, because, by (11), (12) and (16),

\begin{equation*}q_nP(x,\sqrt {k}-\varepsilon _n)=q_nP(x,p_n/q_n)=q_nx^3-p_n(x^2-x)-kq_n=P_n(x).\end{equation*}

4. Proof of Theorem3

We will apply Theorem4 with $k=2$ , integer $n \geq n_0$ and $\alpha _n$ being the real root of $P_n$ defined in (12). Note that the coefficients of $P_n$ are coprime, since so are the integers $p_n$ and $q_n$ . If $P_n$ were reducible, then it would have a rational root, say $\mu _n \in {\mathbb Q}$ . Since two other roots of $P_n$ are complex for $n$ large enough, this root $\mu _n$ must be equal to $\alpha _n$ . So $\alpha _n\gt 0$ is either rational (which will be ruled out below) or $P_n$ is irreducible in ${\mathbb Z}[x]$ . In the latter case, $\alpha _n$ is the cubic algebraic number with height

\begin{equation*}H(\alpha _n)=2q_n,\end{equation*}

because $p_n\lt 2q_n$ . Note that $p_0/q_0=1$ , $p_1/q_1=3/2$ , $p_2/q_2=7/5$ , etc. In fact, since $\sqrt {2}=[1,2,2,2,\ldots ]$ , from the formulas

\begin{equation*}p_{n}=2p_{n-1}+p_{n-2} \quad \text{and} \quad q_n=2q_{n-1}+q_{n-2}\end{equation*}

for $n=2,3,4,\ldots\,$ , where $p_0=q_0=1$ , $p_1=3$ , $q_1=2$ [Reference Steuding16, p. 38]), we obtain

(21) \begin{equation} p_n=\frac {(1+\sqrt {2})^{n+1}+(1-\sqrt {2})^{n+1}}{2} \end{equation}

and

(22) \begin{equation} q_n=\frac {(1+\sqrt {2})^{n+1}-(1-\sqrt {2})^{n+1}}{2\sqrt {2}}. \end{equation}

Multiplying (22) by $\sqrt {2}$ and subtracting (21), we get

(23) \begin{equation} q_n \sqrt {2}-p_n=-(1-\sqrt {2})^{n+1}. \end{equation}

Consider the rational number

(24) \begin{equation} r_n=\frac {2p_n+2q_n}{p_n+2q_n}. \end{equation}

By the above, we see that the number $p_n$ is odd, so the denominator of $r_n$ is odd. The integers $2p_n+2q_n$ and $p_n+2q_n$ are therefore coprime, since so are $p_n+q_n$ and $p_n+2q_n$ . Hence

\begin{equation*}H(r_n)=2(p_n+q_n)\gt 2q_n=H(\alpha _n).\end{equation*}

Fix a small number $\epsilon \gt 0$ . For each integer $H\gt 10$ , take the largest $n \in {\mathbb N}$ for which

(25) \begin{equation} 2(p_n+q_n)\lt H. \end{equation}

Then $H(\alpha _n)\lt H(r_n)\lt H$ . We claim that

(26) \begin{equation} H(\alpha _n)=2q_n\gt (1-\epsilon )(\sqrt {2}-1)^2 H \end{equation}

for $H$ large enough.

Indeed, if (26) does not hold, that is, $2q_n \leq (1-\epsilon )(\sqrt {2}-1)^2H$ , then for $H$ (and so $n$ ) large enough, by (21) and (22), we have

\begin{equation*}2p_n \lt 2(1+\epsilon )\sqrt {2} q_n \leq (1-\epsilon ^2) (\sqrt {2}-1)^2 \sqrt {2} H.\end{equation*}

Consequently,

\begin{equation*}2p_n+2q_n \lt (1-\epsilon ^2) (\sqrt {2}-1)^2 \sqrt {2} H +(1-\epsilon )(\sqrt {2}-1)^2 H\lt (1-\epsilon ^2)(\sqrt {2}-1)H.\end{equation*}

But then, since $n$ is large enough, using (21) and (22), we find that

\begin{equation*}2p_{n+1}+2q_{n+1}\lt (1+\epsilon ^2)(1+\sqrt {2}) (2p_n+2q_n)\lt (1-\epsilon ^4) H\lt H,\end{equation*}

contrary to the choice of $n$ in (25). Since $(\sqrt {2}-1)^2=0.171\ldots \gt 1/6$ , from (25) and (26), we deduce (7) for the pair $(\alpha ,r)=(\alpha _n,r_n)$ .

It remains to prove (8) for the same pair. By (11), (13) with $k=2$ and (24), we deduce

\begin{align*} \alpha _n-r_n &=\sqrt {2}-\frac {\sqrt {2}-1}{\sqrt {2}+1}\varepsilon _n+C_n\varepsilon _n^2-\frac {2p_n/q_n+2}{p_n/q_n+2} \\[3pt]&= \sqrt {2}-(3-2\sqrt {2})\varepsilon _n+C_n\varepsilon _n^2-\frac {2(\sqrt {2}-\varepsilon _n)+2}{2+\sqrt {2}-\varepsilon _n}, \end{align*}

where $(1-\delta )\xi _2 \lt C_n\lt (1+\delta )\xi _2$ . Therefore, the product

\begin{equation*}(\alpha _n-r_n-C_n\varepsilon _n^2)(2+\sqrt {2}-\varepsilon _n)\end{equation*}

is equal to

\begin{equation*}\sqrt {2}(2+\sqrt {2}-\varepsilon _n)-(3-2\sqrt {2})\varepsilon _n (2+\sqrt {2}-\varepsilon _n)-2\sqrt {2}-2+2\varepsilon _n=(\sqrt {2}-1)^2\varepsilon _n^2.\end{equation*}

(The choice of $r_n$ in (24) is such that here the constant term and the term with $\varepsilon _n$ both cancel out.) For fixed $\delta \gt 0$ and $n$ large enough, this implies

(27) \begin{equation} (1-\delta )\frac {(\sqrt {2}-1)^3}{\sqrt {2}}\varepsilon _n^2\lt \alpha _n-r_n-C_n\varepsilon _n^2\lt (1+\delta )\frac {(\sqrt {2}-1)^3}{\sqrt {2}}\varepsilon _n^2. \end{equation}

By (14), we have

\begin{equation*}\xi _2=\frac {3\cdot 2 -4 \sqrt {2}+1}{\sqrt {2}(\sqrt {2}+1)^3}=\frac {(3\sqrt {2}-1)(\sqrt {2}-1)}{\sqrt {2}(\sqrt {2}+1)^3}=(3-1/\sqrt {2})(\sqrt {2}-1)^4,\end{equation*}

and hence

\begin{equation*}\xi _2+ \frac {(\sqrt {2}-1)^3}{\sqrt {2}}=\bigg (3-\frac {1}{\sqrt {2}}\bigg )(\sqrt {2}-1)^4+\frac {(\sqrt {2}-1)^3}{\sqrt {2}}=4(\sqrt {2}-1)^4.\end{equation*}

Consequently,

\begin{align*} C_n\varepsilon _n^2+(1+\delta )\frac {(\sqrt {2}-1)^3}{\sqrt {2}}\varepsilon _n^2 & \lt (1+\delta )(\xi _2+\frac {(\sqrt {2}-1)^3}{\sqrt {2}})\varepsilon _n^2 \\[3pt]& =4(1+\delta )(\sqrt {2}-1)^4 \varepsilon _n^2. \end{align*}

It is also clear that

\begin{equation*}C_n\varepsilon _n^2+(1-\delta )\frac {(\sqrt {2}-1)^3}{\sqrt {2}}\varepsilon _n^2\gt (1-\delta ) \xi _2 \varepsilon _n^2+(1-\delta )\frac {(\sqrt {2}-1)^3}{\sqrt {2}}\varepsilon _n^2\gt 0.\end{equation*}

Therefore, from (27), it follows that

\begin{equation*}0\lt \alpha _n-r_n\lt 4(1+\delta )(\sqrt {2}-1)^4 \varepsilon _n^2.\end{equation*}

Observe that, by (23), we have

\begin{equation*}\varepsilon _n=\sqrt {2}-\frac {p_n}{q_n}=\frac {q_n \sqrt {2}-p_n}{q_n}=\frac {-(1-\sqrt {2})^{n+1}}{q_n}.\end{equation*}

Thus, selecting small $\delta =\epsilon \gt 0$ and using (22), we derive that

(28) \begin{equation} 0\lt \alpha _n-r_n \lt \frac {4(1+\epsilon )(\sqrt {2}-1)^{2n+6}}{q_n^2} \lt \frac {(1+2\epsilon )(\sqrt {2}-1)^{4}}{2q_n^4}. \end{equation}

Using (26), we see that the latter expression is

\begin{equation*}\frac {8(1+2\epsilon ) (\sqrt {2}-1)^4}{(2q_n)^4}\lt \frac {8(1+2\epsilon ) (\sqrt {2}-1)^4}{(1-\epsilon )^4 (\sqrt {2}-1)^8 H^4}\lt \frac {8(1+7\epsilon )(\sqrt {2}+1)^4}{H^4}.\end{equation*}

This is less than $272/H^4$ due to $8(\sqrt {2}+1)^4=271.764\ldots\,$ , which proves inequality (8) for the pair $(\alpha ,r)=(\alpha _n,r_n)$ . This completes the proof of Theorem3 under the assumption that $\alpha _n$ is irrational. However, if $\alpha _n$ were rational with minimal (linear) polynomial $Q_n \in {\mathbb Z}[x]$ , then, by the multiplicativity of Mahler’s measure and (10), we would get

\begin{equation*}H(\alpha _n)=M(\alpha _n) =M(Q_n)\leq M(P_n) \leq 2H(P_n) =4q_n.\end{equation*}

But then the denominators of the rational numbers $\alpha _n$ and $r_n$ (see (24)) do not exceed $4q_n$ , so $|\alpha _n-r_n|$ is either zero or greater than $1/(4q_n)^2$ , which contradicts (28) for $n$ large enough. The proof of Theorem3 is thus completed.

In fact, inserting $r_n$ as in (24) into the polynomial

\begin{equation*}P_n(x)=q_n(x^3-2)-p_n(x^2-x)\end{equation*}

and multiplying by $(p_n+2q_n)^3$ , we find that $(p_n+2q_n)^3 P_n(r_n)$ equals

\begin{equation*} q_n((2p_n+2q_n)^3-2(p_n+2q_n)^3) -p_n^2(2p_n+2q_n)(p_n+2q_n), \end{equation*}

which is $-2(p_n^2-2q_n^2)^2$ . Thus

(29) \begin{equation} (p_n+2q_n)^3P_n(r_n)=-2(p_n^2-2q_n^2)^2. \end{equation}

From (23), mapping $\sqrt {2}$ into $-\sqrt {2}$ , we get

\begin{equation*}-q_n \sqrt {2}-p_n=-(1+\sqrt {2})^{n+1}.\end{equation*}

Multiplying this equality with (23) yields $p_n^2-2q_n^2=({-}1)^{n+1}$ . Thus, by (29),

\begin{equation*}(p_n+2q_n)^3P_n(r_n)=-2.\end{equation*}

Using this identity, we can give an alternative proof of the theorem. Indeed, from

\begin{equation*}q_n(r_n-\alpha _n)(r_n-\beta _n)(r_n-\gamma _n)(p_n+2q_n)^3=-2,\end{equation*}

where $\alpha _n, \beta _n$ and $\gamma _n=\overline {\beta _n}$ are the roots of $P_n$ , it follows that

(30) \begin{equation} q_n^4 (\alpha _n-r_n)(r_n^2-(\beta _n+\gamma _n)r_n+\beta _n\gamma _n)(p_n/q_n+2)^3=2. \end{equation}

The root $\alpha _n$ of $P_n$ tends to $\sqrt {2}$ as $n \to \infty$ , and $\beta _n$ and $\gamma _n$ tend to $i\sqrt [4]{2}$ and to $-i\sqrt [4]{2}$ or vice versa. Also, $r_n \to \sqrt {2}$ as $n \to \infty$ and $p_n/q_n \to \sqrt {2}$ as $n \to \infty$ . Thus

\begin{equation*}(r_n^2-(\beta _n+\gamma _n)r_n+\beta _n \gamma _n)(p_n/q_n+2)^3 \to (2+\sqrt {2})^4 \quad \text{as} \quad n \to \infty .\end{equation*}

For each $\varepsilon \gt 0$ and $n$ large enough, this combined with (30) yields

\begin{equation*}2-\varepsilon \lt q_n^4 (\alpha _n-r_n) (2+\sqrt {2})^4 \lt 2+\varepsilon .\end{equation*}

Therefore,

\begin{equation*}0\lt \alpha _n-r_n\lt \frac {(1+\varepsilon /2)(\sqrt {2}-1)^4}{2q_n^4},\end{equation*}

which implies (28) with $\varepsilon =4\epsilon$ . This concludes the argument as above.