Proof. Let $f(x)$ be a polynomial of degree $d\geq 1$ with integer coefficients. Suppose that $f(a_p)\equiv 0\,\,\pmod {\,\,p}$ holds for all sufficiently large primes $p$ . Since $f(n)=O(n^d)$ for positive integers $n$ , it follows that $|f(a_p)|<p$ for all sufficiently large $p$ . The condition $f(a_p)\equiv 0\,\,\pmod {\,\,p}$ then implies that $f(a_p) = 0$ . This contradicts the fact that $a_p\to \infty $ .

Proof. Let $f(x)$ be a nonzero polynomial with integer coefficients, and suppose that $f(\alpha )=0$ . Then $f(b_p)\equiv 0 \,\pmod {\,p}$ holds for all but finitely many $p\in S$ . Let $R$ denote the set of all integer roots of $f(x)$ . It is clear that $R$ is a finite set. For sufficiently large real numbers $X$ , define

$$ \begin{align*} S(X):=\{p\leq X : p\in S,\ b_p\not\in R,\ p\mid f(b_p)\}. \end{align*} $$

Then, by the assumption, $\#S(X)\gg X^{\epsilon '}$ holds.

For each $p \in S(X)$ , by the assumption, we have $b_p = O(X^{\epsilon })$ . Hence, the number of integers of the form $f(b_p)$ , as $p$ ranges over $S(X)$ , is at most $O(X^{\epsilon })$ . For each such integer (note that $f(b_p)\neq 0$ since $b_p \not \in R$ ), the number of its prime divisors is at most $O(\log X)$ . This follows from the fact that, for a positive integer $n$ , the number of its prime divisors satisfies $\omega (n)=O(\log n)$ , together with the estimate $f(b_p)=O(X^{O(1)})$ . Since every element of $S(X)$ coincides with one of these prime divisors, it follows that $\#S(X)=O(X^{\epsilon }\log X)$ . Therefore, we have $X^{\epsilon '} \ll X^{\epsilon }\log X$ . Since $\epsilon <\epsilon '$ , this yields a contradiction for sufficiently large $X$ .

Proof. Let $f(x)$ be a nonzero polynomial with integer coefficients, and suppose that $f(\alpha ) = 0$ . Let $R$ denote the set of all integer roots of $f(x)$ . Choose a sufficiently large odd integer $r$ , coprime to $N$ , such that $b_{j,n} \not \in R$ for all $1 \le j \le d$ and all $n \ge r$ . For sufficiently large real numbers $X$ , define

$$ \begin{align*} P(X) := \bigg\{p \in [X,2X] : p \equiv 1\ (\mathrm{mod}\ r), \ p \equiv c\ (\mathrm{mod}\ N), \ \operatorname{\mathrm{ord}}_p(q) \mid \frac{p-1}{r}\bigg\}. \end{align*} $$

By our assumptions, for every $p\in P(X)$ , there exists an index $1\le j\le d$ such that $a_p=~b_{j,I_p(q)}\bmod {p}$ and

(2-1) $$ \begin{align} p \mid f(b_{j, I_p(q)}) \neq 0. \end{align} $$

Here the nonvanishing follows from the inequality $I_p(q)=(p-1)/\operatorname {\mathrm {ord}}_p(q)\ge r$ and the choice of $r$ .

We first prove that

$$ \begin{align*} \#P(X)\gg\frac{X}{\log X}. \end{align*} $$

Write $q=q_0^k$ , where $k$ is a positive integer and $q_0$ is not a perfect power. Set $m:= r/\gcd (r,k)$ and $K:= \mathbb {Q}(\zeta _{rN},\!\!\sqrt [m]{q_0})$ . Here, $\zeta _{rN}$ is a primitive $rN\,\mathrm{th}$ root of unity. By the Chinese remainder theorem, we may choose an integer $a$ satisfying $a\equiv 1 \,\pmod {\,r}$ and $a\equiv c \,\pmod {\,N}$ . Let $\sigma _a\in \operatorname {\mathrm {Gal}}(\mathbb {Q}(\zeta _{rN})/\mathbb {Q})$ be the element determined by $\sigma _a(\zeta _{rN})=~\zeta _{rN}^a$ . Let $p$ be a prime unramified in $K$ , and let $\mathfrak {P}$ be any prime ideal of $\mathcal {O}_K$ lying above $p$ . Write $\varphi _{\mathfrak {P}}\in \operatorname {\mathrm {Gal}}(K/\mathbb {Q})$ for the Frobenius element at $\mathfrak {P}$ . Under the assumption $p\equiv 1 \,\pmod {\,r}$ , the condition $\operatorname {\mathrm {ord}}_p(q)\mid ({p-1})/{r}$ is equivalent to $q^{({p-1})/{r}}\equiv 1\,\pmod {\,p}$ , which is equivalent to $q_0^{({p-1})/{m}}\equiv 1\,\pmod {\,p}$ , and this is further equivalent to $\varphi _{\mathfrak {P}}(\!\!\sqrt [m]{q_0})=\sqrt [m]{q_0}$ . Therefore, for sufficiently large $X$ , we have

$$ \begin{align*} P(X)=\{p\in [X,2X] : \varphi_{\mathfrak{P}}|_{\mathbb{Q}(\zeta_{rN})}=\sigma_a, \ \varphi_{\mathfrak{P}}(\!\!\sqrt[m]{q_0})=\sqrt[m]{q_0}\}. \end{align*} $$

Since $r$ is odd, $m$ is also odd, and $\mathbb {Q}(\zeta _{rN})\cap \mathbb {Q}(\!\!\sqrt [m]{q_0})=\mathbb {Q}$ . It follows that $x^m-q_0$ is irreducible over $\mathbb {Q}(\zeta _{rN})$ . Therefore,

$$ \begin{align*} \{\tau\in\operatorname{\mathrm{Gal}}(K/\mathbb{Q}) : \tau|_{\mathbb{Q}(\zeta_{rN})}=\sigma_a,\ \tau(\!\!\sqrt[m]{q_0})=\sqrt[m]{q_0}\} \end{align*} $$

is nonempty, and in fact consists of a single element. Hence, the Chebotarev density theorem is applicable.

On the other hand, we decompose $P(X)$ into three subsets as follows, and estimate each of them separately:

$$ \begin{align*} P_1(X) &:= \{p \in P(X) : \operatorname{\mathrm{ord}}_p(q) \le \sqrt{X}/\log X\},\\ P_2(X) &:= \{p \in P(X) : I_p(q) \le \sqrt{X}/\log X\},\\ P_3(X) &:= \{p \in P(X) : \operatorname{\mathrm{ord}}_p(q), I_p(q)> \sqrt{X}/\log X\}. \end{align*} $$

First, for $p \in P_1(X)$ , we have $p \mid (q^{\operatorname {\mathrm {ord}}_p(q)} - 1)$ , and hence

$$ \begin{align*} \prod_{p \in P_1(X)} p \le \prod_{n \le \sqrt{X}/\log X} (q^n-1) \le \exp \bigg(\log q \sum_{n \le \sqrt{X}/\log X} n \bigg) = \exp\bigg(O \bigg(\frac{X}{(\log X)^2}\bigg)\bigg). \end{align*} $$

Note that this inequality argument does not hold when $q=1$ . Since $p\ge X$ , it follows that

$$ \begin{align*} \#P_1(X)\ll\frac{X}{(\log X)^3}. \end{align*} $$

Second, the number of integers of the form $f(b_{j, I_p(q)})$ , as $p$ ranges over $P_2(X)$ , is at most $O(\!\sqrt {X}/\log X)$ . By the same argument as in the proof of Lemma 2.5, the number of prime divisors of each $f(b_{j, I_p(q)})$ is at most $O(\log |b_{j, I_p(q)}|) = O(\!\sqrt {X}/\log X)$ . Hence, by (2-1), we obtain

$$ \begin{align*} \#P_2(X)\ll\frac{X}{(\log X)^2}. \end{align*} $$

We estimate $\#P_3(X)$ by using Ford’s result. We use the notation $H(x,y,z)$ and $H(x,y,z;P_{\lambda })$ from [Reference Ford9]. If $p\in P_3(X)$ , then $\operatorname {\mathrm {ord}}_p(q)\mid p-1$ and

$$ \begin{align*} \frac{\sqrt{X}}{\log X}<\operatorname{\mathrm{ord}}_p(q)\leq 2\sqrt{X}\log X. \end{align*} $$

Therefore,

$$ \begin{align*} \#P_3(X)\leq H\bigg(2X,\frac{\sqrt{X}}{\log X}, 2\sqrt{X}\log X; P_{-1}\bigg). \end{align*} $$

By [Reference Ford9, Theorem 6] we have

$$ \begin{align*} H\bigg(2X,\frac{\sqrt{X}}{\log X}, 2\sqrt{X}\log X; P_{-1}\bigg)\ll \frac{H\bigg(2X,\frac{\sqrt{X}}{\log X}, 2\sqrt{X}\log X\bigg)}{\log X}, \end{align*} $$

and by [Reference Ford9, Theorem 1(v)] we have

$$ \begin{align*} H\bigg(2X,\frac{\sqrt{X}}{\log X}, 2\sqrt{X}\log X\bigg)\ll X\cdot u^{\delta}\bigg(\log\frac{2}{u}\bigg)^{-\frac{3}{2}}, \end{align*} $$

where

$$ \begin{align*} u=\frac{\log z}{\log y}-1\sim \frac{4\log\log X}{\log X} \end{align*} $$

and $\delta = 1-(1+\log \log 2)/\log 2=0.08607\ldots $ . Therefore, we have

$$ \begin{align*} \#P_3(X)\ll\frac{X(\log\log X)^{O(1)}}{(\log X)^{1+\delta}}. \end{align*} $$

Combining the above estimates, we obtain

$$ \begin{align*} \frac{X}{\log X}\ll\#P(X)\le\#P_1(X)+\#P_2(X)+\#P_3(X)\ll\frac{X(\log \log X)^{O(1)}}{(\log X)^{1+\delta}}, \end{align*} $$

which yields a contradiction.

Proof. By Theorem 3.1, we see that Lemma 2.7 applies to the sequences $b_{1,n}=F_{n-1}$ and $b_{2,n}=F_{n+1}$ . To ensure the condition $\operatorname {\mathrm {ord}}_p(q) \not \equiv 0 \,\pmod {\,5}$ , it suffices, for instance, to restrict to primes $p \equiv 4 \,\pmod {\,5}$ . Indeed, if $5 \mid \operatorname {\mathrm {ord}}_p(q)$ , then $5 \mid (p-1)$ , which is a contradiction. The condition $\log |b_{j,n}|=O(n)$ follows from Binet’s formula.

Proof. In [Reference Anzawa and Funakura1, Lemma 2.6], Anzawa and Funakura proved that

$$ \begin{align*} \binom{p-1}{k}_q \equiv \begin{cases} \binom{I_p(q)}{k/\mathrm{ord}_p(q)}\, \pmod{\,p} &\text{if } \mathrm{ord}_p(q) \mid k,\\ 0 \,\pmod{\,p} &\text{otherwise}, \end{cases} \end{align*} $$

under the additional assumption that $q-1 \in \mathbb {Z}_{(p)}^{\times }$ . However, it is easy to see that this also holds when this assumption is not satisfied. Therefore, by (3-1), it follows that

$$ \begin{align*} \begin{aligned} D_{p-1}(q) = \sum_{k=0}^{p-1} q^{k^2} \binom{p-1}{k}_q \equiv \sum_{j=0}^{I_p(q)} \binom{I_p(q)}{j}=2^{I_p(q)} \,\pmod{\,p}. \end{aligned}\\[-44pt] \end{align*} $$

Proof. Using the congruences established in Proposition 3.4, it suffices to apply Lemma 2.7 with the sequence $b_n = 2^n$ and any arithmetic progression you like.

Proof. In fact, this holds for any constants $0<c_1<c_2\leq 1$ . For an elliptic curve $E/\mathbb {Q}$ and an interval $I = [\alpha , \beta ] \subset [0,\pi ]$ , it is known that

$$ \begin{align*} \lim_{X \to \infty} \frac{\#\{p \le X : p\text{ is prime}, \theta_p \in I\}}{\#\{p \le X : p\text{ is prime}\}} = \begin{cases} \displaystyle{\frac{2}{\pi} \int_I \sin^2\theta\, \mathrm{d} \theta} &\text{if } E \text{ is non-CM},\\[10pt] \displaystyle{\frac{\delta_I}{2} + \frac{\beta-\alpha}{2\pi}} &\text{if } E \text{ has CM}, \end{cases} \end{align*} $$

where $a_p = 2\sqrt {p} \cos \theta _p$ and $\delta _I = 1$ if $\pi /2 \in I$ and $\delta _I = 0$ otherwise. The CM case is classical and goes back to Hecke and Deuring; see, for example, Sutherland’s survey [Reference Sutherland26, Section 2.4]. The non-CM case was finally proved by Barnet-Lamb, Geraghty, Harris, and Taylor [Reference Barnet-Lamb, Geraghty, Harris and Taylor3]. Therefore, the set of primes $p$ satisfying $2c_1 \sqrt {p} \le a_p(E) \le 2c_2 \sqrt {p}$ , that is, $c_1 \le \cos \theta _p \le c_2$ , has positive relative density.

Proof of Theorem 4.1.

Let $S$ be the set of primes of positive relative density obtained in Lemma 4.2. For each $p \in S$ , define $b_p := a_p(E)$ . Then $b_p \to \infty $ and $b_p = O(p^{\frac 12})$ . Moreover, since $\# \{p \le X : p \in S\} \gg X/\log X$ , it follows from Lemma 2.5 that $\alpha (E) \in \mathcal {A} \setminus \mathcal {C}_{\mathcal {A}}$ .

Proof. Let $S$ be the set of all primes $p>3$ satisfying $p\equiv 3\,\pmod {\,4}$ , and for each $p\in S$ define $b_p:= h(-p)$ . Then the facts that $b_p\to \infty $ and that $b_p=O(p^{\tfrac 12+\varepsilon })$ for every $\varepsilon>0$ follow from the classical results mentioned above. Since $\#\{p\leq X : p\in S\}\sim \tfrac 12({X}/{\log X})$ , it follows from (5-1) and Lemma 2.5 that $-2\mathscr {B}\in \mathcal {A}\setminus \mathcal {C}_{\mathcal {A}}$ .

Proof. Let $S$ be the set of all primes $p$ satisfying $p\equiv 1\,\pmod {\,4}$ , and for each $p\in S$ define $b_p:= h(-4p)$ . Then the facts that $b_p\to \infty $ and that $b_p=O(p^{\tfrac 12+\varepsilon })$ for every $\varepsilon>0$ follow from the classical results mentioned above. Since $\#\{p\leq X : p\in S\}\sim \tfrac 12({X}/{\log X})$ and the congruence

$$ \begin{align*} \frac{1}{2}E_{({p-1})/{2}}\equiv h(-4p)\,\pmod{\,p} \end{align*} $$

holds [Reference Carlitz6, Equation (1.2)], it follows from Lemma 2.5 that $\tfrac 12\mathscr {E}\in \mathcal {A}\setminus \mathcal {C}_{\mathcal {A}}$ .

Proof. Assume that there exist integers $b$ and $e$ such that $b\mathscr {B}+e\mathscr {E}=0$ . Then, for all sufficiently large primes $p$ , one has

$$ \begin{align*} bB_{({p+1})/{2}}+eE_{({p-1})/{2}}\equiv 0\,\pmod{\,p}. \end{align*} $$

If $p\equiv 1\,\pmod {\,4}$ , then $B_{({p+1})/{2}}=0$ and $E_{({p-1})/{2}}\not \equiv 0\,\pmod {\,p}$ , the latter following from $0<h(-4p)<p$ . Hence, $e\equiv 0\,\pmod {\,p}$ . Since there are infinitely many such primes $p$ , it follows that $e=0$ . Next, consider the case where $p\equiv 3\,\pmod {\,4}$ . Since $0<h(-p)<~p$ , one has $B_{({p+1})/{2}}\not \equiv 0\,\pmod {\,p}$ , and therefore $b\equiv 0\,\pmod {\,p}$ . Again, since there are infinitely many such primes $p$ , we obtain $b=0$ .

Proof. Suppose that $\pi (\boldsymbol {p})\in \mathcal {P}^0_{\mathcal {A}}$ . Then, since $\mathcal {P}^0_{\mathcal {A}}\subset \mathcal {C}_{\mathcal {A}}$ , there exists a nonzero polynomial $f(x)$ with rational coefficients such that $f(\pi (\boldsymbol {p}))=0$ . However, by [Reference Rosen20, Theorem 1.4], any such polynomial $f(x)$ must have a rational root. It follows that there exist an integer $a$ and a positive integer $b$ such that $b\pi (p)-a\equiv 0 \,\pmod {\,p}$ holds for infinitely many primes $p$ . Since $\pi (p)\sim {p}/{\log p}$ , we have $0<b\pi (p)-a<p$ for all sufficiently large $p$ , which is a contradiction.

Question 6.4 (cf. [Reference Bober, Fretwell, Martin and Wooley4, Reference Martin17]).

Does Theorem 6.3 remain true if $f(x)$ is replaced by an arbitrary nonconstant polynomial in one variable with integer coefficients?

Proof. Let $f(x)$ be a nonconstant polynomial with integer coefficients. Suppose that $f(n)\equiv 0\,\pmod {\,p_n}$ holds for all sufficiently large integer $n$ . Since $p_n\sim n\log n$ , it would follow that there exist at most finitely many positive integers $n$ such that $f(n)$ is $n$ -smooth.

Proof. Assume that $f(\pi (\boldsymbol {p}))=0$ . Let $X$ be a sufficiently large real number. Let $l_1(x),\ldots ,l_a(x)$ be the distinct linear factors of $f(x)$ , and let $q_1(x),\ldots ,q_b(x)$ be the distinct irreducible quadratic factors of $f(x)$ . Define $S_g(X):= ~\{n\leq \pi (X) : p_n\mid g(n)\}$ for a polynomial $g(x)$ . Then

(6-1) $$ \begin{align} S_f(X)=\bigcup_{i=1}^a S_{l_i}(X)\cup\bigcup_{j=1}^b S_{q_j}(X). \end{align} $$

By the assumption, $S_f(X)$ is in one-to-one correspondence, up to finitely many exceptions independent of $X$ , with the set of primes not exceeding $X$ . Therefore,

(6-2) $$ \begin{align} \#S_f(X)\sim \pi(X). \end{align} $$

By the same argument as in the proof of Proposition 6.1, we have $\#S_{l_i}(X)=O(1)$ . Let $\varepsilon>0$ be small. Since $ \lim _{n\to \infty }n/p_n=0$ , if $n\in S_{q_j}(X)$ is sufficiently large, then $(p_n,n)\in R_{q_j}(X;0,\varepsilon )$ . Hence,

$$ \begin{align*} \#S_{q_j}(X)\leq \#R_{q_j}(X;0,\varepsilon)+O(1). \end{align*} $$

By Theorem 6.6, we have $\#R_{q_j}(X;0,\varepsilon )\sim \varepsilon \pi (X)$ , and therefore, by the arbitrariness of $\varepsilon $ , it follows that $\#S_{q_j}(X)=o(\pi (X))$ . It then follows from (6-1) that

$$ \begin{align*} \#S_f(X)=o(\pi(X)), \end{align*} $$

which contradicts (6-2).

Proof. Let $F(x,y)\in \mathbb {Z}[x,y]$ be a nonzero polynomial, and suppose that $F(a_p,b_p)\equiv 0\,\pmod {\,p}$ for sufficiently large primes $p$ . Write

$$ \begin{align*} F(x,y)=\sum_{i=0}^dg_i(y)x^i, \end{align*} $$

where $g_i(y)\in \mathbb {Z}[y]$ for $0\leq i\leq d$ and $g_d(y)$ is a nonzero polynomial. Since $b_p\to ~\infty $ , $g_d(b_p)\neq 0$ for sufficiently large primes $p$ . In particular, $|g_d(b_p)a_p^d|\geq a_p^d$ . Let $ k:= ~\max _{0\leq i\leq d}\deg g_i(y)$ . Since $g_i(n)=O(n^k)$ for every $0\leq i\leq d$ and every positive integer $n$ , we have

$$ \begin{align*} \sum_{i=0}^{d-1}g_i(b_p)a_p^i=O(b_p^ka_p^{d-1})=o(a_p^d) \end{align*} $$

by $b_p^k=o(a_p)$ . Therefore, $F(a_p,b_p)\neq 0$ for sufficiently large primes $p$ . On the other hand, since $F(a_p,b_p)=O(b_p^ka_p^d)=o(p)$ , we have $|F(a_p,b_p)|<p$ for sufficiently large primes $p$ . Thus, we have a contradiction.

Proof. Let $F(x,y)\in \mathbb {Z}[x,y]$ be a nonzero polynomial, and suppose that $F(\alpha ,\, \beta )=0$ . Write

$$ \begin{align*} F(x,y)=\sum_{i=0}^d f_i(x)y^i, \end{align*} $$

where $f_i(x)\in \mathbb {Z}[x]$ for $0\leq i\leq d$ and $f_d(x)$ is a nonzero polynomial. For an integer $m$ , if $F(m,y)$ is the zero polynomial in $y$ , then it must be the case that $f_d(m)=0$ . It follows that there are only finitely many integers $m$ for which $F(m,y)$ is the zero polynomial in $y$ . Therefore, for some $n$ , the polynomial $g(y) := F(a^{\prime }_n,y)$ is not the zero polynomial in $y$ . In this case, by the assumption, $g(b^{\prime }_p)\equiv 0\,\pmod {\,p}$ holds for infinitely many primes $p$ . From the assumption on $\beta $ , it follows that for such primes $p$ , if $p$ is sufficiently large, then $0<|g(b^{\prime }_p)|<p$ , which leads to a contradiction.

Proof. Since $-\log _{\mathcal {A}}(\alpha ^{-1}) = \log _{\mathcal {A}}(-\alpha ) = \log _{\mathcal {A}} \alpha $ , there is no loss of generality in assuming that $\alpha> 1$ . Suppose for a contradiction that $\log _{\mathcal {A}}(\alpha ) = a/b \in \mathbb {Q}^\times $ with $a \neq 0$ , $b>0$ , and $\gcd (a,b) = 1$ . Then for all sufficiently large primes $p \ge N$ , we have

$$ \begin{align*} q_p(\alpha) \equiv a b^{-1} \,\pmod{\,p}. \end{align*} $$

Write $\alpha = u/v$ with $\mathrm {gcd}(u, v) = 1$ . Note that $u>v>0$ . Let $\ell $ be a sufficiently large prime (such that $\ell> u, |a|, b, N$ ) and consider the value of the homogenization of the $\ell $ th cyclotomic polynomial

$$ \begin{align*} \Phi_\ell(u,v) := v^{\ell-1} \frac{\alpha^\ell-1}{\alpha -1} = \frac{u^\ell - v^\ell}{u-v} = u^{\ell-1} + u^{\ell-2} v + \cdots + v^{\ell-1} \in \mathbb{Z}. \end{align*} $$

First, by the lifting-the-exponent lemma, any prime divisor $p$ of $u-v$ does not divide $\Phi _\ell (u,v)$ , since $p \nmid \ell $ . Let $p$ be any prime divisor of $\Phi _\ell (u,v)$ . Since $u$ and $v$ are coprime, we have $p\nmid v$ . Hence, $\alpha ^\ell \equiv 1 \,\pmod {\,p}$ and $\alpha \not \equiv 1 \,\pmod {\,p}$ , so that $\operatorname {\mathrm {ord}}_p(\alpha )=\ell $ . In particular, such a prime $p$ satisfies $p\equiv 1 \,\pmod {\,\ell }$ , and therefore $p>\ell >N$ .

We next show that $\Phi _\ell (u,v)$ is square-free. For each prime $p\mid \Phi _\ell (u,v)$ , write

$$ \begin{align*} \Phi_\ell(u,v) = p t_p. \end{align*} $$

From $\alpha ^\ell = 1 + v^{1-\ell } (\alpha -1) pt_p$ , we obtain

$$ \begin{align*} \ell q_p(\alpha) \equiv q_p(\alpha^\ell) = \frac{(1 + v^{1-\ell} (\alpha-1) pt_p)^{p-1} - 1}{p} \equiv - v^{-\ell} (u-v) t_p \,\pmod{\,p}. \end{align*} $$

By our assumption, we get

(8-1) $$ \begin{align} (u-v) b t_p + a \ell v^{\ell} \equiv 0 \,\pmod{\,p}. \end{align} $$

If $p \mid t_p$ , then $a\ell v^\ell \equiv 0 \,\pmod {\,p}$ , which is impossible since $p> \ell $ and $\ell $ is sufficiently large (note that $a\neq 0$ ). Thus $p \nmid t_p$ , and hence $\Phi _\ell (u,v)$ is square-free.

Now we define

(8-2) $$ \begin{align} T_\ell:= (u-v) b \sum_{p \mid \Phi_\ell(u,v)} \frac{\Phi_\ell(u,v)}{p} + a\ell v^\ell \in \mathbb{Z}. \end{align} $$

Then (8-1) implies that for each prime $p\mid \Phi _\ell (u,v)$ , $T_\ell \equiv 0\,\pmod {\,p}$ . Since $\Phi _\ell (u,v)$ is square-free, it follows that

(8-3) $$ \begin{align} \Phi_\ell(u,v) \mid T_\ell. \end{align} $$

Since $\Phi _\ell (u,v)$ is square-free, we can write $\Phi _\ell (u,v) = p_1 \cdots p_r$ with distinct primes $p_1 < \cdots < p_r$ . As observed above, each $p_j \equiv 1 \,\pmod {\,\ell }$ , so we may write $p_j = \ell m_j+~1$ with $m_1 < \cdots < m_r$ . In particular, $m_j \ge j$ . Since $p_j> \ell $ , we have $\ell ^r<\Phi _\ell (u,v)<\ell u^{\ell -1}<~\ell ^\ell $ , which implies $1 \le r < \ell $ . Moreover,

$$ \begin{align*} \sum_{p\mid\Phi_\ell(u,v)} \frac{1}{p} = \sum_{j=1}^r \frac{1}{\ell m_j +1} \le \sum_{j=1}^r \frac{1}{\ell j+1} < \frac{H_r}{\ell} < \frac{1+\log \ell}{\ell}, \end{align*} $$

where $H_n = \sum _{j=1}^n 1/j$ is the $n\,\mathrm{th}$ harmonic number. Hence, for sufficiently large $\ell $ , since $\alpha> 1$ , we obtain

$$ \begin{align*} \sum_{p\mid\Phi_\ell(u,v)}\frac{1}{p}<\frac{1}{2(u-v)b} \quad \text{and} \quad 2|a|<\frac{\Phi_\ell(u,v)}{\ell v^\ell}, \end{align*} $$

which implies

(8-4) $$ \begin{align} |T_\ell|<\Phi_\ell(u,v). \end{align} $$

Combining (8-3) and (8-4), we conclude that $T_\ell = 0$ .

By the definition of $T_\ell $ in (8-2), this yields

$$ \begin{align*} (u-v)b\sum_{p\mid\Phi_\ell(u,v)}\frac{\Phi_\ell(u,v)}{p}=-a\ell v^\ell. \end{align*} $$

For each $p_i$ , since $p_i\equiv 1 \,\pmod {\,\ell }$ , we have

$$ \begin{align*} t_{p_i}=\frac{\Phi_{\ell}(u,v)}{p_i}=\prod_{1\le j\leq r, \ j\neq i} p_j \equiv 1\,\pmod{\,\ell}. \end{align*} $$

Therefore, by reducing modulo $\ell $ , we have

$$ \begin{align*} (u-v)br\equiv 0 \,\pmod{\,\ell}. \end{align*} $$

However, by construction, the prime $\ell $ is chosen to be larger than both $u-v>0$ and $b>0$ , and moreover $1\leq r < \ell $ , which yields a contradiction.

Proof. As a generalization of [Reference Kaneko, Matsusaka and Seki12, Theorem 3.2], we have

$$ \begin{align*} G_{\mathcal{A}}(k;x) &= (-1)^{k-1} \sum_{j=0}^k (-1)^j \binom{k}{j} (x+j+1) \log_{\mathcal{A}}(x+j+1)\\ &= (-1)^{k-1} \log_{\mathcal{A}}(h_k(x)) \end{align*} $$

by [Reference Matsusaka, Miyazaki and Yara18, Theorem 3.8]. Here, we set

$$ \begin{align*} h_k(x):=\prod_{j=0}^k (x+j+1)^{(-1)^j (x+j+1)\binom{k}{j}}. \end{align*} $$

We can check that $h_k(-k-2-x)=h_k(x)^{(-1)^{k+1}}$ and if $x\geq 0$ then $h_k(x)$ is rational and

$$ \begin{align*} \log(h_k(x))=(k-2)!\int_{[0,1]^k}\frac{\mathrm{d}t_1\cdots\mathrm{d}t_k}{(x+1+t_1+\cdots+t_k)^{k-1}}>0. \end{align*} $$

Therefore, the claim follows from Corollary 8.2.