AN EFFECTIVE BOUND FOR GENERALISED DIOPHANTINE m-TUPLES

Abstract

For $k\geq 2$ and a nonzero integer n, a generalised Diophantine m-tuple with property $D_k(n)$ is a set of m positive integers $S = \{a_1,a_2,\ldots , a_m\}$ such that $a_ia_j + n$ is a kth power for $1\leq i< j\leq m$. Define $M_k(n):= \text {sup}\{|S| : S$ having property $D_k(n)\}$. Dixit et al. [‘Generalised Diophantine m-tuples’, Proc. Amer. Math. Soc. 150(4) (2022), 1455–1465] proved that $M_k(n)=O(\log n)$, for a fixed k, as n varies. In this paper, we obtain effective upper bounds on $M_k(n)$. In particular, we show that for $k\geq 2$, $M_k(n) \leq 3\,\phi (k) \log n$ if n is sufficiently large compared to k.

1 Introduction

Given a nonzero integer n, we say a set of natural numbers $S=\{a_1,a_2,\ldots , a_m\}$ is a Diophantine m-tuple with property $D(n)$ if $a_ia_j +n$ is a perfect square for $1\leq i<j\leq m$ . Diophantus first studied such sets of numbers and found the quadruple $\{1,33,68,105\}$ with property $D(256)$ . The first $D(1)$ -quadruple $\{1,3,8,120\}$ was discovered by Fermat, and this was later generalised by Euler who found an infinite family of quadruples with property $D(1)$ , namely,

$$ \begin{align*} \{a, b, a + b + 2r, 4r(r + a)(r + b) \}, \end{align*} $$

where $ab + 1 = r^2$ . In fact, any $D(1)$ -triple can be extended to a Diophantine quadruple [1]. In 1969, using Baker’s theory of linear forms in the logarithms of algebraic numbers and a reduction method based on continued fractions, Baker and Davenport [2] proved that Fermat’s example is the only extension of $\{1,3,8\}$ with property $D(1)$ . In 2004, Dujella [10], using similar methods, proved that there are no $D(1)$ -sextuples and there are only finitely many $D(1)$ -quintuples, if any. The nonexistence of $D(1)$ -quintuples was finally settled in $2019$ by He et al. in [15].

In general, there are $D(n)$ -quintuples for $n\neq 1$ . For example,

$$ \begin{align*}\{1, 33, 105, 320, 18240\}\quad \text{and} \quad \{5, 21, 64, 285, 6720\}\end{align*} $$

are Diophantine quintuples satisfying property $D(256)$ . There are also examples of $D(n)$ -sextuples, but no $D(n)$ -septuple is known. So, it is natural to study the size of the largest m-tuple with property $D(n)$ . Define

$$ \begin{align*} M_n := \text{sup}\{|S| : S \text{ satisfies property } D(n)\}. \end{align*} $$

In 2004, Dujella [9] showed that

$$ \begin{align*} M_n \leq C \log |n|, \end{align*} $$

where C is an absolute constant. He also showed that for $n>10^{100}$ , one can choose $C=8.37$ . This constant was improved by Becker and Murty [3], who showed that for any n,

(1.1) $$ \begin{align} M_n \leq 2.6071 \log |n| + \mathcal{O}\left(\frac{\log |n|}{(\log\log |n|)^2}\right). \end{align} $$

Our goal is to study this problem when squares are replaced by higher powers.

Definition 1.1 (Generalised Diophantine m-tuples)

Fix a natural number $k\geq 2$ . A set of natural numbers $S=\{a_1,a_2,\ldots , a_m\}$ satisfies property $D_k(n)$ if $a_ia_j +n$ is a kth power for $1\leq i<j\leq m$ .

For each nonzero integer n, define

$$ \begin{align*} M_k(n):= \text{sup}\{|S| : S \text{ satisfies property }D_k(n)\}. \end{align*} $$

For $k\geq 3$ and $m\geq 3$ , we can apply the celebrated theorem of Faltings [12] to deduce that a superelliptic curve of the form

$$ \begin{align*} y^k = f(x)=(a_1 x +n)(a_2 x +n) (a_3 x+n)(a_4 x + n)~\cdots~ (a_m x +n) \end{align*} $$

has only finitely many rational points and a fortiori, finitely many integral points. Therefore, a set S satisfying property $D_k(n)$ must be finite. When $n=1$ , Bugeaud and Dujella [6] showed that

$$ \begin{align*} M_{3}(1) \leq 7, \quad M_{4}(1) \leq 5, \quad M_{k}(1) \leq 4 ~ \text{ for}~ 5 \leq k \leq 176, \quad M_{k}(1) \leq 3~ \text{ for}~ k \geq 177. \end{align*} $$

In other words, the size of $D_k(1)$ -tuples is bounded by $3$ for large enough k. In the general case, for any $n \neq 0$ and $k \geq 3$ , Bérczes et al. [5] obtained upper bounds for $M_k(n)$ . In particular, they showed that for $k\geq 5$ ,

$$ \begin{align*} M_k(n) \leq 2|n|^5 + 3. \end{align*} $$

Dixit et al. [8] improved these bounds on $M_k(n)$ for large n and a fixed k. Define

$$ \begin{align*} M_k(n;L):=\text{sup}\{|S \cap [|n|^L,\infty)|: S \text{ satisfies property } D_k(n)\}. \end{align*} $$

Then, for $k\geq 3$ , as $n\to \infty $ ,

(1.2) $$ \begin{align} M_k(n,L)\ll_{k,L} 1~ \text{ for }L\geq 3 \quad \text{and}\quad M_k(n) \ll_k \log n. \end{align} $$

The purpose of this paper is to make the implied constants in (1.2) explicit. In [8], the bounds for $M_k(n)$ were proved under the further assumption that $n>0$ . As remarked in [8], this assumption is not necessary, but an argument was not provided. We begin by proving the bounds (1.2) for all nonzero integers n.

Theorem 1.2. Let $k \geq 3$ be an integer. Then the following statements hold as $|n| \to \infty $ .

1. (1) For $L \geq 3$ ,

   $$ \begin{align*}M_k(n, L) \ll 1,\end{align*} $$
   where the implied constant depends on k and L, but is independent of n.

2. (2) Moreover,

   $$ \begin{align*}M_k(n) \ll \log |n|,\end{align*} $$
   where the implied constant depends on k.

We next state our main theorem, which is an effective version of Theorem 1.2.

Theorem 1.3. Let $k \geq 3$ be a positive integer. Then the following statements hold.

1. (a) For $L \geq 3$ ,

   (1.3) $$ \begin{align} M_k(n, L) \leq 2^{28} \log (2k) \log (2 \log (2k)) + 14. \end{align} $$

2. (b) Suppose n and k vary such that as $|n|\to \infty $ and $k = o(\log \log |n|)$ , then

   $$ \begin{align*} M_k(n) \leq 3 \, \phi(k)\, \log |n| + \mathcal{O} \bigg(\frac{ (\phi(k))^2\log |n|}{\log \log |n|}\bigg), \end{align*} $$
   where $\phi (n)$ denotes the Euler totient function.

Remark 1.4. (a) It is possible to replace $14$ on the right-hand side of (1.3) with a smaller positive integer for large values of k.

(b) For a fixed $k>2$ , Theorem 1.3(b) gives

$$ \begin{align*} M_k(n) \leq 3 \, \phi(k) \, \log |n| + \mathcal{O}\bigg(\frac{\log |n|}{\log\log |n|}\bigg) \quad \mbox{as } |n|\to \infty. \end{align*} $$

For $k=2$ , this upper bound is very close to the best known upper bound due to Becker and Murty which is given by (1.1).

(c) Theorem 1.3 holds in a slightly more general setting for Diophantine tuples with property $D_k(n)$ in the ring of integers of the kth cyclotomic field $\mathbb {Q}(\zeta _k)$ . In that case, we replace the Legendre symbol by the power residue symbol and follow the same method as in the proof of Theorem 1.3.

2 Preliminaries

In this section, we develop the necessary tools to prove our main theorems.

2.1 Gallagher’s larger sieve

In 1971, Gallagher [13] discovered an elementary sieve inequality which he called the larger sieve. We refer the reader to [7] for the general discussion and record the result in a form applicable to our context.

Theorem 2.1. Let N be a natural number and $\mathcal {S}$ a subset of $\{1,2,\ldots , N\}$ . Let ${\mathcal P}$ be a set of primes. For each prime $p \in {\mathcal P}$ , let $\mathcal {S}_p=\mathcal {S} \pmod {p}$ . For $1<Q\leq N$ ,

$$ \begin{align*} |\mathcal{S}|\leq \frac{\sum_{p\leq Q, p\in {\mathcal P}} \log p - \log N}{\sum_{p\leq Q, p \in {\mathcal P}} \dfrac{\log p}{|\mathcal{S}_p|}-\log N}, \end{align*} $$

where the summations are over primes $p\leq Q, p \in {\mathcal P}$ , and the inequality holds provided the denominator is positive.

2.2 A quantitative Roth’s theorem

There are several quantitative results counting exceptions in Roth’s celebrated theorem on Diophantine approximations. We will use the following result due to Evertse [11]. For an algebraic number $\xi $ of degree r, we define the (absolute) height by

$$ \begin{align*}H(\xi):= \bigg( a\prod_{i=1}^r \max(1, |\xi^{(i)}|)\bigg)^{1/r}, \end{align*} $$

where $\xi ^{(i)}$ for $1\leq i\leq r$ are the conjugates of $\xi $ (over $\mathbb {Q}$ ) and a is the positive integer such that

$$ \begin{align*}a\prod_{i=1}^r (x-\xi^{(i)}) \end{align*} $$

has rational integer coefficients with greatest common divisor equal to 1.

Theorem 2.2. Let $\alpha $ be a real algebraic number of degree r over $\mathbb {Q}$ and $0 <\kappa \leq 1$ . The number of rational numbers $p/q$ satisfying $\max (|p|,|q|) \geq \max (H(\alpha ), 2)$ and

$$ \begin{align*} \bigg|\alpha - \frac{p}{q} \bigg| \leq \frac{1}{\max (|p|,|q|)^{2+\kappa}} \end{align*} $$

is at most

$$ \begin{align*}2^{25}\kappa^{-3}\log (2r) \log (\kappa^{-1} \log (2r)).\end{align*} $$

2.3 Vinogradov’s theorem

The following bound on character sums was proved by Vinogradov (see [16]).

Lemma 2.3. Let $\chi \pmod q$ be a nontrivial Dirichlet character and n an integer such that $(n,q)=1$ . If $\mathcal {A} \subseteq (\mathbb {Z}/q\mathbb {Z})^*$ and $\mathcal {B} \subseteq (\mathbb {Z}/q\mathbb {Z})^* \cup \{0\}$ , then

$$ \begin{align*} \sum_{a\in\mathcal{A}} \sum_{b\in\mathcal{B}} \chi(ab + n) \leq \sqrt{q |\mathcal{A}| |\mathcal{B}|}. \end{align*} $$

The original method of Vinogradov gives the bound on the right-hand side of the inequality as $ \sqrt {2q |\mathcal {A}| |\mathcal {B}|}$ . However, the above bound holds and a short proof can be found in [3, Proposition 2.5].

2.4 Bounds for primes in arithmetic progression

Let $Q,k,a$ be positive integers with $(a,k)=1$ . Denote by $\theta (Q;k,a)$ the sum of the logarithms of the primes ${p \equiv a \pmod k}$ with $p \leq Q$ , that is,

$$ \begin{align*} \theta(Q;k,a) := \sum_{\substack{p \equiv a \bmod k \\ p \text{ prime} \leq Q}} \log p. \end{align*} $$

The following bound on $\theta (Q;k,a)$ was obtained by Bennet et al. in [4, Theorem 1.2].

Theorem 2.4. For $k \geq 3$ and $(a, k) = 1$ ,

$$ \begin{align*} \bigg |\theta (Q; k, a) - \frac{Q}{\phi(k)} \bigg| < \frac{1}{160} \frac{Q}{\log Q} \end{align*} $$

for all $Q \geq Q_0 (k)$ , where

$$ \begin{align*} Q_0(k) = \begin{cases} 8 \cdot 10^9 & \text{if } 3 \leq k \leq 10^5, \\ \exp(0.03\sqrt{k}\log^3 k) & \text{if }k> 10^5. \end{cases} \end{align*} $$

2.5 Gap principle

The next two lemmas are variations of a gap principle of Gyarmati [14].

Lemma 2.5 [8, Lemma 2.4]

Let $k\geq 2$ . Suppose that $a,b,c,d$ are positive integers such that $a<b$ and $c<d$ . Suppose further that

$$ \begin{align*}ac+n, \quad bc+n, \quad ad+n, \quad bd+n \end{align*} $$

are perfect kth powers. Then,

$$ \begin{align*}bd \geq k^k n^{-k} (ac)^{k-1}. \end{align*} $$

An immediate corollary of this lemma shows that ‘large’ elements of any set with property $D_k(n)$ have ‘super-exponential growth’.

Corollary 2.6 [8, Corollary 2]

Let $k\geq 3$ and $m\geq 5$ . Suppose that $n^3 \leq a_1 < a_2 < \cdots < a_m$ and the set $\{ a_1, a_2,\ldots , a_m\}$ has property $D_k(n)$ . Then $a_{2+3j} \geq a_2^{(k-1)^{j}}$ provided $1\leq j \leq (m-2)/3$ .

A modification of the proof of Lemma 2.5 yields a gap principle for negative values of n.

Lemma 2.7. For $n> 0$ and natural numbers $a,b,c,d$ such that $n^3 \leq a < b < c < d$ ,

$$ \begin{align*} (ac-n)(bd-n) \geq \frac{abcd}{2}. \end{align*} $$

Proof. Since $(ac-n)(bd-n) = abcd-n(ac+bd)+n^2$ , it is enough to prove that

$$ \begin{align*} \frac{abcd}{2} \geq n(ac+bd) - n^2. \end{align*} $$

Also, since $a \geq n^3$ and $c> n^3$ , for all cases other than $n=1, a=1, b=2, c=3$ ,

$$ \begin{align*} abcd &\geq 4n bd\\ &\geq 2nbd +2nac\\ &\geq 2nbd+2nac-2n^2, \end{align*} $$

where the first inequality is obvious as $a \geq n^3$ and $c \geq n^3+2$ . This gives the desired result. For the case $n=1, a=1, b=2, c=3$ , since $d> c$ , clearly

$$ \begin{align*} 2n(ac+bd) - 2n^2 = 4+4d < 6d = abcd.\\[-33pt] \end{align*} $$

We are now ready to prove the following analogue of Lemma 2.5.

Lemma 2.8. Let $n> 0$ and $k \geq 2$ . Suppose that $a,b,c,d$ are positive integers such that $n^3 \leq a < b < c < d$ . Suppose further that $ac - n, bc-n, ad-n, bd-n$ are perfect kth powers. Then,

$$ \begin{align*}bd \geq k^k 2^{-k} n^{-k} (ac)^{k-1}.\end{align*} $$

Proof. Since $(b-a)(d-c)>0$ , we have $bd + ac> ad + bc$ and it is easily seen that

$$ \begin{align*} (ad-n)(bc-n)> (ac-n)(bd-n). \end{align*} $$

As $(ac-n)(bd-n)$ and $(ad-n)(bc-n)$ are both perfect kth powers,

$$ \begin{align*} (ad-n)(bc-n) &\geq [((ac-n)(bd-n))^{1/k} + 1]^k\\ &\geq (ac-n)(bd-n) + k((ac-n)(bd-n))^{k-1/k}\\ &\geq (ac-n)(bd-n) + k\bigg(\frac{abcd}{2}\bigg)^{k-1/k}, \end{align*} $$

where the last inequality follows from Lemma 2.7. Thus,

$$ \begin{align*} -n(ad+bc) \geq -n(ac+bd) + k\bigg(\frac{abcd}{2}\bigg)^{k-1/k}. \end{align*} $$

Since $bd> ad + bc - ac > 0$ , we have $bd+ac-ad-bc < bd$ and hence,

$$ \begin{align*} nbd> k\bigg(\frac{abcd}{2}\bigg)^{k-1/k}. \end{align*} $$

Therefore,

$$ \begin{align*} bd \geq k^k 2^{1-k} n^{-k} (ac)^{k-1} \geq k^k 2^{-k} n^{-k} (ac)^{k-1}, \end{align*} $$

which proves the lemma.

This enables us to prove super-exponential growth for large elements of a set with $D_k(n)$ , when $n<0$ .

Corollary 2.9. Let $k \geq 3$ . If $n^3 \leq a < b < c < d < e$ are natural numbers such that the set $\{a,b,c,d,e\}$ has property $D_k(-n)$ , then $e \geq b^{k-1}$ .

Proof. From Lemma 2.8,

$$ \begin{align*} ce \geq k^k 2^{-k} n^{-k} (bd)^{k-1} \geq k^k 2^{-k} n^{-k} (bc)^{k-1}. \end{align*} $$

Therefore,

$$ \begin{align*} e \geq b^{k-1} c^{k-2} n^{-k} 2^{-k} \geq b^{k-1} n^{2k-6} 2^{-k} \geq b^{k-1}. \\[-36pt] \end{align*} $$

Using induction on the previous corollary, we deduce the following corollary.

Corollary 2.10. Let $k \geq 3$ and $m \geq 5$ . Suppose that $n^3 \leq a_1 < a_2 < \cdots < a_m$ and the set $\{a_1, a_2, \ldots , a_m\}$ has property $D_k(-n)$ . Then we have $a_{2+3j} \geq a_2^{(k-1)^j}$ provided $1 \leq j \leq (m-2)/3$ .

3 Proof of the main theorems

3.1 Proof of Theorem 1.2

We first prove Theorem 1.2. The proof follows a similar method to [8].

Let n be a positive integer, $m = M_k (-n)$ and $S = \{a_1, a_2, a_3, \ldots , a_m\}$ be a generalised m-tuple with the property $D_k(-n)$ . Suppose $n^L < a_1 < a_2 < \cdots < a_m$ for some $L \geq 3$ . Consider the system of equations

(3.1) $$ \begin{align} \begin{split} a_1 x -n = u^k, \\ a_2 x -n = v^k. \end{split} \end{align} $$

Clearly, $x=a_i$ for $i \geq 3$ are solutions to this system. Also,

$$ \begin{align*} |a_2 u^k - a_1 v^k| = n(a_2-a_1). \end{align*} $$

Let $\alpha := (a_1/a_2)^{1/k}$ and . Then, we have the following two lemmas analogous to those proved in [8]. The proof of the first lemma is identical to the proof of [8, Lemma 3.1].

Lemma 3.1. Let $k \geq 3$ be odd. Suppose $u, v$ satisfy the system of equations (3.1). Let

Then, for $n> 2^{1/(L-1)}c(k)^{-1/(L-1)}$ ,

$$ \begin{align*} \left|\frac{u}{v} - \alpha\right| \leq \frac{a_2}{2v^k}. \end{align*} $$

Lemma 3.2. Let $(u_i, v_i)$ denote distinct pairs that satisfy the system of equations (3.1) with $v_{i+1}> v_i$ . For $n> 2^{1/(L-1)}c(k)^{-1/(L-1)}$ and $i\geq 14$ ,

$$ \begin{align*} \bigg|\frac{u}{v} - \alpha\bigg| < \frac{1}{v_i^{k-1/2}} \quad \mbox{and} \quad v_i> a_2^4. \end{align*} $$

Proof. From Lemma 3.1, $|{u_i}/{v_i} - \alpha | < {a_2}/{2v_i^k}$ . Thus, we need to show $a_2 < 2v_i^{1/2}$ for $i> 14$ . Since $v_i^k = a_2 a_i - n$ , we have $v_i \geq a_i^{1/k}$ . By Corollary 2.10, $a_{2+3j} \geq a_2^{(k-1)^j}$ , so that $v_{2+3j} \geq a_2^{(k-1)^j/k}$ . We choose a positive integer $j_0$ such that $(k-1)^{j_0}> 4k$ . Since $k \geq 3$ , we can take $j_0=4$ . As $2+3j_0 = 14$ , we have $v_i \geq v_{14}> a_2^4$ for all $i \geq 14$ . This completes the proof.

For larger values of k, the number $14$ in the above lemma can be improved to $2 + 3 j_0$ , where $j_0$ satisfies the condition $(k-1)^{j_0}> 4k$ .

Proof of Theorem 1.2

Now, assume that $(u_1,v_1), (u_2, v_2), \ldots , (u_m,v_m)$ satisfy the system of equations (3.1) with

$$ \begin{align*} v_i> \max(a_2^{1/k}, 2) \geq \max (H(\alpha), 2). \end{align*} $$

By Lemma 3.2, for $14 \leq i \leq m$ ,

$$ \begin{align*} \bigg|\frac{u}{v} - \alpha\bigg| < \frac{1}{v_i^{k-1/2}} \leq \frac{1}{v_i^{2.5}}, \end{align*} $$

as $k \geq 3$ . Since $\alpha =(a_1/a_2)^{1/k} < 1$ and $\max (u_i, v_i) = v_i$ , from Theorem 2.2, the number of such $v_i$ is $\mathcal {O} (\log k \,\log \log k)$ . This proves Theorem 1.2.

3.2 Proof of Theorem 1.3

Let $m = M_k (n)$ and $S = \{a_1, a_2, a_3, \ldots , a_m\}$ be a generalised m-tuple with the property $D_k(n)$ . Suppose $|n|^L < a_1 < a_2 < \cdots < a_m$ for some $L \geq 3$ . We consider the system of equations

(3.2) $$ \begin{align} \begin{split} a_1 x +n = u^k, \\ a_2 x +n = v^k. \end{split} \end{align} $$

As before, $x=a_i$ for $i \geq 3$ are solutions to this system. The statements of Lemmas 3.1 and 3.2 hold for all nonzero integers n. For $n>0$ , this was proved in [8].

Proof of Theorem 1.3(a)

Let $(u_1,v_1), \ldots , (u_m,v_m)$ satisfy the system of equations (3.2) with $v_i> \max (a_2^{1/k}, 2) \geq \max (H(\alpha ), 2)$ . By Lemma 3.2, for $14 \leq i \leq m$ ,

$$ \begin{align*} \bigg |\frac{u_i}{v_i} - \alpha \bigg| \leq \frac{1}{v_i^{k-1/2}} \leq \frac{1}{v_i^{2.5}}, \end{align*} $$

as $k \geq 3$ . Since $\alpha = (a_1/a_2)^{1/k} < 1$ and $\max (u_i, v_i) = v_i$ , applying Theorem 2.2 with $\kappa = 0.5$ shows that the number of $v_i$ satisfying the above inequality is

$$ \begin{align*}2^{25} (0.5)^{-3} \log (2k) \log ((0.5)^{-1} \log (2k)) = 2^{28} \log (2k) \log (2 \log (2k)). \end{align*} $$

So, for $k \geq 3$ , the total number of solutions is at most

$$ \begin{align*} 2^{28} \log (2k) \log (2 \log (2k)) + 14.\\[-37pt] \end{align*} $$

Proof of Theorem 1.3(b)

Let $S = \{a_1, a_2, \ldots , a_m\}$ be a generalised Diophantine m-tuple with property $D_k(n)$ such that each $a_i \leq |n|^3$ . Since $M_k (n;3)$ has a finite bound depending on k, it is enough to prove the statement for $|S|$ . We shall apply Gallagher’s larger sieve with primes $p \leq Q$ satisfying $p \equiv 1 \pmod k$ . Let $\mathcal {P}$ be the set of all primes $p \equiv 1\pmod k$ . For all such primes $p \in \mathcal {P}$ , there exists a Dirichlet character $\chi \, (\mathrm {mod\, } p)$ of order k.

Denote by $S_p$ the image of $S \pmod p$ for a given prime p. For $p\in \mathcal {P}$ , applying Lemma 2.3 with $\mathcal {A} = \mathcal {B} = S_p$ and $\chi \, (\mathrm {mod\, } p)$ a character of order k,

$$ \begin{align*} |S_p|(|S_p| - 1) \leq \sum_{a \in S_p-\{0\}} \sum_{b \in S_p} \chi(ab+n) + |S_p| \leq \sqrt{p} |S_p| + |S_p|. \end{align*} $$

Thus,

$$ \begin{align*} |S_p| \leq \sqrt{p}+2. \end{align*} $$

Take $N = |n|^3$ . Since $a_i \leq |n|^3$ , applying Theorem 2.1 yields

$$ \begin{align*} |S| \leq \frac{\sum_{p\in \mathcal{P}, p \leq Q } \log p- \log N}{\sum_{p\in \mathcal{P}, p \leq Q } \dfrac{\log p}{|S_p|}-\log N}. \end{align*} $$

By Theorem 2.4,

$$ \begin{align*} \sum_{\substack{p \leq Q \\ p \equiv 1 \mathrm{mod\, } k}} \log p = \frac{Q}{\phi(k)} + \mathcal{O} \bigg(\frac{Q}{\log Q}\bigg), \end{align*} $$

when $Q> Q_0(k)$ . As in Section 2.4, $\theta (Q;k,1) = \sum _{p \leq Q, p \equiv 1 \mathrm {mod\, } k} \log p$ . We take $f(t) = {1}/{(\sqrt t+2)}$ . By partial summation,

(3.3) $$ \begin{align} \sum_{\substack{p \leq Q \\ p \equiv 1 \mathrm{mod\, } k}} f(p) \log p = \theta (Q;k,1) f(Q) - \int_{2}^{Q} \theta (t;k,1) f'(t) \,dt. \end{align} $$

The right-hand side is equal to

$$ \begin{align*}\frac{Q}{\phi(k) (\sqrt{Q}+2)} + \mathcal{O}\bigg(\frac{\sqrt{Q}}{\log Q}\bigg) + \frac{1}{\phi(k)} \bigg(\int_{2}^{Q} \frac{t}{2(\sqrt{t}+2)^2 \sqrt{t}} \,dt\bigg) + \mathcal{O} \bigg(\int_{2}^{Q} \frac{1}{\sqrt{t} \log t} \,dt\bigg).\end{align*} $$

The three terms above can be estimated as

$$ \begin{align*} \frac{Q}{\phi(k) (\sqrt{Q}+2)} &= \frac{\sqrt{Q}}{\phi(k)} + \mathcal{O}(1),\\ \frac{1}{\phi(k)} \bigg(\int_{2}^{Q} \frac{t}{2(\sqrt{t}+2)^2 \sqrt{t}} \,dt\bigg) &= \frac{\sqrt{Q}}{\phi(k)} + \mathcal{O}(\log Q),\\ \mathcal{O} \bigg(\int_{2}^{Q} \frac{1}{\sqrt{t} \log t} \,dt\bigg) &= \mathcal{O}\bigg(\frac{\sqrt{Q}}{\log Q}\bigg). \end{align*} $$

Putting this together in (3.3) yields

$$ \begin{align*} \sum_{\substack{p \leq Q \\ p \equiv 1 \mathrm{mod\, } k}} \frac{\log p}{\sqrt{p}+2} = \frac{2 \sqrt{Q}}{\phi(k)} + \mathcal{O} \bigg(\frac{\sqrt{Q}}{\log Q}\bigg). \end{align*} $$

Thus,

$$ \begin{align*} |S| \leq \frac{\dfrac{Q}{\phi(k)} + \mathcal{O}\bigg(\dfrac{Q}{\log Q}\bigg) - \log N}{\dfrac{2 \sqrt{Q}}{\phi(k)} + \mathcal{O}\bigg(\dfrac{\sqrt{Q}}{\log Q}\bigg) - \log N}. \end{align*} $$

Choose $Q=(\phi (k) \log N)^2$ . Note that the condition $Q> Q_0(k)$ is the same as

(3.4) $$ \begin{align} \log N> \frac{\exp(0.015 \sqrt{k} (\log k)^3) }{ \phi(k)}. \end{align} $$

Since $k = o(\log \log |n|)$ , (3.4) holds for N large enough. Now, for both the numerator and the denominator, divide by $\log N$ to get

(3.5) $$ \begin{align} |S| \leq \frac{\phi(k)\log N -1 + \mathcal{O} \bigg(\dfrac{Q}{\log N \log Q}\bigg)}{1 + \mathcal{O}\bigg(\dfrac{\sqrt{Q}}{\log N \log Q}\bigg)}. \end{align} $$

Because $k=o(\log \log N)$ , it is easy to see that

$$ \begin{align*} \frac{\sqrt{Q}}{\log N \log Q} = \frac{\phi(k)}{2\log \phi(k) + 2\log \log N} = o(1). \end{align*} $$

Hence, from (3.5),

$$ \begin{align*} |S| &\leq \frac{\phi(k)\log N + \mathcal{O} \bigg(\dfrac{(\phi(k))^2\log N}{\log \log N}\bigg)}{1 + \mathcal{O}\bigg(\dfrac{\phi(k)}{\log \log N}\bigg)}. \end{align*} $$

As $\mathcal {O}({\phi (k)}/{\log \log N}) = o(1) $ , it follows that

$$ \begin{align*} \frac{1}{1 + \mathcal{O}\bigg(\dfrac{\phi(k)}{\log \log N}\bigg)}=1 + \mathcal{O}\bigg(\frac{\phi(k)}{\log \log N}\bigg). \end{align*} $$

So, we obtain

$$ \begin{align*} |S| &\leq \phi(k)\log N + \mathcal{O} \bigg(\frac{(\phi(k))^2\log N}{\log \log N}\bigg). \end{align*} $$

Since $N=|n|^3$ and $M_k(n) := \text {sup}\{|S|\}$ , we conclude that

$$ \begin{align*} M_k(n) \leq 3 \, \phi(k) \, \log |n| + \mathcal{O} \bigg(\frac{(\phi(k))^2 \log |n|}{\log \log |n|}\bigg) \end{align*} $$

as required.