On Cilleruelo's conjecture for the least common multiple of polynomial sequences

TL;DR

This paper investigates Cilleruelo's conjecture on the asymptotic growth of the least common multiple of polynomial sequences, proving a version of the conjecture for most shifts of a fixed polynomial.

Contribution

It establishes a near-universal version of Cilleruelo's conjecture for almost all shifts of a fixed polynomial, expanding understanding of LCM growth in polynomial sequences.

Findings

Proves the conjecture for almost all shifts of a polynomial.

Shows the growth rate of LCM aligns with the conjectured asymptotic.

Extends previous results to a broader class of polynomial shifts.

Abstract

A conjecture due to Cilleruelo states that for an irreducible polynomial $f$ with integer coefficients of degree $d\geq 2$, the least common multiple $L_f(N)$ of the sequence $f(1), f(2), \dots, f(N)$ has asymptotic growth $\log L_f(N)\sim (d-1)N\log N$ as $N\to \infty$. We establish a version of this conjecture for almost all shifts of a fixed polynomial, the range of $N$ depending on the range of shifts.

Equations

$$\log \mathrm{lcm}\{1,2,\dots ,N\}=\psi (N):=\sum _{n\le N}\Lambda (n)$$

$$\log \mathrm{lcm}\{1,2,\dots ,N\}\sim N.$$

$$L_f(N):=\mathrm{lcm}\{f(n):n=1,\dots ,n\}.$$

$$\log L_f(N)\sim (\mathrm{deg}f-1)N\log N,N\to \infty .$$

$$\log L_f(N)=N\log N+b_{f}N+o(N)$$

$$f_a(x)=f_0(x)-a$$

$$L_a(N)=\mathrm{lcm}\{f_0(n)-a:n=1,\dots ,N\}$$

$$T^{\frac{1}{d-1}}<N<\frac{T}{\log T}$$

$$\log L_a(N)\sim (d-1)N\log N.$$

$$P_a(N)=\prod _{n\le N}|f_0(n)-a|$$

$$P_a(N)=\prod _p{p}^{{\alpha }_p(a;N)}$$

$$\log P_a(N)=dN\log N+O(N).$$

$$L_a(N)=\prod _p{p}^{{\beta }_p(N)}$$

$$D(a)=\mathrm{disc}(f_0(x)-a)$$

$$\log L_a(N)=dN\log N-{\mathrm{Bad}}_N(a)-{\Delta }_N(a)-N{C}_N(a)+O(N)$$

$${\mathrm{Bad}}_N(a)=\sum _{\begin{array}{c}p\le N\\ p\mid D(a)\end{array}}{\alpha }_p(N)\log p$$

$${\Delta }_N(a)=\sum _{N<p\ll N^d}({\alpha }_p(N)-{\beta }_p(N))\log p$$

$$C_N(a)=\sum _{\begin{array}{c}p\le N\\ p\nmid D(a)\end{array}}\frac{\log p}{p-1}\rho (a;p)$$

$$C_N(a)\sim \log N.$$

$${\mathrm{Bad}}_N(a)\ll N(\log \log N{)}^{1+o(1)}.$$

$${\Delta }_N(a)\ll N(\log \log N{)}^{1+o(1)}.$$

$$⟨|C_N(a)-\log N{|}^2⟩\ll (\log \log N{)}^2$$

$$⟨{\mathrm{Bad}}_N(a)⟩\ll N\log \log N$$

$$⟨{\Delta }_N(a)⟩\ll N\log \log N.$$

$$D(a)=\mathrm{disc}(f_0(x)-a)$$

$$\log L_a(N)=d\log N-{\mathrm{Bad}}_N(a)-N{C}_N(a)-{\Delta }_N(a)+O(N)$$

$$P_a(N):=\prod _{n\le N}|f_a(n)|$$

$$P_a(N)=\prod _p{p}^{{\alpha }_p(N)}$$

$${\alpha }_p(N)=\sum _{n\le N}{\nu }_p(f_a(n))$$

$$L_a(N)=\prod _p{p}^{{\beta }_p(N)}$$

Full text

1. Introduction

1.1. Background

It is a well known and elementary fact that the least common multiple of all integers $1,2,\dots,N$ is exactly given by

[TABLE]

with $\Lambda(n)$ being the von Mangoldt function, and hence by the Prime Number Theorem,

[TABLE]

For a polynomial $f\in{\mathbb{Z}}[X]$, set

[TABLE]

The goal is to understand the asymptotic growth of $\log L_{f}(N)$ as $N\to\infty$.

In the linear case $\deg f=1$, we still have $\log L_{f}(N)\sim c_{f}N$ from the Prime Number Theorem in arithmetic progressions, see e.g. [1]. A similar growth occurs for products of linear polynomials, see [4], and for any polynomial with non-negative integer coefficients, there is a lower bound $\log L_{f}(N)\gg N$ [3]. However, in the case of irreducible polynomials higher degree, Cilleruelo [2] conjectured that the growth is faster than linear, precisely:

Conjecture 1.1**.**

If $f$ is an irreducible polynomial with $\deg f\geq 2$, then

[TABLE]

Cilleruelo proved Conjecture 1.1 for quadratic polynomials. Moreover, in that case there is a secondary main term

[TABLE]

see also [7]. No other case of Conjecture 1.1 is known to date. We do know that for any irreducible $f$ of degree $d\geq 3$, we have an upper bound $\log L_{f}(N)\lesssim(d-1)N\log N$ and one can prove $\log L_{f}(N)\gg N\log N$, as came up in a discussion with James Maynard [5].

We will show that Conjecture 1.1 holds for almost all $f$ in a suitable sense.

1.2. General setup

We fix a polynomial $f_{0}(x)\in{\mathbb{Z}}[x]$ of degree $d\geq 3$, which we assume is monic (or more generally, primitive - no prime divides all coefficients) and for $a\in{\mathbb{Z}}$ we set

[TABLE]

It is known that they are generically irreducible. Set

[TABLE]

We want to show that

Theorem 1.2**.**

For almost all $|a|\leq T$, and all $N$ satisfying

[TABLE]

we have

[TABLE]

Remark**.**

What one would like to show is that (1) holds for all $N>N_{0}(a)$, for all but $o(T)$ values of $|a|\leq T$. At this time we do not know how to do this.

1.3. Plan

Let

[TABLE]

We write down the prime power factorization

[TABLE]

If $T\ll N^{d-1}$ then $\alpha_{p}(a;N)=0$ for $p\gg N^{d}$, and (Lemma 2.3)

[TABLE]

We also write the prime power factorization of $L_{a}(N)$ as

[TABLE]

Let

[TABLE]

be the discriminant of $f_{0}(x)-a$. It is a polynomial in $a$ of degree $d-1$, with integer coefficients.

We show (Proposition 2.2)

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

where $\rho(a;d)=\#\{n\bmod d:f_{0}(n)-a=0\bmod d\}$.

We will show that for almost all $|a|\leq T$, with $N\log N<T<N^{d-1}$, we have

[TABLE]

[TABLE]

[TABLE]

Inserting these into (2) will prove Theorem 1.2.

To prove (3), (4), (5) we use averaging: Denoting by $\left\langle\bullet\right\rangle$ the average over all $|a|\leq T$ such that $f_{0}(x)-a$ is irreducible, we show that for $N\log N<T<N^{d-1}$,

[TABLE]

[TABLE]

[TABLE]

Noting that $\Delta_{N}(a),\operatorname{Bad}_{N}(a)\geq 0$ are non-negative, we obtain (3), (4), (5) from the Chebysehv/Markov inequality.

Remark**.**

In the deterministic case ($a$ fixed, $N\to\infty$), the quantities $\operatorname{Bad}_{N}$ and $C_{N}$ can be handled easily, as in the quadratic case $d=2$, see [2]. It is the quantity $\Delta_{N}(a)$ which we do not know how to show is $o(N\log N)$ individually (though the upper bound $O(N\log N)$ is easy). This is why we need to average over $a$. However, letting $a$ grow with $N$ introduces new problems, in particular for the study of $C_{N}$, which may need GRH to overcome individually. The results (6) and (7) for random $a$ are much easier and this is the method that we use.

Acknowledgements. We thank Shaofang Hong and Guoyou Qian for introducing the problem during a visit to Chengdu in 2017, and Lior Bary Soroker and James Maynard for discussions. The research was supported by the European Research Council under the European Union’s Seventh Framework Programme (FP7/2007-2013) / ERC grant agreements n${}^{\text{o}}$ 320755 and 786758.

2. Background

2.1. Generic irreducibility

Fix $f_{0}(x)\in{\mathbb{Z}}[x]$ monic, of degree $d\geq 2$. It is known that $f_{0}(x)-a$ is generically irreducible, in fact (see [9, §9.7])

Lemma 2.1**.**

Fix $f_{0}(x)\in{\mathbb{Z}}[x]$ of degree $d\geq 2$. Then the number of $|a|\leq T$ for which $f_{0}(x)-a$ is reducible is $O(\sqrt{T})$.

This is sharp in this generality, since for even degree $d=2m$, for the polynomial $f_{0}(x)=x^{2m}$ we have $x^{2m}-a$ is reducible whenever $a=b^{2}$ is a perfect square.

Denote

[TABLE]

the discriminant of $f_{a}(x)$, which is a polynomial in $a$ of degree $\leq d-1$ with integer coefficients (depending on the coefficients of $f_{0}$). We assume that $a$ is such that $f_{0}(x)-a$ is irreducible, and therefore $D(a)$ is not zero, i.e. $f_{a}$ has no multiple roots.

Examples:

i) $f_{0}(x)=x^{3}$, then $\operatorname{disc}(f_{0}(x)-a)=\operatorname{disc}(x^{3}-a)=-27a^{2}$.

ii) $f_{0}(x)=x^{3}-3x$ then $\operatorname{disc}(f_{0}(x)-a)=-27(a-2)(a+2)$.

2.2. A decomposition

Proposition 2.2**.**

For $|a|\leq N^{d-1}$ such that $f_{0}(x)-a$ is irreducible, we have

[TABLE]

For $a\in{\mathbb{Z}}$ such that $f_{a}(x)=f_{0}(x)-a$ is irreducible, let

[TABLE]

which is nonzero since $f_{a}$ has no rational roots, and write the prime power decomposition as

[TABLE]

so that

[TABLE]

where $\nu_{p}(m):=\max(k\geq 0:p^{k}\mid m)$.

Following Cilleruelo [2], we want to relate $\log L_{a}(N)$ to $\log P_{a}(N)$, which is clearly bigger. We write the prime power decomposition of $L_{a}(N)$ as

[TABLE]

with

[TABLE]

Using the prime factorization of $L_{a}(N)$ and $P_{a}(N)$ we have

[TABLE]

where we have separated out the contribution of primes $p\leq N$, and the larger ones. We further break off the contribution of primes $p\leq N$ which divide the discriminant $D(a)=\operatorname{disc}(f_{a})$, by setting

[TABLE]

and abbreviate the contribution of big primes $p>N$ as

[TABLE]

Note that $\operatorname{Bad}_{N},\Delta_{N}\geq 0$ are both non-negative. We obtain an expression

[TABLE]

2.3. The quantity $P_{a}(N)$

Lemma 2.3**.**

For $f_{0}(x)\in{\mathbb{Z}}[x]$ monic of degree $d$, and $|a|\ll N^{d-1}$ so that $f_{0}(x)-a$ is irreducible, we have

[TABLE]

Proof.

Write

[TABLE]

Since we assume $f_{0}(x)-a$ is irreducible, non of the factors $f_{0}(n)-a$ can vanish so that $\log P_{a}(N)$ is well defined. If $f_{0}(x)=x^{d}+c_{d-1}x^{d-1}+\dots$, we have for $n\leq N$

[TABLE]

Consider first $n$’s satisfying $N/\log N<n\leq N$, for which we use (recall $|a|\leq N^{d-1}$)

[TABLE]

so that

[TABLE]

For $1\leq n\leq N/\log N$, we just use $1\leq|f_{0}(n)-a|\ll N^{d}$ so that $0\leq\log|f_{0}(n)-a|\ll\log N$, and

[TABLE]

Hence

[TABLE]

as claimed. ∎

2.3.1. Dealing with $\beta_{p}(N)$

For $a$ such that $f_{0}(x)-a$ is irreducible, we have

[TABLE]

because

[TABLE]

and since $f_{0}(n)-a\neq 0$ for all $n$, if $p^{k}\mid f_{0}(n)-a\neq 0$ then

[TABLE]

Hence since $|a|\lll N^{d-1}$,

[TABLE]

and hence the contribution of primes $p\leq N$ to (10) is

[TABLE]

2.3.2. Dealing with $\alpha_{p}(N)$

Using Hensel’s lemma, it is easy to see that ([6] see also [2, Lemma 4]):

Lemma 2.4**.**

For $p\nmid D(a)=\operatorname{disc}(f_{0}(x)-a)$, we have

[TABLE]

where $\rho(a;p)=\#\{n\bmod p:f_{0}(n)-a=0\bmod p\}$.

Consequently, we find that in (10),

[TABLE]

where

[TABLE]

Therefore we have proven Proposition 2.2.

3. Bounding $\operatorname{Bad}_{N}$ almost surely

Recall that we defined

[TABLE]

(we assume that $f_{0}(x)-a$ is irreducible).

We denote the averaging operator over $|a|\leq T$ such that $f_{0}(x)-a$ is irreducible by

[TABLE]

The number of $|a|\leq T$ for which $f_{0}(x)-a$ is reducible is $O(\sqrt{T})$ (Lemma 2.1), so that

[TABLE]

Proposition 3.1**.**

If $T\geq N$ but $\log T\ll\log N$ then

[TABLE]

Proof.

We separate out the contribution $B_{1}(a)$ of $k=1$ and the contribution $B_{2}(a)$ of the remaining $k\geq 2$:

[TABLE]

where

[TABLE]

and

[TABLE]

We will show that

[TABLE]

and that

[TABLE]

proving Proposition 3.1

We first show that

[TABLE]

which suffices for (13) since $\log|D(a)|\ll\log T\ll\log N$.

Indeed, for $p\leq N$ we have

[TABLE]

where

[TABLE]

which we see by dividing the interval $[1,N]$ into consecutive intervals of length $p$.

Since $f_{0}(x)$ is a monic polynomial of degree $d$, it is nonzero modulo $p$ and still of degree $\leq d$, hence $\rho(a;p)\leq d$. Thus

[TABLE]

We use:111This is standard.

Lemma 3.2**.**

For $k>1$

[TABLE]

Proof.

Indeed, splitting the sum into small primes $p\leq\log k$, and the rest (where the summands are at most $\log\log k/\log k$), we get

[TABLE]

since the number of distinct prime divisors of $k$ is $\ll\log k/\log\log k$. ∎

Therefore

[TABLE]

and obtain

[TABLE]

Next we bound the mean value of $B_{2}(a)$

[TABLE]

Now if $f_{0}(x)-a$ is irreducible, then $f_{0}(n)-a\neq 0$ and so if $p^{k}\mid f_{0}(n)-a$ with $n\leq N$ then $k\ll\log N/\log p$, so we restrict the summation to $2\leq k\ll\log N/\log p$. Moreover, given $n$, the condition $f_{0}(n)=a\bmod p^{k}$ determines $a$ modulo $p^{k}$, so there are $\ll T/p^{k}+1$ choices for $a$. Hence we may bound

[TABLE]

we have

[TABLE]

and

[TABLE]

Altogether we find

[TABLE]

which is $O(N)$ if $T\geq N$. ∎

4. Averaging $\Delta_{N}(a)$

Let

[TABLE]

Then clearly $\Delta_{N}\geq 0$, and we want to show

Proposition 4.1**.**

Assume that $T\geq N\log N$, but $\log T\ll\log N$. Then

[TABLE]

4.1. Preparations

Let

[TABLE]

which, given $n$, is a (nonzero) polynomial in $m$, of degree $\leq d-1$. If $f_{0}$ is monic then so is $G(m,n)$ so its degree is exactly $d-1$.

Lemma 4.2**.**

There is some $C_{1}=C_{1}(f_{0})$ so that if $m,n\geq 1$ and $\max(m,n)>C_{1}$ then $G(m,n)\neq 0$.

Proof.

We have

[TABLE]

and if $j\geq 2$ then for $n=\max(m,n)$,

[TABLE]

while

[TABLE]

so that (assuming $f_{0}$ monic, so $c_{d}=1$)

[TABLE]

which is clearly positive once $n$ is sufficiently large in terms of the coefficients $c_{1},\dots,c_{d-1}$ of $f_{0}$. ∎

Lemma 4.3**.**

There is some $C(d)>0$ so that for all $|a|\leq N^{d}$, such that $f_{a}(x)=f_{0}(x)-a$ is irreducible, we have $\alpha_{p}(N)\leq C(d)$ if $p>N$. Moreover $\alpha_{p}(N)=0$ unless $p\ll N^{d}+|a|$.

Proof.

We have by definition

[TABLE]

Since we assume that $f_{a}(x)=f_{0}(x)-a$ is irreducible, hence has no rational zeros, we must have, if $p\mid f_{a}(n)$, that $p\leq|f_{a}(n)|\ll N^{d}+|a|\ll N^{d}$ uniformly in $|a|\leq T$ (recall $T\leq N^{d}$). Hence $\alpha_{p}(N)=0$ for $p\gg N^{d}$.

Given $n$ so that $p\mid f_{a}(n)$, with $p>N$, we claim that there are at most $d$ such integers:

[TABLE]

Indeed, for any $c\in{\mathbb{Z}}/p{\mathbb{Z}}$, the number of solutions $m\bmod p$ of $f_{a}(m)=c\bmod p$ is at most $d$, and since $p>N$, this certainly applies to those $m\leq N$ which solve $f_{a}(m)=c$ with $c=f_{a}(n)$.

Moreover, if $p>N$, the maximal $k$ so that $p^{k}\mid f_{0}(n)-a$ for some $n\leq N$ is, because we assume $f_{a}(n)\neq 0$,

[TABLE]

because we assume that $|a|\leq T$ with $\log T\ll\log N$.

Therefore

[TABLE]

∎

4.2. A preliminary bound on $\Delta_{N}(a)$

Lemma 4.4**.**

If $a$ is such that $f_{0}(x)-a$ has no rational zeros, and $\log|a|\ll\log N$, then

[TABLE]

Proof.

We have $\alpha_{p}(N)\neq\beta_{p}(N)$ if and only if there are two distinct integers $m,n\leq N$ so that $p\mid f_{a}(m)$ and $p\mid f_{a}(n)$. Using Lemma 4.3, we see that $\alpha_{p}(N)-\beta_{p}(N)=O_{d}(1)$ for $p>N$, and hence applying a union bound we obtain, if $a$ is such that $f_{a}(x)$ has no rational zeros,

[TABLE]

Note that if $p\mid f_{a}(m)$ and $p\mid f_{a}(n)$ then $p\mid f_{a}(m)-f_{a}(n)=(m-n)G(m,n)$ and so since $p\nmid m-n$ (because $1\leq n-m\leq N-1<p$), we must have $p\mid G(m,n)$. Thus

[TABLE]

We break off the terms corresponding to $G(m,n)=0$. According to Lemma 4.2, the condition $G(m,n)=0$ forces $m,n\leq C_{1}$ to be bounded. Hence the contribution of such pairs to (15) is bounded by

[TABLE]

Note that $0<|f_{0}(m)-a|\ll|a|+1$ if $m\leq C_{1}$ (we assume that $a$ is such that $f_{0}(x)-a$ has no rational zeros, hence $f_{0}(m)-a\neq 0$, and hence the number of primes $p>N$ dividing $f_{0}(m)-a$ is at most $\ll\log|a|/\log N$. Hence the contribution of pairs $m<n$ with $G(m,n)=0$ to (15) is at most $\ll\log|a|$. Thus

[TABLE]

Finally, the assumption $\log|a|\ll\log N$ gives (14). ∎

4.3. Proof of Proposition 4.1

Now to average over $|a|\leq T$ (such that $f_{0}(x)-a$ is irreducible). Using (14), noting that $\log|a|\ll\log T\ll\log N$ gives

[TABLE]

Given $1\leq m<N$, and $N<p\ll N^{d}$, the number of $|a|\leq T$ with $a=f_{0}(m)\bmod p$ is $\ll T/p+1$. Hence

[TABLE]

To treat the sum $II$, we note if $m,n\leq N$, then $|G(m,n)|\leq C(f_{0})N^{d-1}$ and so there are at most $d-1$ distinct primes $p>N$ which divide $G(m,n)$ (which we assume is non-zero), and for these $\log p\ll\log N$. Therefore

[TABLE]

which is $O(N)$ if $T>N\log N$.

To treat the sum $I$, we separate the prime sum into primes with $N<p\leq N\log N$ and the remaining large primes $N\log N<p\ll N^{d-1}$ to get

[TABLE]

We treat the sum over small primes by switching the order of summation

[TABLE]

Now given $m$, the congruence $G(m,n)=0\bmod p$ (if solvable) determines $n\bmod p$ up to $d-1$ possibilities, since $G(m,n)$ is a monic polynomial of degree $d-1$ in $n$, and since $n\leq N<p$ it means that $n$ is determined as an integer up to $d-1$ possibilities. Hence

[TABLE]

and the sum over small primes is bounded by

[TABLE]

on using Mertens’ theorem.

The sum over large primes is treated by using $\log p/p\ll 1/N$ for $p>N\log N$, giving

[TABLE]

Now given $1\leq m<n\leq N$ with $G(m,n)\neq 0$, there are at most $d-1$ primes $p>N\log N$ dividing $G(m,n)\ll N^{d-1}$, so that the contribution of large primes is bounded by

[TABLE]

This gives $I\ll N\log\log N$, and hence

[TABLE]

as claimed. ∎

5. Almost sure behaviour of $C_{N}$

5.1.

Let $f\in{\mathbb{Z}}[x]$ be an irreducible polynomial, and let $\rho_{f}(p)$ be the number of distinct roots of the polynomial $f$ modulo a prime $p$. It is well known that for fixed $f$, the mean value of $\rho_{f}(p)$ over all primes is $1$ [6]:

[TABLE]

We write

[TABLE]

where $\sigma_{f}(p)$ is a fluctuating quantity, having mean zero.

Now fix $f_{0}(x)=x^{d}+c_{d-1}x^{d-1}+\dots+c_{1}x\in{\mathbb{Z}}[x]$, a monic polynomial of degree $d$, and for $a\in{\mathbb{Z}}$ set

[TABLE]

Write $\rho(a;p)=\rho_{f_{a}}(p)$, $\sigma(a;p)=\sigma_{f_{a}}(p)$. Note that $0\leq\rho(a;p)\leq d$.

We write

[TABLE]

where

[TABLE]

and

[TABLE]

By Mertens’ theorem

[TABLE]

The contribution $E_{N}(a)$ of primes dividing the discriminant $D(a)=\operatorname{disc}(f_{0}(x)-a)$ can be bounded individually, for $|a|\leq T\ll N^{d}$, using Lemma 3.2 (assuming $D(a)\neq 0$)

[TABLE]

Since $D(a)$ is a polynomial of degree $d-1$ in $a$, and $|a|\leq T\ll N^{d}$, we find

[TABLE]

which is negligible relative to the main term. Hence

[TABLE]

In the following part, we will establish the following upper bound on the second moment of $D_{N}(a)$:

Proposition 5.1**.**

For $T\geq N\log N$, the second moment of $D_{N}(a)$ satisfies:

[TABLE]

Using the triangle inequality and Cauchy-Schwartz, we obtain

Proposition 5.2**.**

[TABLE]

As a consequence, we deduce our main objective for this section:

Proposition 5.3**.**

For almost all $|a|\leq T$ (with $N\log N\leq T\ll N^{d-1}$)

[TABLE]

5.2. Proof of Proposition 5.1

Proof.

Expanding, we have

[TABLE]

The diagonal contribution $p=q$ gives

[TABLE]

Now note that

[TABLE]

is uniformly bounded. This is because the polynomial $f_{0}(x)-a$ is monic of degree $d$, hence has at most $d$ zeros modulo $p$, so that $0\leq\rho(a;p)\leq d$ and so $-1\leq\sigma(a;p)\leq d-1$. Thus we obtain a bound for the diagonal sum

[TABLE]

For the off-diagonal terms, we use

Lemma 5.4**.**

For distinct primes $p\neq q$,

[TABLE]

Therefore, given Lemma 5.4, we obtain

[TABLE]

which is $O(1)$ if $T\geq N\log N$, proving Proposition 5.1. ∎

5.3. Proof of Lemma 5.4

For the argument, it will be important to have $a$ run over an interval. So we first remove the restriction in the averaging, that $f_{0}(x)-a$ is irreducible. Since $-1\leq\sigma(a;p)\leq d-1$, this introduces an error bounded by

[TABLE]

and so

[TABLE]

We express $\rho(a;p)$ as an exponential sum:

[TABLE]

The term $t=0$ contributes the main term of $1$, and we obtain the following expression for $\sigma(a;p)=\rho(a;p)-1$:

[TABLE]

where $e(z):=e^{2\pi iz}$. Set

[TABLE]

Using (16), we have on switching orders of summation

[TABLE]

Weil’s bound [10, 8] shows that there is a constant $c(d)>0$, so that all primes $p$ and $b$ coprime to $p$

[TABLE]

In fact for any $f_{0}\in{\mathbb{Z}}[x]$ with $f_{0}(x)$ primitive of degree $d$, if $p>d$ then $|S_{f_{0}}(b,p)|\leq(d-1)\sqrt{p}$.

Hence we find

[TABLE]

where we have used that if $p\neq q$ are distinct primes, then as $t$ and $s$ vary over all invertible residues modulo $p$ (resp., modulo $q$), $tq+sp\bmod pq$ covers all invertible residues modulo $pq$ exactly once.

We sum the geometric progression

[TABLE]

where $||\alpha||=\operatorname{dist}(\alpha,{\mathbb{Z}})$. We may take $1\leq m<pq/2$ and then the bound is $\ll pq/m$. This will give

[TABLE]

proving Lemma 5.4. ∎