1 Introduction
Let
$K = \mathbb {Q}(\sqrt {D})$
be the quadratic field corresponding to a fundamental discriminant D, and denote by
$\mathrm {Cl}_{D}$
its class group. For a prime p, the p-rank of a finite abelian group G is defined as
$\mathrm {r}_{p}(G) = \dim _{\mathbb {F}_{p}}(G / G^{p})$
; for brevity, we write
$r_{p}(D) = \mathrm {r}_{p}(\mathrm {Cl}_{D})$
. Class groups are among the most fundamental objects in algebraic number theory, yet many basic questions about their structure remain open. A central problem is to understand, for an odd prime p and a nonnegative integer r, how often the equality
$r_{p}(D) = r$
occurs as D runs over fundamental discriminants.
In 1984, Cohen and Lenstra [Reference Cohen, Lenstra and Jager7] proposed a celebrated heuristic that predicts a precise answer. Their conjecture states that for
$D> 0$
, the probability that
$r_{p}(D) = r$
equals
$$\begin{align*}p^{-r^{2}-r}\prod_{j=r+2}^{\infty}\bigl(1-p^{-j}\bigr)\prod_{j=1}^{r}\bigl(1-p^{-j}\bigr)^{-1}, \end{align*}$$
while for
$D < 0$
it is
$$\begin{align*}p^{-r^{2}}\prod_{j=r+1}^{\infty}\bigl(1-p^{-j}\bigr)\prod_{j=1}^{r}\bigl(1-p^{-j}\bigr)^{-1}. \end{align*}$$
A concise exposition of the Cohen–Lenstra heuristic for
$r_p(D)=0$
can be found in Chapter 5, Section 10 of [Reference Cohen6]. The conjecture is still open for every odd prime p and every nonnegative integer r. (The case
$p=2$
is more classical and can be handled via Gauss’s theory of genera.)
Research on the special case of the
$3$
-rank began even before the heuristic was formulated. In 1971, Davenport and Heilbronn [Reference Davenport and Heilbronn8] obtained a landmark asymptotic formula for the number of
$3$
-torsion elements in
$\mathrm {Cl}_{D}$
. Hartung [Reference Hartung9] proved the existence of infinitely many imaginary quadratic fields with
$r_{3}(D) = 0$
in 1974.
The theorem of Davenport and Heilbronn stimulated a rich line of further investigations. For example, by refining their method, Nakagawa and Horie [Reference Nakagawa and Horie12] extended Hartung’s result for discriminants subject to arbitrary congruence conditions. Later work [Reference Bhargava, Shankar and Tsimerman2, Reference Bhargava, Taniguchi and Thorne3, Reference Taniguchi and Thorne14] provided the secondary main term and sharpened the error term in the Davenport–Heilbronn asymptotic.
Building on the result of Nakagawa and Horie as a key tool, a considerable body of literature has since been devoted to the simultaneous vanishing of the
$3$
-rank for infinite families of quadratic fields. Byeon [Reference Byeon4] initiated this direction by showing that for any fixed square-free integer m there exist infinitely many fundamental discriminants
$D> 0$
, such that
$r_{3}(D)=r_{3}(mD)=0$
. Subsequently, Ito [Reference Ito11] proved that for any triple of square-free positive integer
$m_{1}, m_{2}, m_{3}$
, one can find infinitely many
$D> 0$
, with positive lower density satisfying
$\gcd (m_{1}m_{2}m_{3}, D) = 1$
and
$r_{3}(m_{1}D) = r_{3}(m_{2}D) = r_{3}(m_{3}D) = 0$
. In a different vein, Chattopadhyay and Saikia [Reference Chattopadhyay and Saikia5] showed that for any integer t, there exists a set of fundamental discriminants
$D> 0$
of positive density for which
$r_{3}(D)=r_{3}(D + 4t)=0$
. On the other hand, Iizuka [Reference Iizuka10] studied a slight variant of this problem and constructed an infinite family of pairs of imaginary quadratic fields
$\mathbb {Q}(\sqrt {D})$
and
$\mathbb {Q}(\sqrt {D+1})$
whose class numbers are divisible by 3.
We observe that existing work mainly addresses two separate types of quadratic field families: those of the form
$\mathbb {Q}(\sqrt {mD})$
examined by Byeon and Ito, and translated families of the form
$\mathbb {Q}(\sqrt {D + t})$
examined by Chattopadhyay and Saikia. The more general combined case
$\mathbb {Q}(\sqrt {mD + t})$
, and more broadly, families given by
$\mathbb {Q}(\sqrt {f(D)})$
for a polynomial f, has received little attention. This article is devoted to studying the
$3$
-rank problem for families of quadratic fields of the form
$\mathbb {Q}(\sqrt {f(D)})$
, where f is a general polynomial. To state our results precisely, we first introduce the following definition.
A pair of positive integers
$(A, B)$
is good if the following conditions hold:
-
(i) For every odd prime p dividing
$\operatorname {gcd}(A, B)$
, we have
$p^{2} \mid B$
but
$p^{2} \nmid A$
. -
(ii) When
$2 \mid B$
, either-
•
$A \equiv 1\ \pmod 4$
and
$B \equiv 0\ \pmod 4$
, or -
•
$A \equiv 8,12\ \pmod {16}$
and
$B \equiv 0\ \pmod {16}$
.
-
Theorem 1.1 Let
$n\ge 1$
,
$r\ge 0$
, and
$s\ge 0$
be integers satisfying
$$\begin{align*}4+r+3 s\le3^{n+1}\text{ and }r+s\ge1.\end{align*}$$
Suppose that we have
$r+s$
linear polynomials
$$ \begin{align*}f_{i}(x)=m_{i} x+n_{i},1 \le i \le r+s,\end{align*} $$
where the coefficients
$m_i, n_i $
are positive integers. If all pairs
$\left ( n_{i} , m_{i}\right )$
are good for
$ 1\le i\le r+s$
, then there exist infinitely many fundamental discriminants
$D>0$
, such that the following hold:
-
(1)
$r_3(-f_i(D))<n \text { for all } 1\le i\le s, \text {if } s\ge 1, r=0 ;$
-
(2)
$r_3(f_i(D))<n \text { for all } 1\le i\le r, \text {if } r\ge 1, s=0;$
-
(3)
$r_3(f_i(D))<n \text { and } r_3( -f_{r+j}(D)) <n \text { for any } 1\le i\le r, 1\le j\le s, \text {if } r\ge 1, s\ge 1.$
If we take
$r=0, s=3^{n}-2$
, and
$f_{i}(x)=x+i, 1 \le i \le s$
or
$r=3^{n+1}-4, s=0$
and
$f_{i}(x)=x+i, 1 \le i \le r$
in Theorem 1.1, then we obtain the following corollary.
Corollary 1.2 Let n be a positive integer. Then we have the following:
-
(1) There exist infinitely many fundamental discriminants
$D>0$
, such that for all
${1 \le i \le 3^{n}-2}$
,
$$\begin{align*}r_3(-D-i)<n. \end{align*}$$
-
(2) There exist infinitely many fundamental discriminants
$D>0$
, such that for all
${1 \le i \le 3^{n+1}-4}$
,
$$\begin{align*}r_3(D+i)<n. \end{align*}$$
A natural question is whether the linear polynomials in Theorem 1.1 can be replaced by polynomials of degree
$>1$
. For example, given positive integers k and n, one may ask if there are infinitely many fundamental discriminants
$D>0$
, such that for all
$1 \le i \le k$
,
$$\begin{align*}r_3(D^2+i)<n?\end{align*}$$
The method used to prove Theorem 1.1 is insufficient to address this question. However, by applying Bergelson and Leibman’s theorem on polynomial extensions of van der Waerden’s and Szemerédi’s theorems [Reference Bergelson and Leibman1], we can establish the following result:
Theorem 1.3 Let
$k, n$
be two positive integers and
$g_1(x), g_2(x), \ldots ,g_k(x)\in \mathbb {Q}[x]$
be polynomials that take integer values at the integers and have no constant terms. Then there exist positive integers
$a,d$
, such that for each
$1\le i\le k$
, the integer
$a+g_i(d)$
is a fundamental discriminant and
$$\begin{align*}r_3\bigl(a+g_i(d)\bigr) < n. \end{align*}$$
2 Preliminaries
In this section, we gather the required preliminaries. It is well known that an integer D is a fundamental discriminant of a quadratic field if either D is a square-free integer with
$D \equiv 1\ \pmod 4 $
or
$D=4m$
for some square-free integer
$m \equiv 2, 3\ \pmod 4 $
. Throughout, we denote by D a fundamental discriminant.
For any positive real number
$X $
, we denote by
$S_{1}(X)$
the set of positive fundamental discriminants
$0<D<X$
, and by
$S_{-1}(X)$
the set of negative fundamental discriminants
$-X<D<0$
. Let
$m\ge 1$
and
$N \ge 1$
be integers. For
$\lambda =\pm 1,$
we define
$$\begin{align*}S_\lambda(X, m, N):=\left\{D: D\in S_\lambda(X), D \equiv m\quad\pmod N\right\}. \end{align*}$$
The following theorem is due to Nakagawa and Horie [Reference Nakagawa and Horie12], which is a refinement of Davenport–Heilbronn theorem [Reference Davenport and Heilbronn8].
Theorem 2.1 (Nakagawa and Horie)
Suppose that the pair
$(m,N)\in \mathbb {N}^2$
is good. Then we have
$$\begin{align*}\sum_{D \in S_{\lambda}(X, m, N)} 3^{r_{3}(D)} \sim c(\lambda) \sum_{D \in S_{\lambda}(X, m, N)}1, \end{align*}$$
where
$c(\lambda )=\frac {4}{3}$
if
$\lambda =1$
, and
$2$
if
$\lambda =-1$
. Moreover,
$$\begin{align*}\sum_{D \in S_{\lambda}(X, m, N)}1\sim \frac{3X}{\pi^2 \varphi(N)}\prod_{p|N}\frac{q}{1+p},\end{align*}$$
where
$\varphi (\cdot )$
is the Euler’s totient function,
$q=4$
if
$p=2$
, and
$q=p$
otherwise.
Remark 2.2 If
$(6m,N)=1$
, then the second term and the error term in the Nakagawa–Horie theorem were given by Taniguchi and Thorne [Reference Taniguchi and Thorne14, Theorem 1.6], that is,
$$\begin{align*}\sum_{\substack{0\le D\le X\\D\equiv m\quad\pmod N }} 3^{r_{3}(D)} =\frac{4X}{\pi^2 N}\prod_{p|N}\frac{1}{1-p^{-2}} +cX^{\frac{5}{6}}+O(X^{18/23}N^{20/23}), \end{align*}$$
where c is a constant depending on N. It seems that this theorem also holds for any good pair
$(m,N)$
.
Proposition 2.3 Suppose that the pair
$(m,N)\in \mathbb {N}^2$
is good. Then for any
$n \in \mathbb {N}$
, we have
$$ \begin{align} \liminf_{X \rightarrow \infty} \frac{\sharp\left\{D \in S_{\lambda}(X, m, N) \mid r_{3}(D) <n \right\}}{\sharp S_{\lambda}(X, m, N)} \ge\frac{3^n-c(\lambda) }{3^{n}-1}, \end{align} $$
where
$c(\lambda )=\frac {4}{3}$
if
$\lambda =1$
, and
$2$
if
$\lambda =-1$
.
Proof We observe that
$$\begin{align*}\begin{aligned} \sum_{D \in S_{\lambda}(X, m, N)} 3^{r_{3}(D)} & =\sum_{\substack{D \in S_{\lambda}(X, m, N) \\ r_{3}(D)<n}} 3^{r_{3}(D)}+\sum_{\substack{D \in S_{\lambda}(X, m, N) \\ r_{3}(D) \geq n}} 3^{r_{3}(D)} \\ & \ge \sum_{\substack{D \in S_{\lambda}(X, m, N) \\ r_{3}(D)<n}} 1+3^n\sum_{\substack{D \in S_{\lambda}(X, m, N) \\ r_{3}(D) \geq n}} 1\\ &=\sum_{\substack{D \in S_{\lambda}(X, m, N) \\ r_{3}(D)<n}} 1+3^n\left(\sum_{\substack{D \in S_{\lambda}(X, m, N) }} 1-\sum_{\substack{D \in S_{\lambda}(X, m, N) \\ r_{3}(D) < n}} 1\right)\\ &=3^n\sum_{\substack{D \in S_{\lambda}(X, m, N) }} 1+(1-3^n)\sum_{\substack{D \in S_{\lambda}(X, m, N) \\ r_{3}(D) < n}} 1. \end{aligned} \end{align*}$$
Thus,
$$\begin{align*}\sum_{\substack{D \in S_{\lambda}(X, m, N) \\ r_{3}(D) < n}} 1 \ge \frac{3^n}{3^n-1}\sum_{\substack{D \in S_{\lambda}(X, m, N) }} 1-\frac{1}{3^n-1}\sum_{D \in S_{\lambda}(X, m, N)} 3^{r_{3}(D)}.\end{align*}$$
Dividing
$\sum _{\substack {D \in S_{\lambda }(X, m, N) }}1$
and applying Theorem 2.1, we complete the proof.
3 Proof of Theorem 1.1
In this section, we will prove Theorem 1.1. We begin by introducing some notations. For each i with
$1 \le i \le r+s$
, let us denote by
$u_{i}$
the highest power of
$2$
dividing
$m_i+n_i$
, that is,
$2^{u_i}||m_{i}+n_{i}$
. Put
$$\begin{align*}u=\max _{1 \le i \le r+s}\left\{u_{i}\right\}\end{align*}$$
and denote
$$\begin{align*}B=2^{u+2}.\end{align*}$$
It follows that for any fundamental discriminant
$D \equiv 1\ \pmod B$
, we have
$$\begin{align*}f_i(D)=m_iD+n_i\equiv m_i+n_i\quad\pmod B,\end{align*}$$
which implies that
$$ \begin{align}\frac{f_i(D)}{2^{u_i}}\equiv1\quad\pmod2.\end{align} $$
For convenience, we define
$$\begin{align*}\widetilde{f}_{i}(x):=\frac{f_{i}(x)}{2^{u_{i}}}=\frac{m_{i} x+n_{i}}{2^{u_{i}}},\qquad i=1, 2, \ldots, r+s. \end{align*}$$
Therefore, by (2) we see that
$\widetilde {f}_{i}(D)\in \mathbb {Z}$
and is odd for
$D\equiv 1\ \pmod B$
. Let us define
$l_i$
for
$1\le i\le r+s$
as
$$\begin{align*}l_{i}:= \begin{cases}1, & \text { if } \widetilde{f}_{i}(1) \equiv 1 \quad\pmod 4 , \\ 4, & \text { if } \widetilde{f}_{i}(1) \equiv 3 \quad\pmod 4.\end{cases} \end{align*}$$
Note that if
$\widetilde {f}_{i}(D)$
is square-free, then
$l_{i} \widetilde {f}_{i}(D)$
is a fundamental discriminant.
Lemma 3.1 For
$1\le i \le r+s$
, the pair
$( l_{i} \widetilde {f}_{i}(1), 4l_{i} m_{i})$
is good.
Proof Let p be an odd prime satisfying
$p\mid \operatorname {gcd}(l_{i} \widetilde {f}_{i}(1), 4 l_{i} m_{i})$
. Since
$p\mid m_i$
,
$p\mid l_{i}\widetilde {f}_{i}(1)=\frac {l_i(m_i+n_i)}{2^{u_i}}$
, we have
$p\mid n_i$
. Because
$(n_i,m_i)$
is good, we find that
$p^2|m_i$
and
$p^2\nmid n_i$
. This implies that
$p^2\mid 4l_im_i$
and
$p^2\nmid l_{i}\widetilde {f}_{i}(1)$
.
Now we verify that the pair
$(l_{i} \widetilde {f}_{i}(1), 4 l_{i} m_{i})$
satisfies (ii) of the definition of “good.” Since
$\widetilde {f}_{i}(1)$
is odd, we have two cases to be discussed according to
$\widetilde {f}_{i}(1) \equiv 1 \text { or }3\ \pmod 4$
. If
$\widetilde {f}_{i}(1) \equiv 1\ \pmod 4 $
, then
$l_{i}=1$
by definition. In this case, we have
$l_{i} \widetilde {f}_{i}(1) \equiv 1\ \pmod 4$
and
$4 l_{i} m_{i} \equiv 0\ \pmod 4$
. If
$\widetilde {f}_{i}(1) \equiv 3\ \pmod 4$
, then
$l_{i}=4$
. Thus,
$l_{i} \widetilde {f}_{i}(1) \equiv 12\ \pmod { 16} $
and
$4 l_{i} m_{i} \equiv 0\ \pmod { 16}$
. Thus, we conclude that the pair
$(l_{i} \widetilde {f}_{i}(1), 4 l_{i} m_{i})$
is good.
Let
$\mu (n)$
denote the usual Möbius function. For
$1 \le i \le r+s$
, we define
$$\begin{align*}T_{i}(X):=\left\{0<D< X: D\equiv 1\quad\pmod{2^{u_{i}+2}}, \mu(D)\mu\left(\widetilde{f}_{i}(D)\right) \neq 0\right\} \end{align*}$$
and
$$\begin{align*}T(X):=\left\{0<D< X: D\equiv 1\quad\pmod{ {B}}, \mu(D)\prod_{i=1}^{r+s}\mu\left(\widetilde{f}_{i}(D)\right) \neq 0\right\}. \end{align*}$$
Now we are in a position to prove Theorem 1.1.
Proof of Theorem 1.1
Let us denote
$$\begin{align*}A_i=\left\{D \in T(X): \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right) \geqslant n\right\} \end{align*}$$
and
$$\begin{align*}\overline{A_i}=\left\{D \in T(X): \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right) < n\right\}, i=1, 2, \ldots, r+s.\end{align*}$$
By the inclusion–exclusion principle, we have
$$\begin{align*}\begin{aligned} &\sharp\left\{D \in T(X): \max _{1 \le i \le r+s} \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right)<n\right\} \\ & =|\bigcap_{i=1}^{r+s}\overline{A_i}| \\ &\ge\sharp T(X) - \sum_{i=1}^{r+s}|A_i|. \end{aligned} \end{align*}$$
Since
$T(X) \subseteq T_{i}(X)$
, it follows that
$$\begin{align*}A_i\subseteq \left\{D \in T_i(X): \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right)\ge n\right\}.\end{align*}$$
Thus,
$$\begin{align*}\sharp\left\{D \in T(X): \max _{1 \le i \le r+s } \operatorname{rk}_{3}\left(\mathrm{Cl}_{\widetilde{f}_{i}(D)}\right)<n\right\}\end{align*}$$
$$\begin{align*}\ge\sharp T(X) - \sum_{i=1}^{r+s} \sharp\left\{D \in T_i(X): \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right)\ge n\right\}. \end{align*}$$
Dividing both sides by
$\sharp T(X)$
gives
$$ \begin{align} &\quad \frac{\sharp\left\{D \in T(X): \max _{1 \le i \le r+s} \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right)<n\right\}}{\sharp T(X)}\notag \\ & \geqslant 1-\sum_{i=1}^{r+s} \frac{\sharp\left\{D \in T_i(X): \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right) \geqslant n\right\}}{\sharp T(X)}\notag\\ &= 1-\sum_{i=1}^{r+s} \left(1-\frac{\sharp\left\{D \in T_i(X): \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right) <n\right\}}{\sharp T(X)}\right). \end{align} $$
We observe that each linear polynomial
$l_{i} \widetilde {f}_{i}(x)$
gives a bijection
$$\begin{align*}\begin{aligned} l_{i} \widetilde{f}_{i}: T_{i}(X) & \longrightarrow S_{\lambda_i}\left( l_{i}\widetilde{f}_{i}(X) , l_{i} \widetilde{f}_{i}(1), 4 l_{i} m_{i}\right), \\ D & \longmapsto \lambda_i l_{i} \widetilde{f}_{i}(D), \end{aligned} \end{align*}$$
where
$$\begin{align*}\lambda_i=\left\{ \begin{array}{ll} 1, &\text{ if }s=0 \text{ and } 1\le i\le r; \\ -1, & \text{ if }r=0 \text{ and } 1\le i\le s; \\ 1, &\text{ if }r\ge1,s\ge1 \text{ and } 1\le i\le r; \\ -1, &\text{ if }r\ge1,s\ge1 \text{ and } r+1\le i\le r+s. \end{array} \right. \end{align*}$$
It follows immediately that
$$\begin{align*}\sharp T_{i}(X)=\sharp S_{\lambda_i}\left( l_{i} \widetilde{f}_{i}(X), l_{i} \widetilde{f}_{i}(1), 4 l_{i} m_{i}\right). \end{align*}$$
Lemma 3.1 tells us that the pair
$( l_{i} \widetilde {f}_{i}(1), 4 l_{i} m_{i})$
is good. Since
$\mathrm {Cl}_{l_i\widetilde {f}_{i} (D)}=\mathrm {Cl}_{\widetilde {f}_{i}(D)}$
, we apply Proposition 2.3 for (3) and find that
$$ \begin{align*} &\quad \liminf_{X\rightarrow\infty} \frac{\sharp\left\{D \in T_{i}(X): \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right)< n\right\}}{\sharp T(X)}\notag \\ & \ge\liminf_{X\rightarrow\infty} \frac{\sharp\left\{D \in T_{i}(X): \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right)< n\right\}}{\sharp T_{i}(X)}\notag \\ &= \liminf_{X\rightarrow\infty} \frac{\sharp\left\{D \in S_{\lambda_i}\left( l_{i} \widetilde{f}_{i}(X), l_{i} \widetilde{f}_{i}(1), 4 l_{i} m_{i}\right): \operatorname{rk}_{3}\left({D}\right)< n\right\}}{\sharp S_{\lambda_i}\left( l_{i} \widetilde{f}_{i}(X), l_{i} \widetilde{f}_{i}(1), 4 l_{i} m_{i}\right)}\notag\\ &\ge\frac{3^n-c(\lambda_i) }{3^n-1}. \end{align*} $$
Inserting this inequality into (3), we obtain
$$ \begin{align} &\liminf_{X \rightarrow \infty} \frac{\sharp\left\{D \in T(X): \max _{1 \le i \le r+s} \operatorname{rk}_{3}\left({\widetilde{f}_{i}(D)}\right)<n\right\}}{\sharp T(X)}\notag\\ &\geqslant 1-\sum_{i=1}^{r+s} \frac{c\left(\lambda_i\right)-1}{3^{n}-1}\notag\\ &=1-\frac{\frac{1}{3}r}{3^{n }-1}-\frac{s}{3^{n}-1}\notag\\ &\ge \frac{1}{3^{n+1}-3}. \end{align} $$
To finalize the proof of the theorem, it is sufficient to show that the cardinality of
$T(X)$
has a positive density as X tends to infinity. For this purpose, we invoke a theorem established by Tsang [Reference Tsang15].
Theorem 3.2 (Tsang)
For positive integer t, let
$f_{i}(x)$
for
$1 \le i \le t$
be irreducible integral polynomials with degree not exceeding l, where
$(l \geqslant 3)$
. Let
$$ \begin{align} \Omega(l):=\prod_{p}\left(1-\omega(p) p^{1-l}\right), \end{align} $$
where for each prime
$p, \omega (p)$
denotes the number of solutions for
$f_{i}(x)\equiv 0\ \pmod {p^{l-1}}$
. If
$N(X)$
is the number of integers
$n \le X$
such that the values
$f_{i}(n)$
are all
$(l-1)$
-th power-free, then we have
$$\begin{align*}N(X)=\Omega(l) X+O\left(\frac{X}{(\log X)^{ 1-\frac{2}{l+1}}}\right). \end{align*}$$
The O-constant here depends on t as well as on the coefficients of the given polynomials.
Note that
$$\begin{align*}T(X)= \left\{n \le \frac{X-1}{B}: \mu(Bx+1)\prod_{i=1}^{r+s}\mu(\widetilde{f}_i(Bx+1)) \neq 0\right\}. \end{align*}$$
Applying Theorem 3.2 with
$l=3$
for irreducible polynomials
$Bx+1, \widetilde {f}_i(Bx+1), i=1,2,\ldots ,r+s$
, we obtain the asymptotic formula
$$ \begin{align} \sharp T(X)=\frac{\Omega(3)}{B} X+O\left(\frac{X}{\sqrt{\log X}}\right). \end{align} $$
We show that
$\Omega (3)\neq 0$
. To see this, let us first consider the case where
$p=2$
. We have
$Bx+1\equiv 1\ \pmod 4$
and
$$\begin{align*}\widetilde{f}_i(Bx+1)\equiv\widetilde{f}_i(1)\quad\pmod 4.\end{align*}$$
Since
$\widetilde {f}_i(1)$
is odd for all
$1\le i\le r+s$
, we conclude that
$w(2)=0$
. Suppose that p is an odd prime. We have three cases to solve the congruence equation
$$\begin{align*}\widetilde{f}_i(Bx+1)=\frac{ m_{i} Bx+m_i+n_{i} }{4^{u_{i}}}\equiv0\quad\pmod{p^2}.\end{align*}$$
Let
$N_i$
be the number of solutions of the equation above. It is clear that
$N_i=1$
if
$p^2\nmid m_1$
or
$p||m_i, p\mid n_i$
. If
$p||m_i, p\nmid n_i$
, then
$N_i=0$
. If
$p^2|m_i^2, p\nmid n_i$
, then
$N_i=0$
. If
$p^2|m_i^2, p\mid n_i$
, then the condition
$(m_i,n_i)$
is good implies that
$p^2\nmid n_i$
, so
$N_i=0$
. Thus, we find that
$w(p)\le 1$
for all
$p\ge 3$
. Therefore, we conclude that
$\Omega (3)\neq 0$
and the cardinality of
$ T(X)$
has a positive density.
Now combining (4) and (6), we find that
$$\begin{align*}\liminf _{X \rightarrow \infty} \frac{\sharp\left\{D \in T(X): \max _{1 \le i \le r+s} \operatorname{rk}_{3}\left({f_{i}(D)}\right)<n\right\}}{X} \geqslant \frac{\Omega(3)}{\left(3^{n+1}-3\right) B}>0. \end{align*}$$
This implies that there are infinitely many fundamental discriminants D satisfying
$\mathrm {rk}_{3}\left ({f_{i}(D)}\right )<n$
simultaneously for all
$1 \le i \le r+s$
. This completes the proof of Theorem 1.1.
4 Proof of Theorem 1.3
Now we apply the theorem of Bergelson and Leibman [Reference Bergelson and Leibman1, Theorem B] to prove Theorem 1.3. Bergelson–Leibman theorem states that if
$g_1(x), g_2(x), \ldots ,g_k(x)$
are polynomials with rational coefficients taking integer values at the integers and have no constant terms, then for every subset A of
$\{1, 2, \ldots ,n\}$
having a positive upper Banach density, that is,
$$\begin{align*}\lim_{N-M\rightarrow\infty}\sup_M\frac{ |A\cap\{M+1,\ldots,N\}|}{N-M}>0,\end{align*}$$
there exist positive integers
$a, d$
, such that
$a+g_1(d), a+g_2(d), \ldots , a+g_k(d)$
all belong to A. For a good pair
$(m,N)$
, let
$$\begin{align*}A=\left\{D \in S_{\lambda}(X, m, N) \mid r_{3}(D) <n \right\}.\end{align*}$$
Since a set having a positive lower limit implies that it has a positive upper Banach density, Theorem 1.3 follows immediately from Proposition 2.3 and Bergelson–Leibman theorem. In fact, a quantitative bound of Bergelson–Leibman theorem have been obtained recently in [Reference Shao and Wang13], which shows that for a positive constant c, any set
${A \subseteq \{1, 2, \ldots ,n\}}$
of density at least
$n/(\log n)^c$
contains a nontrivial polynomial progression of the form
$a+g_1(d), a+g_2(d),\ldots ,a+g_k(d)$
.
Acknowledgments
The authors gratefully acknowledge the anonymous referee for his/her valuable comments that immensely improved the results as well as the presentation of the article.
