Calcification in human vessels and valves: from pathological point of view

Yu Sato; Hiroyuki Jinnouchi; Atsushi Sakamoto; Anne Cornelissen; Masayuki Mori; Rika Kawakami; Kenji Kawai; Renu Virmani; Aloke V. Finn; Yu Sato; Hiroyuki Jinnouchi; Atsushi Sakamoto; Anne Cornelissen; Masayuki Mori; Rika Kawakami; Kenji Kawai; Renu Virmani; Aloke V. Finn

doi:10.3934/molsci.2020009

AIMS Molecular Science

2020, Volume 7, Issue 3: 183-210. doi: 10.3934/molsci.2020009

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Review Special Issues

Calcification in human vessels and valves: from pathological point of view

1.
CVPath Institute, Gaithersburg, MD
2.
University of Maryland, School of Medicine, Baltimore, MD, USA

Received: 27 March 2020 Accepted: 29 May 2020 Published: 02 June 2020

Vascular and valvular calcification are commonly encountered in clinical medicine and a greater understanding of their significance and pathophysiology remain a subject of immense importance. In the coronary arteries, vascular calcification burden correlates with the severity of luminal stenosis and atherosclerotic plaque burden. While in progressive lesions, the presence of coronary calcification is not binary but rather depends on the type of calcification. Racial and gender differences, and comorbidities like diabetes mellitus and chronic kidney disease, all affect the presence and severity of calcification. The peripheral arteries of the lower extremities are affected by both medial calcification and intimal calcification, and the former barely contributes to luminal stenosis. The character of atherosclerosis differs between above-knee and below-knee lesions. Valvular calcification generally occurs on the aortic valve leaflets, and pathologic findings range from minimal fibrocalcific changes in early disease to end-stage lesions characterized by fibrotic thickening and nodular calcification. Valvular calcification is similar to atherosclerotic changes, in terms of lipid deposition, inflammation, osteogenic differentiation of valvular interstitial cells, and oxidative stress. However, the mechanisms of vascular and valvular calcification are still not well understood. A deeper understanding of vascular and valvular calcification is needed in order to develop effective anti-calcification therapies and to improve outcomes in these patients.

Keywords:

Citation: Yu Sato, Hiroyuki Jinnouchi, Atsushi Sakamoto, Anne Cornelissen, Masayuki Mori, Rika Kawakami, Kenji Kawai, Renu Virmani, Aloke V. Finn. Calcification in human vessels and valves: from pathological point of view[J]. AIMS Molecular Science, 2020, 7(3): 183-210. doi: 10.3934/molsci.2020009

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Abstract

1. Introduction

Let $q, m, n\in \mathbf{Z}^{+}$ with $q > 2$ and $m > n\geq 1$ . For any $u, v\in\mathbf{Z}$ , we are concerned with the two-term exponential sums

$\mathcal{G}(u,v,m, n;q) = \sum\limits_{j = 1}^{q} \mathrm{e}_{q}\left(uj^m+vj^n\right),$

where $\mathrm{e}_{q}(x) = \mathrm{exp}\left(2\pi \mathrm{i}x/ q\right)$ and $\mathrm{i}^2 = -1$ .

For convenience, the following letters and symbols are commonly used in this paper and should be interpreted in the following sense unless otherwise stated.

● $\chi$ is Dirichlet character.

● $\chi_{k}$ is $k$ -order Dirichlet character.

● $\phi(a)$ is Euler function.

● $\alpha$ is uniquely determined by $4p = \alpha^2+27\beta^2$ and $\alpha\equiv 1\bmod 3$ .

● $\tau(\chi)$ is Gauss sums defined by

$\tau(\chi) = \sum\limits_{s = 1}^{q}\chi(s) \mathrm{e}_{q}\left(s\right).$

The mean value calculation and upper bound estimation of exponential sums has always been a classical problem in analytic number theory. As a special kind of exponential sums, Gauss sums have had an important effect on both cryptography and analytic number theory. Analytic number theory and cryptography will benefit greatly from any significant advancements made in this area. In this paper, we will estimate and calculate the fourth power mean value of two-term exponential sums weighted by a character $\chi_{3}$ . In this field, many scholars have investigated the results of $\mathcal{G}(u, v, m, n; q)$ in various forms, and obtained many meaningful results, see ^{[3,5,7,8,11,15]}. For instance, Zhang and Zhang ^[9] obtained the power mean about $\mathcal{G}(u, v, 3, 1; p)$

$\begin{eqnarray} \sum\limits_{u = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3+vi\right)\right|^4 = \left\{ \begin{array}{ll} 2p^3-p^2 & \text{ if}\quad 3\nmid p-1 , \nonumber\\ 2p^3-7p^2 & \text{ if}\quad 3\mid p-1 ,\nonumber \end{array}\right. \end{eqnarray}$

where $p$ is an odd prime and $v$ is not divisible by $p$ .

Wang and Zhang ^[6] obtained the eighth power mean of $\mathcal{G}(u, v, 3, 1; p)$

$\begin{eqnarray} \sum\limits_{u = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3+vi\right)\right|^8 = \left\{ \begin{array}{ll} 7(2p^5-3p^4) & \mathrm{ if}\quad 6\mid p-5 , \nonumber\\ 14p^5-75p^4-8p^3\alpha^2 & \mathrm{ if}\quad 6\mid p-1 .\nonumber \end{array}\right. \end{eqnarray}$

In addition, Zhang and Han ^[12] shown the power mean of $\mathcal{G}(1, v, 3, 1; p)$

$\begin{eqnarray} \sum\limits_{v = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(i^3+vi\right)\right|^6 = 5p^4-8p^3-p^2, \end{eqnarray}$

(1.1)

where $p$ is an odd prime with $3\nmid \phi(p)$ .

But if $3\mid \phi(p)$ , whether there exists an exact formula for (1.1). Consider the mean of the simplest

$\begin{eqnarray} \sum\limits_{v = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(i^3+vi\right)\right|^4. \end{eqnarray}$

(1.2)

It is worth mentioning that, Zhang and Zhang ^[10] studied the power mean of the exponential sums weight by $\chi_{2}$ , one has the identities

$\begin{eqnarray} \sum\limits_{u = 1}^{p-1}\chi_{2}(u)\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3+i\right)\right|^4 = \left\{ \begin{array}{ll} p^2(\zeta+3) & \mathrm{ if}\quad 6\mid p-5 , \nonumber\\ p^2(\zeta-3) & \mathrm{ if}\quad 6\mid p-1 ,\nonumber \end{array}\right. \end{eqnarray}$

where $\zeta = \sum_{t = 1}^{p-1}\left(\frac{t-1+\overline{t}}{p}\right)$ with $\zeta\in\mathbf{Z}$ satisfies inequality $|\zeta|\leq2\sqrt{p}$ .

Cao and Wang (see Lemma 3 in ^[2]) proved the following conclusion, that is, if $p$ is a prime with $3\mid \phi(p)$ , then for any $\chi_{3}\bmod p$ , one has the identity

$\begin{eqnarray*} \sum\limits_{u = 1}^{p-1}\chi_{3}(u)\left(\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3+i\right)\right)^4 = \big(\overline{\chi_{3}}(3)-3p-\overline{\chi_{3}}(3)p\big) \tau^2\left(\overline{\chi_{3}}\right)-\alpha p\tau(\chi_{3}). \end{eqnarray*}$

Unfortunately, this lemma is incorrect, there is a calculation error in it. It is precisely because of the computational error in this lemma that the main result in the whole text is wrong.

The following year, Zhang and Meng ^[16] studied the power mean of $\mathcal{G}(u, 1, 3, 1; p)$ weighted by $\chi_{2}$ . In this paper, We intend to correct the error in ^[2] and give a correct conclusion. At the same time, as an application, we give an exact result for (1.2). That is, it will prove these two conclusions:

Theorem 1. If $p$ is a prime with $3\mid \phi(p)$ , then we have

$\begin{eqnarray*} \sum\limits_{u = 1}^{p-1}\chi_{3}(u)\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3+i\right)\right|^4 = -\alpha p\tau(\chi_{3})-3p\cdot\tau^2(\overline{\chi_{3}}). \end{eqnarray*}$

Theorem 2. If $p$ is a prime with $3\mid \phi(p)$ , then we have

$\begin{eqnarray*} \sum\limits_{v = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(i^3+vi\right)\right|^4 = 2p^3-p^2-3pA^2_k-p\alpha A_k, \end{eqnarray*}$

where $A_k = \omega^k\left[{\frac{\alpha p}{2}+\left(\left(\frac{\alpha p}{2}\right)^{2}-p^{3}\right)^{\frac{1}{2}}}\right]^{\frac{1}{3}}+\omega^{-k}\left[{\frac{\alpha p}{2}-\left(\left(\frac{\alpha p}{2}\right)^{2}-p^{3}\right)^{\frac{1}{2}}}\right]^{\frac{1}{3}}$ , $k = 1, 2$ or $3$ is dependent on $p$ , and $\omega = \frac{-1+\sqrt{3}i}{2}$ .

Corollary 1. If $p$ is a prime with $3\mid \phi(p)$ , then we have the asymptotic formula

$\begin{eqnarray*} &&\sum\limits_{v = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(i^3+vi\right)\right|^4 = 2p^3+O\left(p^2\right). \end{eqnarray*}$

Corollary 2. If $p$ is a prime with $3\mid \phi(p)$ , then for any integer $l$ , we have recursive formula

$\begin{eqnarray*} &&V_l(p) = \sum\limits_{u = 1}^{p-1}\vartheta^{l}(u)\left|\sum\limits_{j = 1}^{p} \mathrm{e}_{p}\left(uj^3+j\right)\right|^4\\ & = &\alpha p\sum\limits_{u = 1}^{p-1}\vartheta^{l-3}(u)\left|\sum\limits_{j = 1}^{p} \mathrm{e}_{p}\left(uj^3+j\right)\right|^4+3p\sum\limits_{u = 1}^{p-1}\vartheta^{l-2}(u)\left|\sum\limits_{j = 1}^{p} \mathrm{e}_{p}\left(uj^3+j\right)\right|^4\\ & = & \alpha p V_{l-3}(p)+ 3pV_{l-2}(p), \end{eqnarray*}$

when $l$ take $1$ – $3$ , the following equations hold

$\begin{eqnarray*} V_1(p) = \sum\limits_{u = 1}^{p-1}\vartheta(u)\left|\sum\limits_{j = 1}^{p} \mathrm{e}_{p}\left(uj^3+j\right)\right|^4& = &-5\alpha p^2,\\ V_2(p) = \sum\limits_{u = 1}^{p-1}\vartheta^{2}(u)\left|\sum\limits_{j = 1}^{p} \mathrm{e}_{p}\left(uj^3+j\right)\right|^4& = &4p^4-20p^3-\alpha^2p^2,\\ V_3(p) = \sum\limits_{u = 1}^{p-1}\vartheta^{3}(u)\left|\sum\limits_{j = 1}^{p} \mathrm{e}_{p}\left(uj^3+j\right)\right|^4& = &2\alpha p^4-22\alpha p^3, \end{eqnarray*}$

where $\vartheta(u) = \sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3\right)$ .

In fact, with the third-order linear recursive formula in Corollary 2 and its three initial values $V_1(p)$ , $V_2(p)$ and $V_3(p)$ , we can easily give the general term formula for the sequence $\{V_l(p)\}$ .

Corollary 3. If $p$ is a prime with $3\mid \phi(p)$ , then we have

$\begin{eqnarray*} \sum\limits_{u = 1}^{p-1}\left|\frac{ \sum\limits_{j = 1}^{p} \mathrm{e}_{p}\left(uj^3+j\right)}{ \sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3\right)}\right|^4 = \frac{54p^3}{\alpha^4}-\frac{p^2}{\alpha^2}-\frac{27p^2}{\alpha^4}+\frac{2p}{\alpha^2}-1. \end{eqnarray*}$

2. Some lemmas

Before starting our proofs of main results, we present the proofs of several key equations in preparation for the next chapter. The properties of Gauss sums and reduced (complete) residue systems are used repeatedly in the proof. In addition, we will refer to the basic contents of number theory in references ^[1] and ^[14].

Lemma 1. If $p$ is a prime with $3\mid \phi(p)$ , then

$\begin{eqnarray} \tau^3(\chi_{3})+ \tau^3\left(\overline{\chi_{3}}\right) = \alpha p. \end{eqnarray}$

(2.1)

Proof. This is consequence of ^[4] or ^[13], herein we omit it. □

Lemma 2. If $p$ is a prime with $3\mid \phi(p)$ , then

$\begin{eqnarray*} \sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3-1\right) = p(\alpha-3)+3 \tau^3\left(\overline{\chi_{3}}\right). \end{eqnarray*}$

Proof. Recall that $\tau(\chi_{3})\cdot \tau\left(\overline{\chi_{3}}\right) = p$ and (2.1), we have

$\begin{eqnarray*} &&\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3-1\right)\\ & = &\frac{1}{\tau(\chi_{3})}\sum\limits_{t = 1}^{p-1}\chi_{3}(t)\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p} \mathrm{e}_{p}\left(t\left(i^3+j^3-s^3-1\right)\right)\nonumber\\ & = & \frac{1}{\tau(\chi_{3})}\sum\limits_{t = 1}^{p-1}\chi_{3}(t) \mathrm{e}_{p}\left(-t\right) \left(\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(it^3\right)\right)^2 \left(\sum\limits_{s = 1}^{p} \mathrm{e}_{p}\left(-st^3\right)\right) \nonumber\\ & = & \frac{1}{\tau(\chi_{3})}\sum\limits_{t = 1}^{p-1}\chi_{3}(t) \mathrm{e}_{p}\left(-t\right) \left( 1+ \sum\limits_{i = 1}^{p-1} \left(1+ \chi_{3}(i)+ \overline{\chi_{3}}(i)\right) \mathrm{e}_{p}\left(it\right)\right)^3 \nonumber\\ & = &\frac{1}{\tau(\chi_{3})}\sum\limits_{t = 1}^{p-1}\chi_{3}(t) \mathrm{e}_{p}\left(-t\right) \big( \overline{\chi_{3}}(t)\cdot\tau(\chi_{3}) +\chi_{3}(t)\cdot\tau\left(\overline{\chi_{3}}\right)\big)^3 \nonumber\\ & = &\frac{1}{\tau(\chi_{3})}\sum\limits_{t = 1}^{p-1}\chi_{3}(t) \mathrm{e}_{p}\left(-t\right)\left[\tau^3(\chi_{3})+ \tau^3\left(\overline{\chi_{3}}\right)+ 3p\big( \overline{\chi_{3}}(t)\cdot\tau(\chi_{3}) +\chi_{3}(t)\cdot\tau\left(\overline{\chi_{3}}\right)\big)\right]\nonumber\\ & = &\alpha p + \frac{3p}{\tau(\chi_{3})}\cdot \tau(\chi_{3})\cdot \sum\limits_{t = 1}^{p-1} \mathrm{e}_{p}\left(-t\right)+\frac{3p}{\tau(\chi_{3})}\cdot \tau\left(\overline{\chi_{3}}\right)\cdot \sum\limits_{t = 1}^{p-1}\chi_{3}^2(t) \mathrm{e}_{p}\left(-t\right)\nonumber\\ & = &p(\alpha-3)+ \frac{3p\cdot \tau^2\left(\overline{\chi_{3}}\right)}{\tau(\chi_{3})} = p(\alpha-3)+3 \tau^3\left(\overline{\chi_{3}}\right). \end{eqnarray*}$

This completes the proof. □

Lemma 3. If $p$ is a prime with $3\mid \phi(p)$ , then

$\tau(\overline{\chi_{3}}\chi_{2}) = \frac{\overline{\chi_{3}}(2)\cdot\tau^2(\chi_{3})\cdot\tau(\chi_2)}{p}.$

Proof. Recall that $\tau(\chi_{3})\cdot\tau\left(\overline{\chi_{3}}\right) = p$ , we obtain

$\begin{eqnarray} \sum\limits_{i = 1}^{p}\chi_{3}(i^2-1)& = &\sum\limits_{i = 1}^{p}\chi_{3}(i^2+2i)\\ & = &\frac{1}{\tau(\overline{\chi_{3}})}\sum\limits_{j = 1}^{p-1}\overline{\chi_{3}}(j)\sum\limits_{i = 1}^{p-1}\chi_{3}(i) \mathrm{e}_{p}\big(j(i+2)\big)\\ & = &\frac{\tau(\chi_{3})}{\tau(\overline{\chi_{3}})}\sum\limits_{j = 1}^{p-1}\chi_{3}(j) \mathrm{e}_{p}\left(2j\right)\\ & = &\frac{\overline{\chi_{3}}(2)\cdot\tau^2(\chi_{3})}{\tau(\overline{\chi_{3}})}\\ & = &\frac{\overline{\chi_{3}}(2)\cdot\tau^3(\chi_{3})}{p}. \end{eqnarray}$

(2.2)

From another perspective, we have

$\begin{eqnarray} \sum\limits_{i = 1}^{p}\chi_{3}(i^2-1)& = &\frac{1}{\tau(\overline{\chi_{3}})}\sum\limits_{j = 1}^{p-1}\overline{\chi_{3}}(j)\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(j(i^2-1)\right)\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \\ & = &\frac{1}{\tau(\overline{\chi_{3}})}\sum\limits_{j = 1}^{p-1}\overline{\chi_{3}}(j) \mathrm{e}_{p}\left(-j\right)\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(i^2j\right)\\ & = &\frac{1}{\tau(\overline{\chi_{3}})}\sum\limits_{j = 1}^{p-1}\overline{\chi_{3}}(j) \mathrm{e}_{p}\left(-j\right)\left[1+\sum\limits_{i = 1}^{p-1}\big(1+\chi_{2}(i)\big) \mathrm{e}_{p}\left(ij\right)\right]\\ & = &\frac{1}{\tau(\overline{\chi_{3}})}\sum\limits_{j = 1}^{p-1}\overline{\chi_{3}}(j) \mathrm{e}_{p}\left(-j\right)\sum\limits_{i = 1}^{p-1}\chi_{2}(i) \mathrm{e}_{p}\left(ij\right)\\ & = &\frac{\tau(\chi_{2})}{\tau(\overline{\chi_{3}})}\sum\limits_{j = 1}^{p-1}\overline{\chi_{3}}\chi_{2}(j) \mathrm{e}_{p}\left(-j\right)\\ & = &\frac{\chi_{2}(-1)\cdot\tau(\chi_{2})\cdot\tau(\overline{\chi_{3}}\chi_{2})}{\tau(\overline{\chi_{3}})}\\ & = &\frac{\chi_{2}(-1)\cdot\tau(\chi_{2})\cdot\tau(\overline{\chi_{3}}\chi_{2})\cdot\tau(\chi_{3})}{p}. \end{eqnarray}$

(2.3)

Combining (2.2) and (2.3), we determine the relationship equation between $\tau(\overline{\chi_{3}}\chi_{2})$ , $\tau(\chi_{3})$ and $\tau(\chi_2)$

$\tau(\overline{\chi_{3}}\chi_{2}) = \frac{\overline{\chi_{3}}(2)\cdot\tau^2(\chi_{3})\cdot\tau(\chi_2)}{p}.$

This completes the proof. □

Lemma 4. If $p$ is a prime with $3\mid \phi(p)$ , then

$\begin{eqnarray*} \mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}}_{i+j-s-1\equiv 0\;{\rm mod}\; p }\overline{\chi_{3}}\left(i^3+j^3-s^3-1\right) = \frac{-\overline{\chi_{3}}(3)\cdot\tau^3\left(\overline{\chi_{3}}\right)}{p}. \end{eqnarray*}$

Proof. Using Lemma 3, we have

$\begin{eqnarray*} &&\mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}}_{i+j-s-1\equiv 0{\rm mod} p}\overline{\chi_{3}}\left(i^3+j^3-s^3-1\right)\nonumber\\ & = &\mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}}_{i+j\equiv 1\;{\rm mod}\; p }\overline{\chi_{3}}\left(i^3+j^3 +3j^2s +3js^2-1\right)\nonumber\\ & = &\chi_{3}(4) \mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}}_{i+j\equiv 1\;{\rm mod}\; p }\overline{\chi_{3}}\left(4i^3+j^3 +3j\left(2s +j\right)^2-4\right)\nonumber\\ & = &\frac{\chi_{3}(4)}{\tau(\chi_{3})}\mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{t = 1}^{p-1}}_{i+j\equiv 1\;{\rm mod}\; p}\chi_{3}(t) \sum\limits_{s = 1}^{p} \mathrm{e}_{p}\left(t\left(4i^3+j^3 +3js^2-4\right)\right)\nonumber\\ & = &\frac{\chi_{3}(4)}{\tau(\chi_{3})}\mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{t = 1}^{p-1}}_{i+j\equiv 1\;{\rm mod}\; p }\chi_{3}(t) \mathrm{e}_{p}\big(t(4i^3+j^3-4)\big)\left[1+\sum\limits_{s = 1}^{p-1}\big(1+\chi_{2}(s)\big) \mathrm{e}_{p}\left(3jst\right)\right]\nonumber\\ & = &\frac{\chi_{3}(4)}{\tau(\chi_{3})}\mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{t = 1}^{p-1}}_{i+j\equiv 1\;{\rm mod}\; p }\chi_{3}(t) \mathrm{e}_{p}\left(t(4i^3+j^3-4)\right)\chi_2(3jt)\tau(\chi_2)\nonumber\\ & = &\frac{\chi_{3}(4)\cdot \chi_2(3)\cdot \tau(\chi_2)}{\tau(\chi_{3})}\mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}}_{i+j\equiv 1\;{\rm mod}\; p }\chi_2(j) \tau\left(\chi_{3}\chi_2\right) \overline{\chi_{3}}\chi_2\left(4i^3+j^3-4\right)\nonumber\\ & = &\frac{\chi_{3}(4)\cdot \chi_2(3)\cdot \tau(\chi_2)\cdot \tau\left(\chi_{3}\chi_2\right) }{\tau(\chi_{3})}\sum\limits_{i = 1}^{p}\chi_2(1-i)\overline{\chi_{3}}\chi_2\left(3i^3+3i^2-3i-3\right)\nonumber\\ & = &\frac{\overline{\chi_{3}}(6)\cdot \chi_2(-1)\cdot \tau(\chi_2)\cdot \tau\left(\chi_{3}\chi_2\right) }{\tau(\chi_{3})}\sum\limits_{i = 1}^{p-1}\overline{\chi_{3}}\left((i+2)^2i\right)\nonumber\\ & = &\frac{\overline{\chi_{3}}(6)\cdot \chi_2(-1)\cdot \tau(\chi_2)\cdot \tau\left(\chi_{3}\chi_2\right) }{\tau(\chi_{3})} \sum\limits_{i = 1}^{p-1}\overline{\chi_{3}}(i)\chi_{3}\left(i+2\right)\nonumber\\ & = &\frac{\overline{\chi_{3}}(6)\cdot \chi_2(-1)\cdot \tau(\chi_2)\cdot \tau\left(\chi_{3}\chi_2\right) }{\tau(\chi_{3})}\sum\limits_{i = 1}^{p-1}\chi_{3}\left(1+2\cdot \overline{i}\right)\nonumber\\ & = &\frac{\overline{\chi_{3}}(6)\cdot \chi_2(-1)\cdot \tau(\chi_2)\cdot \tau\left(\chi_{3}\chi_2\right) }{\tau(\chi_{3})}\left(-1+ \sum\limits_{i = 1}^{p}\chi_{3}\left(i\right)\right)\nonumber\\ & = &\frac{-\chi_2(-1) \cdot \overline{\chi_{3}}(6)\cdot \tau(\chi_2)\cdot \tau\left(\overline{\chi_{3}}\right)\cdot \tau\left(\chi_{3}\chi_2\right) }{p}\nonumber\\ & = &\frac{-\overline{\chi_{3}}(3)\cdot\tau^3\left(\overline{\chi_{3}}\right)}{p}. \end{eqnarray*}$

This completes the proof. □

Lemma 5. If $p$ is a prime and $3\mid \phi(p)$ , then

$\begin{eqnarray*} \sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3\right) \mathrm{e}_{p}\left(i+j-s\right) = -3p+\overline{\chi_{3}}(3)\cdot\tau^3(\overline{\chi_{3}}). \end{eqnarray*}$

Proof. Note that $\tau(\chi_{3})\cdot\tau\left(\overline{\chi_{3}}\right) = p$ and $\overline{\chi_{3}}\left(i^3\right) = 1$ with $i$ is an integer relatively prime to $p$ . Therefore we have

$\begin{eqnarray*} &&\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3\right) \mathrm{e}_{p}\left(i+j-s\right)\nonumber\\ & = &\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3\right) \mathrm{e}_{p}\left(i+j\right)+\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}(i^3+j^3-1) \mathrm{e}_{p}\left(s(i+j-1)\right)\\ &&-\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\overline{\chi_{3}}(i^3+j^3-1)\nonumber\\ & = &\sum\limits_{i = 1}^{p}\overline{\chi_{3}}(i^3) \mathrm{e}_{p}\left(i\right)+\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\overline{\chi_{3}}\left(i^3+1\right) \mathrm{e}_{p}\left(j(i+1)\right)-\sum\limits_{i = 1}^{p}\overline{\chi_{3}}(i^3+1)\\ &&+p\mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}}_{i+j\equiv1\;{\rm mod}\; p}\overline{\chi_{3}}(i^3+j^3-1)-\frac{1}{\tau(\chi_{3})}\sum\limits_{t = 1}^{p-1}\chi_{3}(t)\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p} \mathrm{e}_{p}\left(t(i^3+j^3-1)\right)\nonumber\\ & = &\sum\limits_{i = 1}^{p-1} \mathrm{e}_{p}\left(i\right)+p\mathop{\sum\limits_{i = 1}^{p}}_{i+1\equiv0{\rm mod} p}\overline{\chi_{3}}\left(i^3+1\right)-1-\sum\limits_{i = 1}^{p-1}\big(1+\chi_{3}(i)+\overline{\chi_{3}}(i)\big)\overline{\chi_{3}}(i+1)\\ &&+p\mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}}_{i+j\equiv0{\rm mod} p}\overline{\chi_{3}}(i^3+j^3+3j^2+3j)-\frac{1}{\tau(\chi_{3})}\sum\limits_{t = 1}^{p-1}\chi_{3}(t) \mathrm{e}_{p}\left(-t\right)\left(\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(it^3\right)\right)^{2}\nonumber\\ & = &-1-\sum\limits_{i = 1}^{p}\overline{\chi_{3}}(i+1)-\sum\limits_{i = 1}^{p-1}\chi_{3}(i)\overline{\chi_{3}}(i+1)-\sum\limits_{i = 1}^{p-1}\overline{\chi_{3}}(i)\overline{\chi_{3}}(i+1)+p\sum\limits_{j = 1}^{p}\overline{\chi_{3}}(3j^2+3j)\\ &&-\frac{1}{\tau(\chi_{3})}\sum\limits_{t = 1}^{p-1}\chi_{3}(t) \mathrm{e}_{p}\left(-t\right)\left[1+\sum\limits_{i = 1}^{p-1}\big(1+\chi_{3}(i)+\overline{\chi_{3}}(i)\big) \mathrm{e}_{p}\left(it\right)\right]^{2}\nonumber\\ & = &-1-\sum\limits_{i = 1}^{p-1}\overline{\chi_{3}}(1+\overline{i})-\sum\limits_{i = 1}^{p-1}\overline{\chi_{3}}(i^2+i)+p\overline{\chi_{3}}(3)\sum\limits_{j = 1}^{p-1}\overline{\chi_{3}}(j^2+j)\\ &&-\frac{1}{\tau(\chi_{3})}\sum\limits_{t = 1}^{p-1}\chi_{3}(t) \mathrm{e}_{p}\left(-t\right)\left(\overline{\chi_{3}}(t)\tau(\chi_{3})+\chi_{3}(t)\tau(\overline{\chi_{3}})\right)^{2}\nonumber\\ & = &-\frac{1}{\tau(\chi_{3})}\left(\tau^2(\chi_{3})\sum\limits_{t = 1}^{p-1}\overline{\chi_{3}}(t) \mathrm{e}_{p}\left(-t\right)+\tau^2(\overline{\chi_{3}})\sum\limits_{t = 1}^{p-1} \mathrm{e}_{p}\left(-t\right)+2p\sum\limits_{t = 1}^{p-1}\chi_{3}(t) \mathrm{e}_{p}\left(-t\right)\right)\nonumber\\ &&+\left(-1+p\overline{\chi_{3}}(3)\right)\frac{1}{\tau(\chi_{3})}\sum\limits_{s = 1}^{p-1}\chi_{3}(s)\sum\limits_{i = 1}^{p-1}\overline{\chi_{3}}\left(i\right) \mathrm{e}_{p}\left(s(i+1)\right)\\ & = &-\frac{1}{\tau(\chi_{3})}\left(\tau^2(\chi_{3})\cdot\tau(\overline{\chi_{3}})-\tau^2(\overline{\chi_{3}})+2p\cdot\tau(\chi_{3})\right)-\frac{\tau^2(\overline{\chi_{3}})}{\tau(\chi_{3})}+p\cdot\overline{\chi_{3}}(3)\cdot\frac{\tau^2(\overline{\chi_{3}})}{\tau(\chi_{3})}\\ & = &\overline{\chi_{3}}(3)\cdot\tau^3(\overline{\chi_{3}})-3p. \end{eqnarray*}$

This proves Lemma 5. □

3. Proofs of main results

Proof of Theorem 1. Recall that $\overline{\chi_{3}}\left(i^3\right) = 1$ with $(i, p) = 1$ . Hence

$\begin{eqnarray*} &&\sum\limits_{u = 1}^{p-1}\chi_{3}(u)\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3+i\right)\right|^4\nonumber\\ & = &\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\sum\limits_{t = 1}^{p}\sum\limits_{u = 1}^{p-1}\chi_{3}(u) \mathrm{e}_{p}\left(u\left(i^3+j^3-s^3-t^3\right)+i+j-s-t\right)\nonumber\\ & = &\tau(\chi_{3})\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\sum\limits_{t = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3-t^3\right) \mathrm{e}_{p}\left(i+j-s-t\right)\nonumber\\ & = &\tau(\chi_{3})\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\sum\limits_{t = 1}^{p-1}\overline{\chi_{3}}\left(i^3t^3+j^3t^3-s^3t^3-t^3\right) \mathrm{e}_{p}\left(it+jt-st-t\right)\nonumber\\ &&+ \tau(\chi_{3})\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3\right) \mathrm{e}_{p}\left(i+j-s\right)\nonumber\\ & = &\tau(\chi_{3})\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3-1\right)\sum\limits_{t = 1}^{p} \mathrm{e}_{p}\left(t(i+j-s-1)\right)\nonumber\\ &&+\tau(\chi_{3})\sum _{i = 0}^{p-1}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3\right) \mathrm{e}_{p}\left(i+j-s\right)\\ &&-\tau(\chi_{3})\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3-1\right) \nonumber\\ & = &p\tau(\chi_{3}) \mathop{\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}}_{i+j-s\equiv 1\;{\rm mod}\; p }\overline{\chi_{3}}\left(i^3+j^3-s^3-1\right)-\tau(\chi_{3})\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3-1\right) \nonumber\\ &&+\tau(\chi_{3})\sum\limits_{i = 1}^{p}\sum\limits_{j = 1}^{p}\sum\limits_{s = 1}^{p}\overline{\chi_{3}}\left(i^3+j^3-s^3\right) \mathrm{e}_{p}\left(i+j-s\right).\nonumber\\ \end{eqnarray*}$

Applying Lemmas 2, 4 and 5 we obtain

$\begin{eqnarray*} &&\sum\limits_{u = 1}^{p-1}\chi_{3}(u)\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3+i\right)\right|^4\nonumber\\ & = &-\tau(\chi_{3})\cdot\overline{\chi_{3}}(3)\cdot\tau^3\left(\overline{\chi_{3}}\right)-\tau(\chi_{3})\left(p(\alpha-3)+3 \tau^3\left(\overline{\chi_{3}}\right)\right)+\tau(\chi_{3})\left(\overline{\chi_{3}}(3)\cdot\tau^3(\overline{\chi_{3}})-3p\right)\nonumber\\ & = &-\alpha p\cdot\tau(\chi_{3})-3p\cdot\tau^2(\overline{\chi_{3}}).\nonumber \end{eqnarray*}$

□

Proof of Theorem 2. Based on Theorem 1 and the identities obtained in ^[9]

$\begin{eqnarray*} \sum\limits_{u = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(ui^3+vi\right)\right|^4 = \left\{ \begin{array}{ll} 2p^3-p^2 & \text{ if}\quad 3\nmid p-1 , \\ 2p^3-7p^2 & \mathrm{ if\quad 3\mid p-1 .} \end{array}\right. \end{eqnarray*}$

We have

$\begin{eqnarray} &&\sum\limits_{v = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(i^3+vi\right)\right|^4 = \sum\limits_{v = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left((\overline{v}i)^3+i\right)\right|^4\\ & = &\sum\limits_{v = 1}^{p-1}\left(1+\chi_{3}(v)+\overline{\chi_{3}}(v)\right)\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(vi^3+i\right)\right|^4\\ & = &\sum\limits_{v = 1}^{p-1}\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(vi^3+i\right)\right|^4+\sum\limits_{v = 1}^{p-1}\chi_{3}(v)\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(vi^3+i\right)\right|^4\\ &&+\sum\limits_{v = 1}^{p-1}\overline{\chi_{3}}(v)\left|\sum\limits_{i = 1}^{p} \mathrm{e}_{p}\left(vi^3+i\right)\right|^4\\ & = &2p^3-7p^2-\alpha p\tau(\chi_{3})-3p\cdot\tau^2(\overline{\chi_{3}})-\alpha p\tau(\overline{\chi_{3}})-3p\cdot\tau^2(\chi_{3})\\ & = &2p^3-p^2-3p\left(\tau(\chi_{3})+\tau(\overline{\chi_{3}})\right)^2-\alpha p\left(\tau(\chi_{3})+\tau(\overline{\chi_{3}})\right). \end{eqnarray}$

(3.1)

Now we need to determine the value of the real number $\tau(\chi_{3})+\tau(\overline{\chi_{3}})$ in (3.1). For convenience, write the $A = \tau(\chi_{3})+\tau(\overline{\chi_{3}})$ , we construct cubic equation $A^3-3pA-\alpha p = 0$ based on (2.1) and $\tau(\chi_{3})\cdot\tau(\overline{\chi_{3}}) = p$ . According to Cardans formula (formula of roots of a cubic equation), the three roots of the equation are

$\begin{eqnarray*} &A_{1} = \left[{\frac{\alpha p}{2}+\left(\left(\frac{\alpha p}{2}\right)^{2}+\left(-p\right)^{3}\right)^{\frac{1}{2}}}\right]^{\frac{1}{3}}+\left[{\frac{\alpha p}{2}-\left(\left(\frac{\alpha p}{2}\right)^{2}+\left(-p\right)^{3}\right)^{\frac{1}{2}}}\right]^{\frac{1}{3}},\; \; \; \; \; \; \nonumber\\ &A_{2} = \omega\left[\frac{\alpha p}{2}+\left(\left(\frac{\alpha p}{2}\right)^{2}+\left(-p\right)^{3}\right)^{\frac{1}{2}}\right]^{\frac{1}{3}}+\omega^{2}\left[{\frac{\alpha p}{2}-\left(\left(\frac{\alpha p}{2}\right)^{2}+\left(-p\right)^{3}\right)^{\frac{1}{2}}}\right]^{\frac{1}{3}},\nonumber\\ &A_{3} = \omega^{2}\left[{\frac{\alpha p}{2}+\left(\left(\frac{\alpha p}{2}\right)^{2}+\left(-p\right)^{3}\right)^{\frac{1}{2}}}\right]^{\frac{1}{3}}+\omega\left[{\frac{\alpha p}{2}-\left(\left(\frac{\alpha p}{2}\right)^{2}+\left(-p\right)^{3}\right)^{\frac{1}{2}}}\right]^{\frac{1}{3}},\nonumber \end{eqnarray*}$

where $\omega = \frac{-1+\sqrt{3} \mathrm{i}}{2}$ .

It is clear that all $A_k$ $(k = 1, 2\ \text{or}\ 3)$ are real numbers, So $A = A_1, A_2$ or $A_3$ . Therefore, the proof of theorem is complete. □

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors gratefully appreciates the referees and editor for their helpful and detailed comments.

This work is supported by Hainan Provincial Natural Science Foundation of China (123RC473) and Natural Science Foundation of China (12126357).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Abbreviation AV: Aortic valve; CAD: Coronary artery disease; CFA: Common femoral artery; CKD: Chronic kidney disease; CLI: Critical limb ischemia; DM: Diabetes mellitus; FEM-POP: Femoral and popliteal arteries; HbA1c: Hemoglobin A1c; INFRA-POP: Infrapopliteal arteries; MGP: Matrix Gla protein; NF-kB: Nuclear factor kappa beta; OPN: Osteopontin; OPG: Osteoprotegerin; PAD: Peripheral artery disease; RANKL: Receptor activator of NF-B ligand; SFA: Superficial femoral artery; VIC: Valvular interstitial cells; VSMC: Vascular smooth muscle cell;

Conflict of interest

CVPath Institute has received institutional research support from R01 HL141425 Leducq Foundation Grant; 480 Biomedical; 4C Medical; 4Tech; Abbott; Accumedical; Amgen; Biosensors; Boston Scientific; Cardiac Implants; Celonova; Claret Medical; Concept Medical; Cook; CSI; DuNing, Inc; Edwards LifeSciences; Emboline; Endotronix; Envision Scientific; Lutonix/Bard; Gateway; Lifetech; Limflo; MedAlliance; Medtronic; Mercator; Merill; Microport Medical; Microvention; Mitraalign; Mitra assist; NAMSA; Nanova; Neovasc; NIPRO; Novogate; Occulotech; OrbusNeich Medical; Phenox; Profusa; Protembis; Qool; Recor; Senseonics; Shockwave; Sinomed; Spectranetics; Surmodics; Symic; Vesper; W.L. Gore; Xeltis. A.V. F. has received honoraria from Abbott Vascular; Biosensors; Boston Scientific; Celonova; Cook Medical; CSI; Lutonix Bard; Sinomed; Terumo Corporation; and is a consultant to Amgen; Abbott Vascular; Boston Scientific; Celonova; Cook Medical; Lutonix Bard; Sinomed. Anne Cornelissen receives research grants from University Hospital RWTH Aachen. R.V. has received honoraria from Abbott Vascular; Biosensors; Boston Scientific; Celonova; Cook Medical; Cordis; CSI; Lutonix Bard; Medtronic; OrbusNeich Medical; CeloNova; SINO Medical Technology; ReCore; Terumo Corporation; W. L. Gore; Spectranetics; and is a consultant Abbott Vascular; Boston Scientific; Celonova; Cook Medical; Cordis; CSI; Edwards Lifescience; Lutonix Bard; Medtronic; OrbusNeich Medical; ReCore; Sinomededical Technology; Spectranetics; Surmodics; Terumo Corporation; W. L. Gore; Xeltis. The other authors declare no competing interests.

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