Investigating of an immune system-cancer mathematical model with Mittag-Leffler kernel

Necati Özdemir; Esmehan Uçar; Necati Özdemir; Esmehan Uçar

doi:10.3934/math.2020104

AIMS Mathematics

2020, Volume 5, Issue 2: 1519-1531. doi: 10.3934/math.2020104

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Investigating of an immune system-cancer mathematical model with Mittag-Leffler kernel

Necati Özdemir ^,,
Esmehan Uçar

Department of Mathematics, Faculty of Science and Arts, Balıkesir University, Turkey

Received: 29 October 2019 Accepted: 09 January 2020 Published: 21 January 2020
MSC : 34A08, 34A34, 47H10

Cancer that is difficult to treat, is a very common disease today and there are many types of cancer such as lung, colon, stomach. When cancer settles in the body, the immune system tries to resist it. In this study, the mathematical model of the interaction between immune system components and cancer is discussed and is modified by using Atangana-Baleanu derivative. After investigating the existence and uniqueness of the solution of the fractional immune system-cancer model, numerical simulations are given via predictor-corrector scheme.

Keywords:

Citation: Necati Özdemir, Esmehan Uçar. Investigating of an immune system-cancer mathematical model with Mittag-Leffler kernel[J]. AIMS Mathematics, 2020, 5(2): 1519-1531. doi: 10.3934/math.2020104

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Abstract

1. Introduction

The key to solving the general quadratic congruence equation is to solve the equation of the form $x^2\equiv a\mod p$ , where $a$ and $p$ are integers, $p > 0$ and $p$ is not divisible by $a$ . For relatively large $p$ , it is impractical to use the Euler criterion to distinguish whether the integer $a$ with $(a, p) = 1$ is quadratic residue of modulo $p$ . In order to study this issue, Legendre has proposed a new tool-Legendre's symbol.

Let $p$ be an odd prime, the quadratic character modulo $p$ is called the Legendre's symbol, which is defined as follows:

$\left(\frac{a}{p}\right)= \begin{cases}1, & \text { if } a \text { is a quadratic residue modulo } p; \\ -1, & \text { if } a \text { is a quadratic non-residue modulo } p ;\\ 0, & \text { if } p \mid a.\end{cases}$

The Legendre's symbol makes it easy for us to calculate the level of quadratic residues. The basic properties of Legendre's symbol can be found in any book on elementary number theory, such as ^[1,2,3].

The properties of Legendre's symbol and quadratic residues play an important role in number theory. Many scholars have studied them and achieved some important results. For examples, see the ^{[4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21]}.

One of the most representative properties of the Legendre's symbol is the quadratic reciprocal law:

Let $p$ and $q$ be two distinct odd primes. Then, (see Theorem 9.8 in ^[1] or Theorems 4–6 in ^[3])

$\left(\frac{p}{q}\right)\cdot\left(\frac{q}{p}\right) = (-1)^{\frac{(p-1)(q-1)}{4}}.$

For any odd prime $p$ with $p\equiv1\; \rm mod\; 4$ there exist two non-zero integers $\alpha\left(p\right)$ and $\beta\left(p\right)$ such that

$\begin{align*} p = \alpha^2\left(p\right)+\beta^2\left(p\right). \end{align*}$

(1)

In fact, the integers $\alpha\left(p\right)$ and $\beta\left(p\right)$ in the (1) can be expressed in terms of Legendre's symbol modulo $p$ (see Theorems 4–11 in ^[3])

$\alpha\left(p\right) = \frac{1}{2}\sum\limits_{a = 1}^{p-1}\left(\frac{a^3+a}{p}\right)\; \; \; \rm and\; \; \; \it \beta\left(\it p\right) = \frac{\rm1}{\rm 2}\sum\limits_{a = \rm1}^{p-\rm1}\left(\frac {\it a^{\rm3}+\it ra}{\it p}\right),$

where $r$ is any integer, and $\left(r, p\right) = 1$ , $\left(\frac{r}{p}\right) = -1$ , $\left(\frac{\ast}{p}\right) = \chi_2$ denote the Legendre's symbol modulo $p$ .

Noting that Legendre's symbol is a special kind of character. For research on character, Han ^[7] studied the sum of a special character $\chi\left(ma+\bar{a}\right)$ , for any integer $m$ with $\left(m, p\right) = 1$ , then

$\left|\sum\limits_{a = 1}^{p-1}\chi\left(ma+\bar{a}\right)\right|^2 = 2p+\left(\frac{m}{p}\right)\sum\limits_{a = 1}^{p-1}\chi(a)\sum\limits_{b = 1}^{p-1}\left(\frac{b(b-1)(a^2b-1)}{p}\right),$

which is a special case of a general polynomial character sums $\sum_{a = N+1}^{N+M}\chi\left(f\left(a\right)\right)$ , where $M$ and $N$ are any positive integers, and $f\left(x\right)$ is a polynomial.

In ^[8], Du and Li introduced a special character sums $C\left(\chi, m, n, c;p\right)$ in the following form:

$C\left(\chi, m, n, c;p\right) = \sum\limits_{a = 0}^{p-1}\sum\limits_{b = 0}^{p-1}\chi\left(a^2+na-b^2-nb+c\right)\cdot e\left(\frac{mb^2-ma^2}{p}\right),$

and studied the asymptotic properties of it. They obtained

$\begin{equation*} \label{test1} \begin{split} \sum\limits_{c = 1}^{p-1}\left|C\left(\chi, m, n, c;p\right)\right|^{2k} = \left\{ \begin{array}{l} &p^{2k+1}+\frac{k^2-3k-2}{2}\cdot p^{2k}+O\left(p^{2k-1}\right), \; \; \; \; \rm if\; \chi\ \rm is\ \rm the\ \rm Legendre\ \rm symbol\ \rm {modulo} \ \it p \rm; \\ &p^{2k+1}+\frac{k^2-3k-2}{2}\cdot p^{2k}+O\left(p^{2k-1/2}\right), \; \; \rm if\; \chi\ \rm is\ \rm a\ \rm complex\ \rm character\ \rm {modulo} \ \it p.\\ \end{array} \right. \end{split} \end{equation*}$

Recently, Yuan and Zhang ^[12] researched the question about the estimation of the mean value of high-powers for a special character sum modulo a prime, let $p$ be an odd prime with $p\equiv1\; \rm mod\; 6$ , then for any integer $k\geq0$ , they have the identity

$S_k\left(p\right) = \frac{1}{3}\cdot\left[d^k+\left(\frac{-d+9b}{2}\right)^k+\left(\frac{-d-9b}{2}\right)^k\right],$

where

$S_k\left(p\right) = \frac{1}{p-1}\sum\limits_{r = 1}^{p-1}A^k\left(r\right),$

$A\left(r\right) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+r\bar{a}}{p}\right),$

and for any integer $r$ with $(r, p) = 1$ .

More relevant research on special character sums will not be repeated. Inspired by these papers, we have the question: If we replace the special character sums with Legendre's symbol, can we get good results on $p\equiv1\; \rm mod\; 4$ ?

We will convert $\beta\left(p\right)$ to another form based on the properties of complete residues

$\beta\left(p\right) = \frac{1}{2}\sum\limits_{a = 1}^{p-1}\left(\frac{a+n\bar{a}}{p}\right),$

where $\bar{a}$ is the inverse of $a$ modulo $p$ . That is, $\bar{a}$ satisfy the equation $x\cdot a\; \equiv1\mod p$ for any integer $a$ with $(a, p) = 1$ .

For any integer $k\geq0$ , $G\left(n\right)$ and $K_k(p)$ are defined as follows:

$G\left(n\right) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)\; \; \; \rm {and} \; \; \; \it K_{\it k}\left(\it p\right) = \frac{\rm 1}{\it p-\rm 1}\sum\limits_{\it n\rm = 1}^{\it p-\rm 1}\it G^{\it k}\left(\it n\right).$

In this paper, we will use the analytic methods and properties of the classical Gauss sums and Dirichlet character sums to study the computational problem of $K_k\left(p\right)$ for any positive integer $k$ , and give a linear recurrence formulas for $K_k\left(p\right)$ . That is, we will prove the following result.

Theorem 1. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ , then we have

$K_k\left(p\right) = \left(4p+2\right)\cdot K_{k-2}\left(p\right)-8\left(2\alpha^2-p\right)\cdot K_{k-3}\left(p\right)+\left(16\alpha^4-16p\alpha^2+4p-1\right)\cdot K_{k-4}\left(p\right),$

for all integer $k\geq4$ with

$K_0\left(p\right) = 1, \; K_1\left(p\right) = 0, \; K_2\left(p\right) = 2p+1, \; K_3\left(p\right) = -3(4\alpha^2-2p),$

where

$\alpha = \alpha\left(p\right) = \sum\limits_{a = 1}^{\frac{p-1}{2}}\left(\frac{a+\bar{a}}{p}\right).$

Applying the properties of the linear recurrence sequence, we may immediately deduce the following corollaries.

Corollary 1. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then we have

$\frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\frac{1}{1+\sum _{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)} = \frac{16\alpha^2p-28\alpha^2-8p^2+14p} {16\alpha^4-16\alpha^2p+4p-1}.$

Corollary 2. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then we have

$\begin{align*} \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\sum\limits_{m = 0}^{p-1}\left(1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)\right)\cdot e\left(\frac{nm^2}{p}\right) = -\sqrt{p}. \end{align*}$

Corollary 3. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then we have

$\frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\sum\limits_{m = 0}^{p-1}\left[1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)\right]^2\cdot e\left(\frac{nm^2}{p}\right) = \left(4\alpha^2-2p\right)\cdot\sqrt{p}.$

Corollary 4. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 8$ . Then we have

$\begin{align*} \sum\limits_{n = 1}^{p-1}\left(1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)\right)\cdot \sum\limits_{m = 0}^{p-1}e\left(\frac{nm^4}{p}\right) = \sqrt{p}\left(-1+B(1)\right)-p, \end{align*}$

where

$B(1) = \sum\limits_{m = 0}^{p-1}e\left(\frac{m^4}{p}\right).$

If we consider such a sequence $F_k\left(p\right)$ as follows: Let $p$ be a prime with $p\equiv1\; \rm mod\; 8$ , $\chi_4$ be any fourth-order character modulo $p$ . For any integer $k\geq0$ , we define the $F_k\left(p\right)$ as

$F_k\left(p\right) = \sum\limits_{n = 1}^{p-1}\frac{1}{G^{k}(n)},$

we have

$\begin{align*} F_{k}\left(p\right) = &\frac{1}{16\alpha^4-16\alpha^2p+4p-1}F_{k-4}\left(p\right)-\frac{(4p+2)}{16\alpha^4-16\alpha^2p+4p-1}F_{k-2}\left(p\right)\\ &+\frac{4(4\alpha^2-2p)}{16\alpha^4-16\alpha^2p+4p-1}F_{k-1}\left(p\right). \end{align*}$

2. Some lemmas

Lemma 1. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then for any fourth-order character $\chi_4 \; \rm{mod} \; \it p$ , we have the identity

$\tau^2(\chi_4)+\tau^2(\bar{\chi_4}) = 2\sqrt{p}\cdot\alpha,$

where

$\tau(\chi_4) = \sum\limits_{a = 1}^{p-1}\chi_4(a)e\left(\frac{a}{p}\right)$

denotes the classical Gauss sums, $e(y) = e^{2\pi iy}, i^2 = -1$ , and $\alpha$ is the same as in the Theorem 1.

Proof. See Lemma 2.2 in ^[9].

Lemma 2. Let $p$ be an odd prime. Then for any non-principal character $\psi$ modulo $p$ , we have the identity

$\tau(\psi^2) = \frac{\psi^2(2)}{\tau(\chi_2)}\cdot\tau(\psi)\cdot\tau(\psi\chi_2),$

where $\chi_2 = \left(\frac{\ast}{p}\right)$ denotes the Legendre's symbol modulo $p$ .

Proof. See Lemma 2 in ^[12].

Lemma 3. Let $p$ be a prime with $p\equiv 1\; \rm mod\; 4$ , then for any integer $n$ with $\left(n, p\right) = 1$ and fourth-order character $\chi_{4}\; \rm mod \; \it p$ , we have the identity

$\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = -1-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})).$

Proof. For any integer $a$ with $(a, p) = 1$ , we have the identity

$1+\chi_{4}(a)+\chi_2(a)+\bar{\chi_{4}}(a) = 4,$

if $a$ satisfies $a\equiv b^4\; \rm mod \; \it p$ for some integer $b$ with $(b, p) = 1$ and

$1+\chi_{4}(a)+\chi_2(a)+\bar{\chi_{4}}(a) = 0,$

otherwise. So from these and the properties of Gauss sums we have

$\begin{align*} \mathop\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = &\sum\limits_{a = 1}^{p-1}\left(\frac{a^2}{p}\right)\left(\frac{a^4+n}{p}\right)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}\chi_2\left(a^4\right)\chi_2\left(a^4+n\right)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}(1+\chi_{4}(a)+\chi_2(a)+\bar{\chi_{4}}(a))\cdot\chi_2(a)\cdot\chi_2(a+n)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}(1+\chi_{4}(na)+\chi_2(na)+\bar{\chi_{4}}(na))\cdot\chi_2(na)\cdot\chi_2(na+n)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}\chi_2(a)\chi_2(a+1)+\sum\limits_{a = 1}^{p-1}\chi_{4}(na)\chi_2(a)\chi_2(a+1) \end{align*}$

(2)

$\begin{align*} &+\sum\limits_{a = 1}^{p-1}\chi_2(na)\chi_2(a)\chi_2(a+1)+\sum\limits_{a = 1}^{p-1}\bar{\chi_{4}}(na)\chi_2(a)\chi_2(a+1)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}\chi_2(1+\bar{a})+\sum\limits_{a = 1}^{p-1}\chi_{4}(na)\chi_2(a)\chi_2(a+1)\nonumber\\ &+\sum\limits_{a = 1}^{p-1}\chi_2(n)\chi_2(a+1)+\sum\limits_{a = 1}^{p-1}\bar{\chi_{4}}(na)\chi_2(a)\chi_2(a+1)\nonumber. \end{align*}$

Noting that for any non-principal character $\chi$ ,

$\sum\limits_{a = 1}^{p-1}\chi(a) = 0$

and

$\sum\limits_{a = 1}^{p-1}\chi(a)\chi(a+1) = \frac{1}{\tau(\bar{\chi})}\sum\limits_{b = 1}^{p-1}\sum\limits_{a = 1}^{p-1}\bar{\chi}(b)\chi(a)e\left(\frac{b(a+1)}{p}\right).$

Then we have

$\sum\limits_{a = 1}^{p-1}\chi_2(1+\bar{a}) = -1, \; \; \; \sum\limits_{a = 1}^{p-1}\chi_2(a+1) = -1,$

$\begin{align*} \sum\limits_{a = 1}^{p-1}\chi_{4}(a)\chi_2(a)\chi_2(a+1)& = \frac{1}{\tau(\chi_2)}\sum\limits_{b = 1}^{p-1}\sum\limits_{a = 1}^{p-1}\chi_2(b)\chi_{4}(a)\chi_2(a)e\left(\frac{b(a+1)}{p}\right)\nonumber\\ & = \frac{1}{\tau(\chi_2)}\sum\limits_{b = 1}^{p-1}\bar{\chi_{4}}(b)e\left(\frac{b}{p}\right)\sum\limits_{a = 1}^{p-1}\chi_{4}(ab)\chi_2(ab)e\left(\frac{ab}{p}\right) \end{align*}$

(3)

$\begin{align*} & = \frac{1}{\tau(\chi_2)}\cdot\tau(\bar{\chi_{4}})\cdot\tau(\chi_{4}\chi_2)\nonumber. \end{align*}$

For any non-principal character $\psi$ , from Lemma 2 we have

$\begin{align*} \tau(\psi^2) = \frac{\psi^2(2)}{\tau(\chi_2)}\cdot\tau(\psi)\cdot\tau(\psi\chi_2). \end{align*}$

(4)

Taking $\psi = \chi_{4}$ , note that

$\tau(\chi_2) = \sqrt{p}, \ \ \tau(\chi_{4})\cdot\tau(\bar{\chi_{4}}) = \chi_{4}(-1)\cdot p,$

from (3) and (4), we have

$\begin{align*} \sum\limits_{a = 1}^{p-1}\chi_{4}(a)\chi_2(a)\chi_2(a+1)& = \frac{\bar{\chi_{4}}^2(2)\cdot\tau(\chi_{4}^2)\cdot\tau(\chi_2)\cdot\tau(\bar{\chi_{4}})}{\tau(\chi_2)\cdot\tau(\chi_{4})}\nonumber\\ & = \frac{\chi_2(2)\cdot\tau(\chi_2)\cdot\tau^2(\bar{\chi_{4}})}{\tau(\chi_{4})\cdot\tau(\bar{\chi_{4}})}\nonumber\\ & = \frac{\chi_2(2)\cdot\sqrt{p}\cdot\tau^2(\bar{\chi_{4}})}{\chi_4(-1)\cdot p} \end{align*}$

(5)

$\begin{align*} & = \frac{\chi_2(2)\cdot\tau^2(\bar{\chi_{4}})}{\chi_4(-1)\cdot\sqrt{p}}.\nonumber \end{align*}$

Similarly, we also have

$\begin{align*} \sum\limits_{a = 1}^{p-1}\bar{\chi_{4}}(a)\chi_2(a)\chi_2(a+1) = \frac{\chi_2(2)\cdot\tau^2(\chi_{4})}{\chi_4(-1)\cdot\sqrt{p}}. \end{align*}$

(6)

Consider the quadratic character modulo $p$ , we have

$\begin{align*} \left(\frac{2}{p}\right) = \chi_2(2) = \begin{cases} \; \; 1, &\rm{ if\; \ p\equiv\pm1\; mod\; 8};\\ -1, &\rm{ if\; \ p\equiv\pm3\; mod\; 8}.\\ \end{cases} \end{align*}$

(7)

And when $p\equiv1\; \rm mod\; 8$ , we have $\chi_{4}(-1) = 1$ ; when $p\equiv5\; \rm mod\; 8$ , we have $\chi_{4}(-1) = -1.$ Combining (2) and (5)–(7) we can deduce that

$\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = -1-\chi_2(n)+\frac{1}{\sqrt{ p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right).$

This prove Lemma 3.

Lemma 4. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then for any integer $k\geq4$ and $n$ with $(n, p) = 1$ , we have the fourth-order linear recurrence formula

$G^k(n) = (4p+2)\cdot G^{k-2}(n)+8(p-2\alpha^2)\cdot G^{k-3}(n)+[(4\alpha^2-2p)^2-(2p-1)^2]\cdot G^{k-4}(n),$

where

$\alpha = \alpha(p) = \frac{1}{2}\sum\limits_{a = 1}^{p-1}\left(\frac{a^3+a}{p}\right) = \sum\limits_{a = 1}^{\frac{p-1}{2}}\left(\frac{a+\bar{a}}{p}\right),$

$\left(\frac{\ast} {p}\right) = \chi_2$ denotes the Legendre's symbol.

Proof. For $p\equiv1\; \rm mod\; 4$ , any integer $n$ with $\left(n, p\right) = 1$ , and fourth-order character $\chi_{4}$ modulo $p$ , we have the identity

$\chi_{4}^4(n) = \bar{\chi_{4}}^4(n) = \chi_0(n), \ \ \chi_{4}^2(n) = \chi_2(n),$

where $\chi_0$ denotes the principal character modulo $p$ .

According to Lemma 3,

$\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = -1-\chi_2(n)+\frac{1}{\sqrt{ p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right),$

$G(n) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right).\quad\quad\quad\quad\quad\quad\quad\quad\ \$

We have

$\begin{align*} G(n) = -\chi_2(n)+\frac{1}{\sqrt{p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right), \end{align*}$

(8)

$\begin{align*} G^2(n) = &[-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4}))]^2\\ = &1-2\chi_2(n)\cdot\frac{1}{\sqrt{p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right)\\ &+\frac{1}{ p}\cdot\left(\chi_2(n)\cdot\tau^4(\bar{\chi_{4}})+\chi_2(n)\cdot\tau^4(\chi_{4})+2p^2\right)\\ = &1-2\chi_2(n)\cdot\frac{1}{\sqrt{p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right)\\ &+\frac{1}{p}\cdot\left(\chi_2(n)\cdot(\tau^4(\bar{\chi_{4}})+\tau^4(\chi_{4}))+2p^2\right). \end{align*}$

According to Lemma 1, we have

$\left(\tau^2(\chi_{4})+\tau^2(\bar{\chi_{4}})\right)^2 = \tau^4(\bar{\chi_{4}})+\tau^4(\chi_{4})+2p^2 = 4p\alpha^2.$

Therefore, we may immediately deduce

$\begin{align*} G^2(n) = &1-2(\chi_2(n)\cdot\left( G(n)+\chi_2(n)\right)\nonumber\\ &+\frac{1}{p}\left(\chi_2(n)\cdot(\tau^4(\bar{\chi_{4}})+\tau^4(\chi_{4}))+2p^2\right)\nonumber\\ = &1-2\chi_2(n)\cdot\left( G(n)+\chi_2(n)\right) \end{align*}$

(9)

$\begin{align*} &+\frac{1}{p}\cdot\left[\chi_2(n)((\tau^2(\bar{\chi_{4}})+\tau^2(\chi_{4}))^2-2p^2)+2p^2\right]\nonumber\\ = &2p-1-2\chi_2(n)\cdot G(n)+\left(4\alpha^2-2p\right)\cdot\chi_2(n)\nonumber, \end{align*}$

$\begin{align*} \; \; \; \; \; \; \; \; \; \; \; \; \; \; G^3(n) = &[-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4}))]^3\nonumber\\ = &\left(2p-1-2\chi_2(n)\cdot G(n)+\left(4\alpha^2-2p\right)\cdot\chi_2(n)\right)\cdot G(n) \end{align*}$

(10)

$\begin{align*} = &(4\alpha^2-2p)\chi_2(n)\cdot G(n)+(2p+3)G(n)-(4p-2)\chi_2(n) -2(4\alpha^2-2p)\nonumber \end{align*}$

and

$\begin{align*} \left[G^2(n)-(2p-1)\right]^2 = \left[\chi_2(n)\cdot(4\alpha^2-2p)-2\chi_2(n)\cdot G(n)\right]^2, \end{align*}$

which implies that

$\begin{align*} G^4(n) = (4p+2)\cdot G^{2}(n)+8(p-2\alpha^2)\cdot G(n)+[(4\alpha^2-2p)^2-(2p-1)^2]. \end{align*}$

(11)

So for any integer $k\geq4$ , from (8)–(11), we have the fourth-order linear recurrence formula

$\begin{align*} G^k(n)& = G^{k-4}(n)\cdot G^4(n)\\ & = (4p+2)\cdot G^{k-2}(n)+8(p-2\alpha^2)\cdot G^{k-3}(n)+[(4\alpha^2-2p)^2-(2p-1)^2]\cdot G^{k-4}(n). \end{align*}$

This proves Lemma 4.

3. Proof of the theorem

In this section, we will complete the proof of our theorem.

Let $p$ be any prime with $p\equiv 1\; \rm mod\; 4$ , then we have

$\begin{align*} K_0(p) = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^0(n) = \frac{p-1}{p-1} = 1.\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \end{align*}$

(12)

$\begin{align*} K_1(p)& = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^1(n)\nonumber\\ & = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\left(-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\tau^2(\chi_{4}))\right)\\& = 0, \end{align*}$

(13)

$\begin{align*} \; \; \; \; \; K_2(p)& = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^2(n)\nonumber\\ & = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\left(-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\tau^2(\chi_{4}))\right)^2 \\ & = 2p+1,\end{align*}$

(14)

$\begin{align*} \; \; \; \; \; K_3(p)& = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^3(n)\nonumber\\ & = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\left(-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\tau^2(\chi_{4}))\right)^3 \\ & = -3(4\alpha^2-2p).\end{align*}$

(15)

It is clear that from Lemma 4, if $k\geq4$ , we have

$\begin{align*} \; \; \; \; \; K_k(p) = &\frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^k(n)\nonumber\\ = &(4p+2)\cdot K_{k-2}(p)-8\left(2\alpha^2-p\right)\cdot K_{k-3}(p) +(16\alpha^4-16p\alpha^2+4p-1)\cdot K_{k-4}(p). \end{align*}$

(16)

Now Theorem 1 follows (12)–(16). Obviously, using Theorem 1 to all negative integers, and that lead to Corollary 1.

This completes the proofs of our all results.

Some notes:

Note 1: In our theorem, know $n$ is an integer, and $(n, p) = 1$ . According to the properties of quadratic residual, $\chi_2(n) = \pm1$ , $\chi_4(n) = \pm1$ .

Note 2: In our theorem, we only discussed the case $p\equiv1\; \rm mod\; 8$ . If $p\equiv3\; \rm mod\; 4$ , then the result is trivial. In fact, in this case, for any integer $n$ with $(n, p) = 1$ , we have the identity

$\begin{align*} G(n)& = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^4}{p}\right)\cdot\left(\frac{a^4+n}{p}\right)\\ & = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a}{p}\right)\cdot\left(\frac{a+n}{p}\right) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+na}{p}\right)\\ & = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{1+n\bar{a}}{p}\right) = \sum\limits_{a = 0}^{p-1}\left(\frac{1+na}{p}\right) = 0. \end{align*}$

Thus, for all prime $p$ with $p\equiv3\; \rm mod\; 4$ and $k\geq1$ , we have $K_k(p) = 0$ .

4. Conclusions

The main result of this paper is Theorem 1. It gives an interesting computational formula for $K_k(p)$ with $p\equiv1\; \rm mod\; 4$ . That is, for any integer $k$ , we have the identity

$K_k(p) = (4p+2)\cdot K_{k-2}(p)-8\left(2\alpha^2-p\right)\cdot K_{k-3}(p)+(16\alpha^4-16p\alpha^2+4p-1)\cdot K_{k-4}(p).$

Thus, the problems of calculating a linear recurrence formula of one kind special character sums modulo a prime are given.

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 are grateful to the anonymous referee for very helpful and detailed comments.

This work is supported by the N.S.F. (11971381, 12371007) of China and Shaanxi Fundamental Science Research Project for Mathematics and Physics (22JSY007).

Conflict of interest

The authors declare no conflicts of interest.

References

[1]	L. Marsha, K. R. Conroy, J. L. Davis, et al. Atlas Pathophysiology, Lippincott Williams & Wilkins, 2010.
[2]	V. Kumar, A. Abbas, J. Aster, Robbins and cotran pathologic basis of disease, Canada: Elsevier, 2014.
[3]	M. Zanetti, Tapping CD4 T cells for cancer immunotherapy: The choice of personalized genomics, J. Immunol., 194 (2015), 2049-2056. doi: 10.4049/jimmunol.1402669
[4]	D. Cassell, J. Forman, Linked recognition of helper and cytotoxic antigenic determinants for he generation of cytotoxic T lymphocytes, Ann. N. Y. Acad. Sci., 532 (1998), 51-60.
[5]	H. Choudhry, N. Helmi, W. H. Abdulaal, et al. Prospects of IL-2 in cancer immunotherapy, BioMed Res. Int., 2018 (2018), 9056173.
[6]	A. A. Kilbas, H. M. Srivastava, J. J. Trujillo, Theory and applications of fractional differential equations, Elsevier, 2006.
[7]	D. Baleanu, K. Diethelm, E. Scalas, et al. Fractional calculus models and numerical methods, World Scientific, 2012
[8]	N. Özdemir, D. Karadeniz, B. B. Iskender, Fractional optimal control problem of a distributed system in cylindrical coordinates, Phys. Lett. A, 373 (2009), 221-226. doi: 10.1016/j.physleta.2008.11.019
[9]	F. Evirgen, N. Özdemir, Multistage adomian decomposition method for solving NLP problems over a nonlinear fractional dynamical system, J. Comput. Nonlinear Dyn., 6 (2011), 21003.
[10]	F. Evirgen, Analyze the optimal solutions of optimization problems by means of fractional gradient based system using VIM, An International Journal of Optimization and Control: Theories & Applications (IJOCTA), 6 (2016), 75-83.
[11]	Z. Hammouch, T. Mekkaoui, Circuit design and simulation for the fractional-order chaotic behavior in a new dynamical system, Complex Intell. Syst., 4 (2018), 251-260. doi: 10.1007/s40747-018-0070-3
[12]	E. Bonyah, A. Atangana, M. A. Khan, Modeling the spread of computer virus via Caputo fractional derivative and the beta derivative, Asia Pacific Journal on Computational Engineering, 4 (2017), 1-15. doi: 10.1186/s40540-016-0019-1
[13]	N. Özdemir, M. Yavuz, Numerical solution of fractional Black-Scholes equation by using the multivariate pade approximation, Acta Phys. Pol. A., 132 (2017), 1050-1053. doi: 10.12693/APhysPolA.132.1050
[14]	E. Uçar, N. Özdemir, E. Altun, Fractional order model of immune cells influenced by cancer cells, Math. Model. Nat. Phenom., 14 (2019), 308.
[15]	A. Atangana, D. Baleanu, New fractional derivatives with non-local and non-singular kernel: theory and applications to heat transfer model, Therm. Sci., 20 (2016), 763-769. doi: 10.2298/TSCI160111018A
[16]	M. Yavuz, N. Özdemir, H. M. Baskonus, Solutions of partial differential equations using the fractional operator involving Mittag-Leffler kernel, Eur. Phys. J. Plus, 133 (2018), 215.
[17]	V. F. Morales-Delgadoa, J. F. Gomez-Aguilar, M. A. Taneco-Hernandez, et al. Mathematical modeling of the smoking dynamics using fractional differential equations with local and nonlocal kernel, J. Nonlinear Sci. Appl., 11 (2018), 994-1014. doi: 10.22436/jnsa.011.08.06
[18]	N. A. Asif, Z. Hammouch, M. B. Riaz, et al. Analytical solution of a Maxwell fluid with slip effects in view of the Caputo-Fabrizio derivative, Eur. Phys. J. Plus, 133 (2018), 272.
[19]	I. Koca, Analysis of rubella disease model with non-local and non-singular fractional derivatives, Int. J. Optim. Control Theor. Appl. IJOCTA, 8 (2018), 17-25.
[20]	D. Avcı A. Yetim, Analytical solutions to the advection-diffusion equation with the AtanganaBaleanu derivative over a finite domain, J. BAUN Inst. Sci. Technol., 20 (2018), 382-395.
[21]	S. Uçar, E. Uçar, N. Özdemir, et al. Mathematical analysis and numerical simulation for a smoking model with Atangana-Baleanu derivative, Chaos, Solitons & Fractals, 118 (2019), 300-306.
[22]	D. Baleanu, A. Fernandez, On some new properties of fractional derivatives with Mittag-Leffler kernel, Commun. Nonlinear Sci. Numer. Simulat., 59 (2018), 444-462. doi: 10.1016/j.cnsns.2017.12.003
[23]	A. Fernandez, D. Baleanu, H. M. Srivastava, Series representations for fractional-calculus operators involving generalised Mittag-Leffler functions, Commun. Nonlinear Sci. Numer. Simulat., 67 (2019), 517-527. doi: 10.1016/j.cnsns.2018.07.035
[24]	S. Uçar, Existence and uniqueness results for a smoking model with determination and education in the frame of non-singular derivatives, Discrete Continuous Dyn. Syst. Ser. S, in press.
[25]	F. Evirgen, S. Uçar, N. Özdemir, et al. System response of an alcoholism model under the effect of immigration via non-singular kernel derivative, Discrete Continuous Dyn. Syst. Ser. S, in press.
[26]	J. E. Solis-Perez, J. F. Gomez-Aguilar, A. Atangana, A factional mathematical model of breast cancer competition model, Chaos, Solitons and Fractals, 127 (2019), 38-54. doi: 10.1016/j.chaos.2019.06.027
[27]	V. F. Morales-Delgado, J. F. Gomez-Aguilar, K. Saad, et al. Application of the Caputo-Fabrizio and Atangana-Baleanu fractional derivatives to mathematical model of cancer chemotherapy effect, Math. Methods Appl. Sci., 42 (2019), 1167-1193. doi: 10.1002/mma.5421
[28]	P. Vereesha, D. G. Prakasha, H. M. Baskonus, New numerical surfaces to the mathematical model of cancer chemotherapy effect in Caputo fractional derivatives, CHAOS, 29 (2019), 1-13.
[29]	A. Minelli, F. Topputo, F. Bernelli, Controlled drug delivery in cancer immunotherapy: Stability, optimization and monte carlo analysis, SIAM J. Appl. Math., 71 (2011), 2229-2245. doi: 10.1137/100815190
[30]	L. G. De Pillis, A. Radunskaya, A mathematical tumour model with immune resistance and drug therapy: An optimal control approach, Journal of Theoretical Medicine, 3 (2001), 79-100. doi: 10.1080/10273660108833067
[31]	F. Castiglione, B. Piccoli, Cancer immunotheraphy, mathematical modeling and optimal control, J. Theor. Biol., 247 (2007), 723-732. doi: 10.1016/j.jtbi.2007.04.003
[32]	D. Baleanu, A. Jajarmi, M. Hajipour, On the nonlinear dynamical systems within the generalized fractional derivatives with Mittag-Leffler kernel, Nonlinear Dyn., 94 (2018), 397-414. doi: 10.1007/s11071-018-4367-y

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