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The Number and Range of the Solutions of the Nonlinear Diophantine Equation φ(mn)=aφ(m)+bφ(n)+c with Euler Function
DOI: 10.12677/PM.2020.1012140, PDF, 下载: 150  浏览: 274

Abstract: Euler function φ(n) is a very important function in number theory. In this paper, we mainly discuss the necessary and sufficient conditions for the nonlinear Diophantine equation φ(mn)=aφ(m)+bφ(n)+c which has the finite number of positive integer solutions and the range of solutions whenever the number of solutions is limited. Therefore, for a given coefficient a, b, c, searching a given range by a computer, we can give all the positive integer solutions of the equation when the number of solutions is limited.

1. 引言

2. 相关引理

3. 定理及其证明

$\phi \left(mn\right)=a\phi \left(m\right)+b\phi \left(n\right)+c$ (1)

$\phi \left(d\right)\left(d{m}_{1}{n}_{1}-a{m}_{1}-b{n}_{1}\right)=c,$

$d{m}_{1}{n}_{1}-a{m}_{1}-b{n}_{1}={c}_{1},$ (2)

$\left(d{m}_{1}-b\right)\left(d{n}_{1}-a\right)={c}_{1}d+ab.$ (3)

$\left(d{m}_{1}-b\right)\left(d{n}_{1}-a\right)=0.$ (4)

$m\in \left[1,{\left(|c|+|ab|+|b|\right)}^{3/2}+8\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\in \left[1,{\left(|c|+|ab|+|a|\right)}^{3/2}+8\right].$

$d{m}_{1}{n}_{1}-a{m}_{1}-b{n}_{1}=0.$ (5)

$\left(d{m}_{1}-b\right)\left(d{n}_{1}-a\right)=ab.$ (6)

$a<0,b<0$ 时，由欧拉函数的定义可知方程(1)无解。

$a>0,b<0$$a<0,b>0$ 时。由方程(6)可知，当 $d{n}_{1}-a=-1$ 时， ${m}_{1}=\frac{-ab+b}{d}$，由 $\phi \left(m\right)={m}_{1}\phi \left(d\right)$， 得 $\phi \left(m\right)=\frac{\phi \left(d\right)\left(-ab+b\right)}{d}\le |ab|+|b|$，由推论2.7得 $m\in \left[1,{\left(|ab|+|b|\right)}^{3/2}+8\right]$。同理可得 $n\in \left[1,{\left(|ab|+|b|\right)}^{3/2}+8\right]$

$a>0,b>0$ 时。在方程(6)中，当 $d{n}_{1}-a=1$ 时， ${m}_{1}=\frac{ab+b}{d}$，由 $\phi \left(m\right)=\frac{ab+b}{d}\phi \left(d\right)\le |ab|+|b|$，得 $\phi \left(m\right)\in \left[1,|ab|+|b|\right]$，由推论2.7得 $m\in \left[1,{\left(|ab|+|b|\right)}^{3/2}+8\right]$。同理可得 $n\in \left[1,{\left(|ab|+|a|\right)}^{3/2}+8\right]$

${c}_{1}d+ab>0$ 时。由方程(3)得 $d{m}_{1}-b=\frac{{c}_{1}d+ab}{d{n}_{1}-a}$。可知当 $d{n}_{1}-a=\text{1}$ 时， ${m}_{1}$ 取得最大值 $\frac{{c}_{1}d+ab+b}{d}$，故 $\phi \left(m\right)={m}_{1}\phi \left(d\right)$ 取得最大值 $c+\frac{\phi \left(d\right)\left(ab+b\right)}{d}\le |c|+|ab|+|b|$，所以 $\phi \left(m\right)\in \left[1,\left(|c|+|ab|+|b|\right)\right]$，由推论2.7得 $m\in \left[1,{\left(|c|+|ab|+|b|\right)}^{3/2}+8\right]$。同理可得 $n\in \left[1,{\left(|c|+|ab|+|a|\right)}^{3/2}+8\right]$

${c}_{1}d+ab<0$ ( $c\le 0,c+ab<0$ 时，方程(1)有有限个解)，当 $d{n}_{1}-a=-1$ 时， ${m}_{1}=\frac{-{c}_{1}d-ab+b}{d}$，故 $\phi \left(m\right)={m}_{1}\phi \left(d\right)=-c-\frac{\phi \left(d\right)\left(ab-b\right)}{d}\le |c|+|ab|+|b|$，所以 $\phi \left(m\right)\in \left[1,|c|+|ab|+|b|\right]$，由推论2.7得 $m\in \left[1,{\left(|c|+|ab|+|b|\right)}^{3/2}+8\right]$。同理可得 $n\in \left[1,{\left(|c|+|ab|+|a|\right)}^{3/2}+8\right]$。证毕。

1.为了简化证明过程和统一结果，定理3.4中给出的解的范围有些偏大，对于具体给定的系数 $\left(a,b,c\right)$，该范围可适当缩小。如 $\left(a,b,c\right)=\left(-2,-6,20\right)$ 时，具体讨论后解的范围为 $m\in \left[1,\text{140}\right],n\in \left[1,\text{172}\right]$ (实际上，其全部解 $\left(3,4\right),\left(4,3\right)$ )，而定理中给出的解的范围是 $m\in \left[1,\text{242}\right],n\in \left[1,\text{206}\right]$

2.以下情况方程(1)的解数有限(不包括无解)：

1) $c=0,ab\ne 0$

2) $c\ge 0,c+ab>0$，特别当 $c=0,a=b>0$ 时，易知此时 $\left(\text{2}a,2a\right)$ 为方程(1)的一个解；

3) 当 $c\le 0,c+ab<0$

3.以下情况方程(1)无正整数解

1) $c\le 0,a\le 0,b\le 0$

2) $a,b$ 都为偶数，c为奇数；

3) $\left(a,b,c\right)$ 为本原勾股数，即 $\mathrm{gcd}\left(a,b\right)=1$，且满足 ${a}^{2}+{b}^{2}={c}^{2}$

Table 1. Some examples

$\begin{array}{l}\left(x,y\right)=\left(15,41\right),\left(16,41\right),\left(20,41\right),\left(24,41\right),\left(30,41\right),\left(16,55\right),\left(24,55\right),\left(16，75\right),\left(15,82\right),\left(15,88\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(11,17\right),\left(11,32\right),\left(11,34\right),\left(11,40\right),\left(11,48\right),\left(11,60\right),\left(22,17\right),\left(17,15\right),\left(17,16\right),\left(17,20\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(17,24\right),\left(17,30\right),\left(32,15\right)\left(34,15\right),\left(8,70\right),\left(8,78\right),\left(8,90\right),\left(10,52\right),\left(10,56\right),\left(10,72\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(10,78\right),\left(10,84\right),\left(12,70\right),\left(9,48\right),\left(9,60\right),\left(15,24\right),\left(24,15\right),\left(27,12\right),\left(8,52\right),\left(8,84\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(12,52\right),\left(12,56\right),\left(8,56\right),\left(8,72\right),\left(10,20\right),\left(10,30\right).\end{array}$

$\left(x,y\right)=\left(16,2\right),\left(20,2\right),\left(24,2\right),\left(30,2\right).$

$\begin{array}{l}\left(x,y\right)=\left(8,38\right),\left(8,54\right)，\left(10,38\right),\left(10,54\right),\left(12,38\right),\left(15,29\right),\left(15,58\right),\left(16,29\right),\left(20,10\right),\left(20,29\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(24,29\right),\left(30,10\right),\left(30,29\right),\left(48,138\right),\left(53,7\right),\left(53,9\right),\left(53,14\right),\left(53,18\right),\left(60,138\right),\left(75，12\right).\end{array}$

$\begin{array}{l}\left(x,y\right)=\left(13,161\right),\left(13,201\right),\left(13,207\right),\left(13,268\right),\left(13,322\right),\left(13,402\right),\left(13,414\right),\left(21,268\right),\left(26,161\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(26,201\right),\left(26,207\right),\left(36,161\right),\left(161,13\right),\left(201,13\right),\left(207,13\right),\left(268,13\right),\left(322,13\right),\left(402,13\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(414,13\right),\left(268,21\right),\left(161,26\right),\left(201,26\right),\left(207,26\right),\left(161,36\right),\left(22,22\right),\left(33,44\right),\left(44,33\right).\end{array}$

$\begin{array}{l}\left(x,y\right)=\left(87,16\right),\left(87,20\right),\left(116,15\right),\left(16,87\right),\left(20,87\right),\left(15,116\right),\left(8,58\right),\left(10,58\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(12,58\right),\left(58,8\right),\left(58,10\right),\left(58,12\right),\left(21,28\right),\left(28,21\right),\left(14,14\right).\end{array}$

$\begin{array}{l}\left(x,y\right)=\left(73,15\right),\left(73,16\right),\left(73,20\right),\left(73,24\right),\left(73,30\right),\left(91,15\right),\left(91,16\right),\left(91,20\right),\left(91,24\right),\left(91,30\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(95,16\right),\left(95,24\right),\left(111,16\right),\left(111,20\right),\left(117,16\right),\left(117,20\right),\left(135,16\right),\left(135,20\right),\left(146,15\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(148,15\right),\left(152,15\right),\left(182,15\right),\left(31,11\right),\left(31,22\right),\left(31,11\right),\left(62,11\right),\left(17,32\right),\left(17,40\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(17,48\right),\left(17,60\right),\left(32,17\right),\left(40,17\right),\left(48,17\right),\left(60,17\right),\left(13,29\right),\left(13,58\right),\left(21,29\right),\left(21,58\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(26,29\right),\left(28,29\right),\left(36,29\right),\left(42,29\right),\left(11,71\right),\left(11,142\right),\left(22,71\right),\left(74,8\right),\left(74,10\right),\left(74,12\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(76,10\right),\left(114,8\right),\left(114,10\right),\left(126,8\right),\left(126,10\right),\left(16,30\right),\left(30,16\right),\left(76,8\right),\left(108,8\right),\left(76,12\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(16,20\right),\left(20,16\right),\left(20,24\right),\left(24,20\right),\left(35,15\right),\left(35,20\right),\left(35,30\right),\left(45,20\right),\left(70,15\right),\left(16,16\right).\end{array}$

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