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On the Diophantine Equation (75n)x+ (308n)y= (317n)z
DOI: 10.12677/PM.2023.1311348, PDF, HTML, XML, 下载: 300  浏览: 506

Abstract: Let a,b,c be a primitive Pythagogrean triples such that a2+b2=c2. Jesmanowicz conjectured that, for any positive integer n, the Diophantine equation (an)x+(bn)y=(cn)z has only positive integer solution (x,y,z)=(2,2,2). In this paper, by using some methods of number theory,we prove that, for any positive integer n, the Diophantine equation (75n)x+ (308n)y= (317n)z has only positive integer solution (x,y,z)=(2,2,2), that is the Jesmanowicz conjecture is true, when (a,b,c)=(75,308,317).

1. 引言

$\left(a,b,c\right)$ 是本原商高数组。Jesmanowicz [1] 曾猜想：对于任意正整数n，丢番图方程

${\left(na\right)}^{x}+{\left(nb\right)}^{z}={\left(nc\right)}^{z}$ (1)

${\left({m}^{2}-4\right)}^{x}+{\left(4m\right)}^{y}=\left({m}^{2}+4\right),m>1,m\equiv 1\left(\mathrm{mod}2\right)$

${\left(3\left(2k+3\right)\right)}^{x}+{\left(2k\left(k+3\right)\right)}^{y}={\left(2k\left(k+3\right)+9\right)}^{z}$

$\left(a,b,c\right)=\left({2}^{2r}-1,{2}^{r+1},{2}^{2r}+1\right),r\in N$ 时，Jesmanowicz猜想正确；陈凤娟 [5] 证明了当

$\left(a,b,c\right)=\left({p}^{2r}-4,4{p}^{r},{p}^{2r}+4\right),p>3$ ，p为素数， $p\equiv 3,5,7\left(\mathrm{mod}8\right)$ 时，若 $P\left(n\right)|b$$P\left(a\right)|n$ ，则Jesmanowicz猜想正确；孙翠芳 [6] 证明了当 $\left(a,b,c\right)=\left(4{p}^{2r}-1,4{p}^{r},4{p}^{2r}+1\right),p\equiv 3\left(\mathrm{mod}4\right)$ 且p为素数时，若 $P\left(n\right)|b$$P\left(a\right)|n$ ，则Jesmanowicz猜想正确。等等。虽然在很多特殊情况下，Jesmanowicz猜想是正确的，但一般情形仍未解决，目前的结果大都集中在 $n=1$ 的情形，而对于 $n>1$ ，只有为数不多的特殊情形被解决。记

$P\left(n\right)$ 为n的所有素因子的乘积， $\left(\frac{\ast }{\ast }\right)$ 为雅克比符号。

${\left(75n\right)}^{x}+{\left(308n\right)}^{y}={\left(317n\right)}^{z}$ (2)

2. 预备知识

$\left(\frac{a}{p}\right)=\left\{\begin{array}{l}1,\text{ }\text{ }a\text{\hspace{0.17em}}是模\text{\hspace{0.17em}}p\text{\hspace{0.17em}}的二次剩余,\\ -1,\text{ }\text{\hspace{0.17em}}\text{ }a\text{\hspace{0.17em}}是模\text{\hspace{0.17em}}p\text{\hspace{0.17em}}的二次非剩余,\\ 0,\text{ }\text{ }p|a\end{array}$

$\left(\frac{a}{m}\right)=i=\underset{i=1}{\overset{r}{\prod }}\left(\frac{a}{{p}_{i}}\right)$

1) 若 $a\equiv {a}_{1}\left(\mathrm{mod}m\right)$$\left(m,n\right)=1$ ，则 $\left(\frac{a}{m}\right)=\left(\frac{{a}_{1}}{m}\right)$

2) 若 $\left(a,m\right)=\left(a,n\right)=1$ ，则 $\left(\frac{a}{m}\right)\left(\frac{a}{n}\right)=\left(\frac{a}{mn}\right)$

3) 若 $\left(a,m\right)=\left(b,m\right)=1$ ，则 $\left(\frac{a}{m}\right)\left(\frac{b}{m}\right)=\left(\frac{ab}{m}\right)$

$\left(\frac{m}{n}\right)={\left(-1\right)}^{\frac{\left(m-1\right)\left(n-1\right)}{4}}\left(\frac{n}{m}\right)$

${\left(3\left(2k+3\right)\right)}^{x}+{\left(2k\left(k+3\right)\right)}^{y}={\left(2k\left(k+3\right)+9\right)}^{z}$

1) $x>z>y,P\left(n\right)|b$

2) $y>z>x,P\left(n\right)|a$

3. 定理的证明

${317}^{z}-{308}^{y}{n}^{y-z}=\frac{{75}^{x}}{{n}^{z-x}}$ (3)

${317}^{z}-{308}^{y}\cdot {3}^{s\left(y-z\right)}\cdot {5}^{t\left(y-z\right)}=1$ (4)

$\left({317}^{{z}_{1}}-1\right)\left({317}^{{z}_{1}}+1\right)={308}^{y}\cdot {3}^{s\left(y-z\right)}\cdot {5}^{t\left(y-z\right)}={2}^{2y}\cdot {7}^{y}\cdot {11}^{y}\cdot {3}^{s\left(y-z\right)}\cdot {5}^{t\left(y-z\right)}$ 。因为 $\mathrm{gcd}\left({317}^{{z}_{1}}-1,{317}^{{z}_{1}}+1\right)=2$ ，由

$\left\{\begin{array}{l}{317}^{{z}_{1}}-1={2}^{2y-1}\cdot {7}^{y}\cdot {11}^{y}\cdot {3}^{s\left(y-z\right)}\\ {317}^{{z}_{1}}+1=2\cdot {5}^{t\left(y-z\right)}\end{array}$

${2}^{2y-1}\cdot {7}^{y}\cdot {11}^{y}|\left({317}^{2{z}_{1}}-1\right)$ 。但是

${2}^{2y-1}\cdot {7}^{y}\cdot {11}^{y}>{4}^{y-1}\cdot {7}^{y}\cdot {11}^{y}>{\left(4\cdot 7\cdot 11\right)}^{z}={308}^{z}={308}^{4{z}_{1}}>{\left(317+1\right)}^{2{z}_{1}}>{317}^{2{z}_{1}}+1>{317}^{2{z}_{1}}-1$ 这不可能。

${317}^{z}-{308}^{y}\cdot {3}^{s\left(y-z\right)}={5}^{2x}$ (5)

${308}^{y}\cdot {3}^{s\left(y-z\right)}=-{5}^{2x}\left(\mathrm{mod}317\right)$ (6)

$\left(\frac{308}{317}\right)=\left(\frac{4}{317}\right)\left(\frac{7}{317}\right)\left(\frac{11}{317}\right)=\left(\frac{2}{7}\right)\left(\frac{9}{11}\right)=1,\left(\frac{3}{317}\right)=\left(\frac{317}{3}\right)=\left(\frac{2}{3}\right)=-1,\left(\frac{-1}{317}\right)=-1$

$\left\{\begin{array}{l}{317}^{{z}_{1}}-{308}^{{y}_{1}}\cdot {3}^{s\left({y}_{1}-{z}_{1}\right)}=1\text{ }\text{ }\text{ }①\\ {317}^{{z}_{1}}+{308}^{{y}_{1}}\cdot {3}^{s\left({y}_{1}-{z}_{1}\right)}={5}^{2x}\text{ }\text{ }\text{ }②\end{array}$

$②-①$ 得， ${5}^{2x}-1=2\cdot {308}^{{y}_{1}}\cdot {3}^{s\left({y}_{1}-{z}_{1}\right)}$ ，即 $\left({5}^{x}-1\right)\left({5}^{x}+1\right)={2}^{2{y}_{1}+1}\cdot {7}^{{y}_{1}}\cdot {11}^{{y}_{1}}\cdot {3}^{s\left({y}_{1}-{z}_{1}\right)}$ ，假设 ${5}^{x}-1=7t$ ，即

${5}^{x}\equiv 1\left(\mathrm{mod}7\right)$ ，因为 ${\left(\frac{5}{7}\right)}^{x}=\left(\frac{1}{7}\right)=1$$\left(\frac{5}{7}\right)=\left(\frac{7}{5}\right)=\left(\frac{2}{5}\right)=-1$ ，所以 ${\left(-1\right)}^{x}=1$ ，即 $2|x$ 。此时若 ${5}^{x}+1=3t$${5}^{x}\equiv -1\left(\mathrm{mod}3\right)$ ，因为 ${\left(\frac{5}{3}\right)}^{x}=\left(\frac{-1}{3}\right)=-1$$\left(\frac{5}{3}\right)=\left(\frac{2}{3}\right)=1$ ，所以 ${1}^{x}=-1$ ，这不可能。又因为 ${5}^{x}-1<{5}^{x}+1$

$\left\{\begin{array}{l}{5}^{x}-1={2}^{2{y}_{1}}\cdot {11}^{{y}_{1}}\\ {5}^{x}+1={7}^{{y}_{1}}\cdot {3}^{s\left({y}_{1}-{z}_{1}\right)}\end{array}$

$\2|x$ ，于是 $\8|\left({5}^{x}-1\right)$ ，即 $y=2{y}_{1}=2$ ，这与 $1\le x 矛盾。

${317}^{z}-{308}^{y}\cdot {5}^{t\left(y-z\right)}={3}^{x}$ (7)

$\left(\frac{308}{317}\right)=1,\left(\frac{5}{317}\right)=\left(\frac{2}{5}\right)=-1,\left(\frac{-1}{317}\right)=1,\left(\frac{3}{317}\right)=-1$

$\left({317}^{{z}_{1}}-{308}^{{y}_{1}}\cdot {5}^{t\left({y}_{1}-{z}_{1}\right)}\right)\left({317}^{{z}_{1}}+{308}^{{y}_{1}}\cdot {5}^{t\left({y}_{1}-{z}_{1}\right)}\right)={3}^{2{x}_{1}}$

$\left\{\begin{array}{l}{317}^{{z}_{1}}-{308}^{{y}_{1}}\cdot {5}^{t\left({y}_{1}-{z}_{1}\right)}=1\\ {317}^{{z}_{1}}+{308}^{{y}_{1}}\cdot {5}^{t\left({y}_{1}-{z}_{1}\right)}={3}^{2{x}_{1}}\end{array}$

${317}^{z}-{75}^{x}{n}^{x-z}=\frac{{308}^{y}}{{n}^{z-y}}$ (8)

${317}^{z}-{75}^{x}\cdot {2}^{2t\left(x-z\right)}\cdot {7}^{t\left(x-z\right)}\cdot {11}^{t\left(x-z\right)}=1$ (9)

$\left({317}^{{z}_{1}}-1\right)\left({317}^{{z}_{1}}+1\right)={3}^{x}\cdot {5}^{2x}\cdot {2}^{2t\left(x-z\right)}\cdot {7}^{t\left(x-z\right)}\cdot {11}^{t\left(x-z\right)}$

$\left\{\begin{array}{l}{317}^{{z}_{1}}-1={2}^{2t\left(x-z\right)-1}\cdot {3}^{x}\cdot {7}^{t\left(x-z\right)}\cdot {11}^{t\left(x-z\right)}\\ {317}^{{z}_{1}}+1=2\cdot {5}^{2x}\end{array}$

${5}^{2x}-{3}^{x}\cdot {2}^{2t\left(x-z\right)-2}\cdot {7}^{t\left(x-z\right)}\cdot {11}^{t\left(x-z\right)}=1$ 。故 $\left({5}^{x}-1\right)\left({5}^{x}+1\right)={3}^{x}\cdot {2}^{2t\left(x-z\right)-2}\cdot {7}^{t\left(x-z\right)}\cdot {11}^{t\left(x-z\right)}$

$2|x$ 时，则 ${5}^{x}+1=2$ ，这不可能。

${3}^{x}\cdot {7}^{t\left(x-z\right)}-{2}^{2t\left(x-z\right)-4}\cdot {11}^{t\left(x-z\right)}=1$ (10)

$1=-1$ ，这不可能。

${317}^{z}-{75}^{x}\cdot {2}^{2t\left(x-z\right)}\cdot {7}^{t\left(x-z\right)}={11}^{y}$ (11)

(11)式模16得 ${13}^{z}={11}^{y}\left(\mathrm{mod}16\right)$ 。又因为 $4|z$ ，故 $4|y$ 。令 $z=4{z}_{1},y=4{y}_{1}$ ，(11)式整理为

$\left({317}^{{z}_{1}}-{11}^{2{y}_{1}}\right)\left({317}^{2{z}_{1}}+{11}^{2{y}_{1}}\right)={75}^{x}\cdot {2}^{2t\left(x-z\right)}\cdot {7}^{t\left(x-z\right)}$

$\left\{\begin{array}{l}{317}^{2{z}_{1}}+{11}^{2{y}_{1}}=2\cdot {5}^{2x}\\ {317}^{2{z}_{1}}-{11}^{2{y}_{1}}={2}^{2t\left(x-z\right)-1}\cdot {3}^{x}\cdot {7}^{t\left(x-z\right)}\end{array}$

${317}^{z}-{75}^{x}\cdot {2}^{2w\left(x-z\right)}\cdot {11}^{w\left(x-z\right)}={7}^{y}$ (12)

$z=2{z}_{1},y=2{y}_{1}$ ，故 $\left({317}^{{z}_{1}}-{7}^{{y}_{1}}\right)\left({317}^{{z}_{1}}+{7}^{{y}_{1}}\right)={75}^{x}\cdot {2}^{2w\left(x-z\right)}\cdot {11}^{w\left(x-z\right)}$

${317}^{z}-{75}^{x}\cdot {7}^{t\left(x-z\right)}\cdot {11}^{t\left(x-z\right)}={2}^{2y}$ (13)

${317}^{z}-{75}^{x}\cdot {2}^{s\left(x-z\right)}={7}^{y}\cdot {11}^{y}$ (14)

$\left({317}^{{z}_{1}}-{77}^{{y}_{1}}\right)\left({317}^{{z}_{1}}+{77}^{{y}_{1}}\right)={75}^{x}\cdot {2}^{s\left(x-z\right)}$

${317}^{z}-{75}^{x}\cdot {7}^{t\left(z-y\right)}={44}^{y}$ (15)

${317}^{z}-{75}^{x}\cdot {11}^{w\left(z-y\right)}={28}^{y}$ (16)

$z=2{z}_{1},y=2{y}_{1}$ ，(16)式整理为 $\left({317}^{{z}_{1}}-{28}^{{y}_{1}}\right)\left({317}^{{z}_{1}}+{28}^{{y}_{1}}\right)={75}^{x}\cdot {11}^{e\left(z-y\right)}$ ，又因为 $\mathrm{gcd}\left({317}^{{z}_{1}}-{26}^{{y}_{1}},{317}^{{z}_{1}}+{28}^{{y}_{1}}\right)=1$ ，故 ${25}^{x}|\left({317}^{{z}_{1}}-{28}^{{y}_{1}}\right)$${25}^{x}|\left({317}^{{z}_{1}}+{28}^{{y}_{1}}\right)$ 。这与 ${25}^{x}>{25}^{2{z}_{1}}>{\left(317+28\right)}^{{z}_{1}}>{317}^{{z}_{1}}+{28}^{{y}_{1}}>{317}^{{z}_{1}}-{28}^{{y}_{1}}$ 矛盾。

4. 结语

 [1] Jesmanowicz, L. (1955-1956) Several Remarks on Pythagorean Numbers. Wiadom Mat., 1, 196-202. [2] Terai, N. (2014) On Jesmanowicz’ Conjecture Concerning Primitive Pythagorean Triples. Journal of Number Theory, 141, 316-323. https://doi.org/10.1016/j.jnt.2014.02.009 [3] 李双志. 关于商高数的Jesmanowicz猜想[D]: [硕士学位论文]. 重庆: 西南大学, 2011. [4] Miyazaki, T. (2015) A Remark on Jesmanowicz’ Conjecture for Non-Coprimality Case. Acta Mathematica Sinica, English Series, 31, 1225-1260. https://doi.org/10.1007/s10114-015-4491-2 [5] 陈凤娟. 关于丢番图方程 [J]. 数学进展, 2018, 47(3): 1-5. [6] Sun, C.F. and Cheng, Z. (2015) On Jesmanowicz’ Conjecture Concerning Pythagorean Triples. Journal of Mathematical Research with Applications, 35, 143-148. [7] 闵嗣鹤, 严士健. 初等数论[M]. 北京: 高等教育出版社, 2003. [8] 柯召, 孙琦. 数论讲义(上) [M]. 北京: 高等教育出版社, 1987. [9] 邢静静. 关于商高数的Jesmanowicz猜想[D]: [硕士学位论文]. 重庆: 西南大学, 2015. [10] Deng, M.J. (2014) A Note on the Dionphantine Equation . Bulletin of the Australian Mathematical Society, 89, 316-321. https://doi.org/10.1017/S000497271300066X