﻿ 关于不定方程5x(x 1)(x 2)(x 3) = 42y(y 1)(y 2)(y 3) On the Diophantine Equation 5x(x 1)(x 2)(x 3) = 42y(y 1)(y 2)(y 3)

Vol. 13  No. 05 ( 2024 ), Article ID: 87681 , 5 pages
10.12677/aam.2024.135198

On the Diophantine Equation 5x(x + 1)(x + 2)(x + 3) = 42y(y + 1)(y + 2)(y + 3)

Yibao Zhang

School of Mathematics and Statistics, Southwest University, Chongqing

Received: Apr. 23rd, 2024; accepted: May 17th, 2024; published: May 29th, 2024

ABSTRACT

This article uses elementary methods such as congruence formula, recursive sequences, and Pell equation to prove that indefinite Diophantine equation 5x(x + 1)(x + 2)(x + 3) = 42x(x + 1)(x + 2)(x + 3) has a unique positive integer (x, y) = (6, 3). Also, All 20 groups of integer solutions of the equation are found.

Keywords:Diophantine Equation, Congruence Formula, Pell Equation, Positive Integer Solution

1. 引言与结论

$5x\left(x+1\right)\left(x+2\right)\left(x+3\right)=42y\left(y+1\right)\left(y+2\right)\left(y+3\right)$ (1)

2. 预备知识

${\left[5\left({x}^{2}+3x+1\right)\right]}^{2}-210{\left({y}^{2}+3y+1\right)}^{2}=-185$ (2)

${x}_{n}+{y}_{n}\sqrt{210}=±\left(5+\sqrt{210}\right)\left({u}_{n}+{v}_{n}\sqrt{210}\right)=±\left(5+\sqrt{210}\right){\left(29+2\sqrt{210}\right)}^{n}\text{}n\in ℕ$

${\stackrel{¯}{x}}_{n}+{\stackrel{¯}{y}}_{n}\sqrt{210}=±\left(-5+\sqrt{210}\right)\left({u}_{n}+{v}_{n}\sqrt{210}\right)=±\left(-5+\sqrt{210}\right){\left(29+2\sqrt{210}\right)}^{n}\text{}n\in ℕ$

${\left(2y+3\right)}^{2}=±4{y}_{n}+5\text{}n\in ℕ$ (3)

${y}_{n+1}=58{y}_{n}-{y}_{n-1},\text{}{y}_{0}=1,\text{}{y}_{1}=39$ (4)

${u}_{n+1}=58{u}_{n}-{u}_{n-1},\text{}{u}_{0}=1,\text{}{u}_{1}=29$ (5)

${v}_{n+1}=58{v}_{n}-{v}_{n-1},\text{}{v}_{0}=0,\text{}{v}_{1}=2$ (6)

${u}_{2n}={u}_{n}^{2}+210{v}_{n}^{2}=2{u}_{n}^{2}-1,\text{}{v}_{2n}=2{u}_{n}{v}_{n}$ (7)

${y}_{n}={u}_{n}+5{v}_{n}$ (8)

${u}_{n+2km}\equiv {\left(-1\right)}^{k}{u}_{n}\left(\mathrm{mod}{u}_{m}\right)$ (9)

${v}_{n+2km}\equiv {\left(-1\right)}^{k}{v}_{n}\left(\mathrm{mod}{u}_{m}\right)$ (10)

${y}_{n+2km}\equiv {\left(-1\right)}^{k}{y}_{n}\left(\mathrm{mod}{u}_{m}\right)$ (11)

3. 分类讨论

3.1. 当(2y + 3)2 = 4yn + 5时

$\begin{array}{l}\left(\frac{±20{v}_{2m}+5}{{u}_{2m}}\right)=\left(\frac{±20{v}_{2m}+10{u}_{m}^{2}}{{u}_{2m}}\right)=\left(\frac{±40{u}_{m}{v}_{m}+10{u}_{m}^{2}}{{u}_{2m}}\right)\\ =\left(\frac{2}{{u}_{2m}}\right)\left(\frac{5}{{u}_{2m}}\right)\left(\frac{{u}_{m}}{{u}_{2m}}\right)\left(\frac{{u}_{m}±4{v}_{m}}{{u}_{2m}}\right)=\left(\frac{-1}{{u}_{m}}\right)\left(\frac{{u}_{2m}}{{u}_{m}±4{v}_{m}}\right)\\ =\left(\frac{{u}_{m}^{2}+210{v}_{m}^{2}}{{u}_{m}±4{v}_{m}}\right)=\left(\frac{210{v}_{m}^{2}+{\left(±4{v}_{m}\right)}^{2}}{{u}_{m}±4{v}_{m}}\right)\\ =\left(\frac{226}{{u}_{m}±4{v}_{m}}\right)=\left(\frac{2}{{u}_{m}±4{v}_{m}}\right)\left(\frac{113}{{u}_{m}±4{v}_{m}}\right)=-\left(\frac{{u}_{m}±4{v}_{m}}{113}\right)\end{array}$

mod 311，排除 $n\equiv 1,3\left(\mathrm{mod}5\right)$ ，这是因为此时 $4{y}_{n}+5\equiv 161,55\left(\mathrm{mod}311\right)$ ，而161、55为mod 311的平方非剩余，故可排除 $n\equiv 1,3\left(\mathrm{mod}5\right)$ 的情形，剩余 $n\equiv 0,2,4\left(\mathrm{mod}5\right)$ 。为节省篇幅，下面不再重复排除的原因。

mod 41，排除 $n\equiv 1,2,3,5,6\left(\mathrm{mod}8\right)$ ，剩余 $n\equiv 0,4,7,12,15,20,24,32,39\left(\mathrm{mod}40\right)$

mod 661，排除 $n\equiv 3,4,6,8\left(\mathrm{mod}10\right)$ ，剩余 $n\equiv 0,7,12,15,20,32,39\left(\mathrm{mod}40\right)$

mod 239，47279，排除 $n\equiv 7,12,15\left(\mathrm{mod}20\right)$ ，剩余 $n\equiv 0,20,39\left(\mathrm{mod}40\right)$

mod 17，11467，排除 $n\equiv 2,4,5,6,7\left(\mathrm{mod}9\right)$ ，剩余 $n\equiv 0,39,80,100,120,180,199,260,279,280,300,359$ $\left(\mathrm{mod}360\right)$

mod 61，排除 $n\equiv 5\left(\mathrm{mod}15\right)$ ，剩余 $n\equiv 0,120,180,300,359\left(\mathrm{mod}360\right)$

mod 37 ，排除 $n\equiv 12\left(\mathrm{mod}36\right)$ ，剩余 $n\equiv 0,180,359\left(\mathrm{mod}360\right)$

$4{y}_{n}+5\equiv 4{u}_{n}+20{v}_{n}+5\equiv ±20{v}_{2m}+5\left(\mathrm{mod}{u}_{2m}\right)$

$\left\{{u}_{m}±4{v}_{m}\right\}$ 对mod 113的剩余序列周期为56，而 $\left\{{2}^{t}\right\}$ 对mod 56的剩余序列周期为3。

$m=\left\{\begin{array}{l}{2}^{t},\text{}t\equiv 0\left(\mathrm{mod}3\right)\hfill \\ 5×{2}^{t},\text{}t\equiv 1\left(\mathrm{mod}3\right)\hfill \\ 3×{2}^{t},\text{}t\equiv 2\left(\mathrm{mod}3\right)\hfill \end{array}$

Table 1. The situation of u m + 4 v m ( mod 113 )

$m=\left\{\begin{array}{l}{2}^{t},\text{}t\equiv 1,2\left(\mathrm{mod}3\right)\hfill \\ 5×{2}^{t},\text{}t\equiv 0\left(\mathrm{mod}3\right)\hfill \end{array}$

$4{y}_{n}+5=4{y}_{-1+2×\left(4k±1\right)×{3}^{2}×5×{2}^{t}}+5\equiv -4{y}_{-1}+5\left(\mathrm{mod}{u}_{m}\right)$

$m={2}^{t}$

3.2. 当(2y + 3)2 = −4yn + 5时

4. 定理证明

On the Diophantine Equation 5x(x 1)(x 2)(x 3) = 42y(y 1)(y 2)(y 3)[J]. 应用数学进展, 2024, 13(05): 2105-2109. https://doi.org/10.12677/aam.2024.135198

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