scholarly journals 3x + 1 Problem: Iterative Operations Neither Cause Loops nor Diverge to Infinity

2021 ◽  
Author(s):  
Li Jiang
Keyword(s):  
Positive Integer ◽  
General Formula ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Basic Theorem ◽  
The Relationship ◽  
The Way

The 3x+1 problem is a problem of continuous iteration for integers. According to the basic theorem of arithmetic and the way of iteration, we derive a general formula for continuous iteration for odd integers. Through this formula, we can construct a loop iteration equation and obtain the result of the equation: the equation has only one positive integer solution. In addition, this general formula can be converted into a linear indeterminate equation. The process of solving this equation shows that the relationship between the iteration result and the odd number being iterated is linear. Extending this result to all positive even numbers, we get the answer to the 3x + 1 question.

scholarly journals 3x + 1 Problem: Iterative Operations Neither Cause Loops nor Diverge to Infinity

2021 ◽  
Author(s):  
Li Jiang
Keyword(s):  
Positive Integer ◽  
General Formula ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Basic Theorem ◽  
The Relationship ◽  
The Way

The 3x+1 problem is a problem of continuous iteration for integers. According to the basic theorem of arithmetic and the way of iteration, we derive a general formula for continuous iteration for odd integers. Through this formula, we can construct a loop iteration equation and obtain the result of the equation: the equation has only one positive integer solution. In addition, this general formula can be converted into a linear indeterminate equation. The process of solving this equation shows that the relationship between the iteration result and the odd number being iterated is linear. Extending this result to all positive even numbers, we get the answer to the 3x + 1 question.

scholarly journals 3x + 1 problem: Iterative operations neither cause loops nor diverge to infinity

2021 ◽  
Author(s):  
Li Jiang
Keyword(s):  
Positive Integer ◽  
General Formula ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Basic Theorem ◽  
The Relationship ◽  
The Way

Abstract The 3x+1 problem is a problem of continuous iteration for integers. According to the basic theorem of arithmetic and the way of iteration, we derive a general formula for continuous iteration for odd integers. Through this formula, we can construct a loop iteration equation and obtain the result of the equation: the equation has only one positive integer solution. In addition, this general formula can be converted into a linear indeterminate equation. The process of solving this equation shows that the relationship between the iteration result and the odd number being iterated is linear. Extending this result to all positive even numbers, we get the answer to the 3x + 1 question.

scholarly journals On the Diophantine Equation x2 – kxy + ky2 + ly = 0, l = 2n

2017 ◽  
Vol 55 (1) ◽  
pp. 115-118
Author(s):  
Sukrawan Mavecha
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Positive Integer Solution

AbstractWe consider the Diophantine equation x2-kxy+ky2+ ly = 0 for l = 2nand determine for which values of the odd integer k, it has a positive integer solution x and y.

scholarly journals On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4)

2021 ◽  
Vol 6 (10) ◽  
pp. 10596-10601
Author(s):  
Yahui Yu ◽  
◽  
Jiayuan Hu ◽  
Keyword(s):  
Positive Integer ◽  
Unsolved Problem ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Elementary Methods ◽  
Fixed Positive Integer ◽  
Almost All ◽  
Terai’S Conjecture ◽  
Positive Integers

<abstract><p>Let $ k $ be a fixed positive integer with $ k &gt; 1 $. In 2014, N. Terai <sup>[<xref ref-type="bibr" rid="b6">6</xref>]</sup> conjectured that the equation $ x^2+(2k-1)^y = k^z $ has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. This is still an unsolved problem as yet. For any positive integer $ n $, let $ Q(n) $ denote the squarefree part of $ n $. In this paper, using some elementary methods, we prove that if $ k\equiv 3 $ (mod 4) and $ Q(k-1)\ge 2.11 $ log $ k $, then the equation has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. It can thus be seen that Terai's conjecture is true for almost all positive integers $ k $ with $ k\equiv 3 $(mod 4).</p></abstract>

scholarly journals A note on Jeśmanowicz’ conjecture for non-primitive Pythagorean triples

2021 ◽  
Vol 5 (1) ◽  
pp. 115-127
Author(s):  
Van Thien Nguyen ◽  
◽  
Viet Kh. Nguyen ◽  
Pham Hung Quy ◽  
◽  
...  
Keyword(s):  
Positive Integer ◽  
Simple Proof ◽  
Diophantine Equation ◽  
Necessary Conditions ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Pythagorean Triples ◽  
Pythagorean Triple

Let \((a, b, c)\) be a primitive Pythagorean triple parameterized as \(a=u^2-v^2, b=2uv, c=u^2+v^2\), where \(u>v>0\) are co-prime and not of the same parity. In 1956, L. Jesmanowicz conjectured that for any positive integer \(n\), the Diophantine equation \((an)^x+(bn)^y=(cn)^z\) has only the positive integer solution \((x,y,z)=(2,2,2)\). In this connection we call a positive integer solution \((x,y,z)\ne (2,2,2)\) with \(n>1\) exceptional. In 1999 M.-H. Le gave necessary conditions for the existence of exceptional solutions which were refined recently by H. Yang and R.-Q. Fu. In this paper we give a unified simple proof of the theorem of Le-Yang-Fu. Next we give necessary conditions for the existence of exceptional solutions in the case \(v=2,\ u\) is an odd prime. As an application we show the truth of the Jesmanowicz conjecture for all prime values \(u < 100\).

scholarly journals On the Growth of Some Functions Related to z(n)

Mathematics ◽  
2020 ◽  
Vol 8 (6) ◽  
pp. 876 ◽  
Author(s):  
Pavel Trojovský
Keyword(s):  
Positive Integer ◽  
Arithmetic Function ◽  
Fibonacci Sequence ◽  
Upper Bounds ◽  
Integer Solution ◽  
Lower And Upper Bounds ◽  
Positive Integer Solution

The order of appearance z : Z > 0 → Z > 0 is an arithmetic function related to the Fibonacci sequence ( F n ) n . This function is defined as the smallest positive integer solution of the congruence F k ≡ 0 ( mod n ) . In this paper, we shall provide lower and upper bounds for the functions ∑ n ≤ x z ( n ) / n , ∑ p ≤ x z ( p ) and ∑ p r ≤ x z ( p r ) .

Expressions for rational approximations to square roots of integers using Pell's equation

2019 ◽  
Vol 103 (556) ◽  
pp. 101-110
Author(s):  
Ken Surendran ◽  
Desarazu Krishna Babu
Keyword(s):  
Positive Integer ◽  
Continued Fractions ◽  
Integer Solution ◽  
Rational Approximations ◽  
Positive Integer Solution ◽  
Pell’S Equation ◽  
Square Roots ◽  
Integer Solutions

There are recursive expressions (see [1]) for sequentially generating the integer solutions to Pell's equation:p2 −Dq2 = 1, whereDis any positive non-square integer. With known positive integer solutionp1 andq1 we can compute, using these recursive expressions,pnandqnfor alln> 1. See Table in [2] for a list of smallest integer, orfundamental, solutionsp1 andq1 forD≤ 128. These (pn,qn) pairs also formrational approximationstothat, as noted in [3, Chapter 3], match with convergents (Cn=pn/qn) of the Regular Continued Fractions (RCF, continued fractions with the numerator of all fractions equal to 1) for.

scholarly journals The Exponential Diophantine Equation2x+by=cz

2014 ◽  
Vol 2014 ◽  
pp. 1-3 ◽  
Author(s):  
Yahui Yu ◽  
Xiaoxue Li
Keyword(s):  
Positive Integer ◽  
Integer Solution ◽  
Elementary Approach ◽  
Positive Integer Solution ◽  
Integer Solutions ◽  
Positive Integers

Letbandcbe fixed coprime odd positive integers withmin{b,c}>1. In this paper, a classification of all positive integer solutions(x,y,z)of the equation2x+by=czis given. Further, by an elementary approach, we prove that ifc=b+2, then the equation has only the positive integer solution(x,y,z)=(1,1,1), except for(b,x,y,z)=(89,13,1,2)and(2r-1,r+2,2,2), whereris a positive integer withr≥2.

ON THE EXPONENTIAL DIOPHANTINE EQUATION

2014 ◽  
Vol 90 (1) ◽  
pp. 9-19 ◽  
Author(s):  
TAKAFUMI MIYAZAKI ◽  
NOBUHIRO TERAI
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Baker’S Method ◽  
Elementary Methods ◽  
Positive Integers

AbstractLet $m$, $a$, $c$ be positive integers with $a\equiv 3, 5~({\rm mod} \hspace{0.334em} 8)$. We show that when $1+ c= {a}^{2} $, the exponential Diophantine equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(c{m}^{2} - 1)}\nolimits ^{y} = \mathop{(am)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$ under the condition $m\equiv \pm 1~({\rm mod} \hspace{0.334em} a)$, except for the case $(m, a, c)= (1, 3, 8)$, where there are only two solutions: $(x, y, z)= (1, 1, 2), ~(5, 2, 4). $ In particular, when $a= 3$, the equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(8{m}^{2} - 1)}\nolimits ^{y} = \mathop{(3m)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$, except if $m= 1$. The proof is based on elementary methods and Baker’s method.

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