On the solvability of the simultaneous Pell equations x2 − ay2 = 1 and y2 − bz2 = v12

2021 ◽  
pp. 1-12
Author(s):  
Ruiqin Fu ◽  
Hai Yang
Keyword(s):  
Positive Integer ◽  
Diophantine Equations ◽  
Prime Divisors ◽  
Positive Integer Solution ◽  
Pell Equations ◽  
Integer Solutions ◽  
Least Solution ◽  
Simultaneous Pell Equations ◽  
Necessary And Sufficient ◽  
Positive Integers

Let [Formula: see text] be fixed positive integers such that [Formula: see text] is not a perfect square and [Formula: see text] is squarefree, and let [Formula: see text] denote the number of distinct prime divisors of [Formula: see text]. Let [Formula: see text] denote the least solution of Pell equation [Formula: see text]. Further, for any positive integer [Formula: see text], let [Formula: see text] and [Formula: see text], where [Formula: see text] and [Formula: see text]. In this paper, using the basic properties of Pell equations and some known results on binary quartic Diophantine equations, a necessary and sufficient condition for the system of equations [Formula: see text] and [Formula: see text] to have positive integer solutions [Formula: see text] is obtained. By this result, we prove that if [Formula: see text] has a positive integer solution [Formula: see text] for [Formula: see text] or [Formula: see text] according to [Formula: see text] or not, then [Formula: see text] and [Formula: see text], where [Formula: see text] is a positive integer, [Formula: see text] or [Formula: see text] and [Formula: see text] or [Formula: see text] according to [Formula: see text] or not, [Formula: see text] is the integer part of [Formula: see text], except for [Formula: see text]

Complete solutions of the simultaneous Pell’s equations x2 − (a2 − 1)y2 = 1 and y2 − pz2 = 1

2019 ◽  
Vol 15 (05) ◽  
pp. 1069-1074 ◽  
Author(s):  
Hai Yang ◽  
Ruiqin Fu
Keyword(s):  
Positive Integer ◽  
Diophantine Equations ◽  
Integer Solution ◽  
System Of Equations ◽  
Positive Integer Solution ◽  
Elementary Methods ◽  
Integer Solutions ◽  
Positive Integers ◽  
Complete Solutions

Let [Formula: see text] be a positive integer with [Formula: see text], and let [Formula: see text] be an odd prime. In this paper, by using certain properties of Pell’s equations and quartic diophantine equations with some elementary methods, we prove that the system of equations [Formula: see text] [Formula: see text] and [Formula: see text] has positive integer solutions [Formula: see text] if and only if [Formula: see text] and [Formula: see text] satisfy [Formula: see text] and [Formula: see text], where [Formula: see text], [Formula: see text] and [Formula: see text] are positive integers. Further, if the above condition is satisfied, then [Formula: see text] has only the positive integer solution [Formula: see text]. By the above result, we can obtain the following corollaries immediately. (i) If [Formula: see text] or [Formula: see text], then [Formula: see text] has no positive integer solutions [Formula: see text]. (ii) For [Formula: see text], [Formula: see text] has only the positive integer solutions [Formula: see text], [Formula: see text], [Formula: see text], [Formula: see text], [Formula: see text] and [Formula: see text].

scholarly journals A note on the exponential Diophantine equation (A^2n)^x+(B^2n)^y=((A^2+B^2)n)^z

2020 ◽  
Vol 55 (2) ◽  
pp. 195-201
Author(s):  
Maohua Le ◽  
◽  
Gökhan Soydan ◽  
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Upper Bound ◽  
Diophantine Equations ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Exponential Diophantine Equations ◽  
Integer Solutions ◽  
Positive Integers ◽  
Upper Bound For Solutions

Let A, B be positive integers such that min{A,B}>1, gcd(A,B) = 1 and 2|B. In this paper, using an upper bound for solutions of ternary purely exponential Diophantine equations due to R. Scott and R. Styer, we prove that, for any positive integer n, if A >B3/8, then the equation (A2 n)x + (B2 n)y = ((A2 + B2)n)z has no positive integer solutions (x,y,z) with x > z > y; if B>A3/6, then it has no solutions (x,y,z) with y>z>x. Thus, combining the above conclusion with some existing results, we can deduce that, for any positive integer n, if B ≡ 2 (mod 4) and A >B3/8, then this equation has only the positive integer solution (x,y,z)=(1,1,1).

scholarly journals Solution of certain Pell equations

2018 ◽  
Vol 11 (04) ◽  
pp. 1850056 ◽  
Author(s):  
Zahid Raza ◽  
Hafsa Masood Malik
Keyword(s):  
Positive Integer ◽  
Fundamental Solution ◽  
Positive Solutions ◽  
Continued Fraction ◽  
Pell Equation ◽  
Lucas Sequences ◽  
Pell Equations ◽  
Integer Solutions ◽  
Positive Integers

Let [Formula: see text] be any positive integers such that [Formula: see text] and [Formula: see text] is a square free positive integer of the form [Formula: see text] where [Formula: see text] and [Formula: see text] The main focus of this paper is to find the fundamental solution of the equation [Formula: see text] with the help of the continued fraction of [Formula: see text] We also obtain all the positive solutions of the equations [Formula: see text] and [Formula: see text] by means of the Fibonacci and Lucas sequences.Furthermore, in this work, we derive some algebraic relations on the Pell form [Formula: see text] including cycle, proper cycle, reduction and proper automorphism of it. We also determine the integer solutions of the Pell equation [Formula: see text] in terms of [Formula: see text] We extend all the results of the papers [3, 10, 27, 37].

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.

A GENERALIZATION OF A THEOREM OF BUMBY ON QUARTIC DIOPHANTINE EQUATIONS

2006 ◽  
Vol 02 (02) ◽  
pp. 195-206 ◽  
Author(s):  
MICHAEL A. BENNETT ◽  
ALAIN TOGBÉ ◽  
P. G. WALSH
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Diophantine Equations ◽  
Integer Solutions ◽  
Positive Integers

Bumby proved that the only positive integer solutions to the quartic Diophantine equation 3X4 - 2Y2 = 1 are (X, Y) = (1, 1),(3, 11). In this paper, we use Thue's hypergeometric method to prove that, for each integer m ≥ 1, the only positive integers solutions to the Diophantine equation (m2 + m + 1)X4 - (m2 + m)Y2 = 1 are (X,Y) = (1, 1),(2m + 1, 4m2 + 4m + 3).

scholarly journals A note on the Diophantine equation $|a^x-b^y|=c$

2010 ◽  
Vol 107 (2) ◽  
pp. 161
Author(s):  
Bo He ◽  
Alain Togbé ◽  
Shichun Yang
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Integer Solutions ◽  
Positive Integers

Let $a,b,$ and $c$ be positive integers. We show that if $(a,b) =(N^k-1,N)$, where $N,k\geq 2$, then there is at most one positive integer solution $(x,y)$ to the exponential Diophantine equation $|a^x-b^y|=c$, unless $(N,k)=(2,2)$. Combining this with results of Bennett [3] and the first author [6], we stated all cases for which the equation $|(N^k \pm 1)^x - N^y|=c$ has more than one positive integer solutions $(x,y)$.

scholarly journals Fermat's Last Theorem: A Proof by Contradiction

2021 ◽  
Author(s):  
Benson Schaeffer
Keyword(s):  
Positive Integer ◽  
Diophantine Equations ◽  
Algebraic Proof ◽  
Fermat's Last Theorem ◽  
Binomial Expansion ◽  
Proof By Contradiction ◽  
Fermat’S Last Theorem ◽  
Integer Solutions ◽  
Positive Integers ◽  
Fermat's Equation

In this paper I offer an algebraic proof by contradiction of Fermat’s Last Theorem. Using an alternative to the standard binomial expansion, (a+b) n = a n + b Pn i=1 a n−i (a + b) i−1 , a and b nonzero integers, n a positive integer, I show that a simple rewrite of the Fermat’s equation stating the theorem, A p + B p = (A + B − D) p , A, B, D and p positive integers, D < A < B, p ≥ 3 and prime, entails the contradiction, A(B − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 A i−1−j (A + B − D) j−1 # + B(A − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 B i−1−j (A + B − D) j−1 # = 0, the sum of two positive integers equal to zero. This contradiction shows that the rewrite has no non-trivial positive integer solutions and proves Fermat’s Last Theorem. AMS 2020 subject classification: 11A99, 11D41 Diophantine equations, Fermat’s equation ∗The corresponding author. E-mail: [email protected] 1 1 Introduction To prove Fermat’s Last Theorem, it suffices to show that the equation A p + B p = C p (1In this paper I offer an algebraic proof by contradiction of Fermat’s Last Theorem. Using an alternative to the standard binomial expansion, (a+b) n = a n + b Pn i=1 a n−i (a + b) i−1 , a and b nonzero integers, n a positive integer, I show that a simple rewrite of the Fermat’s equation stating the theorem, A p + B p = (A + B − D) p , A, B, D and p positive integers, D < A < B, p ≥ 3 and prime, entails the contradiction, A(B − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 A i−1−j (A + B − D) j−1 # + B(A − D) X p−1 i=2 (−D) p−1−i "X i−1 j=1 B i−1−j (A + B − D) j−1 # = 0, the sum of two positive integers equal to zero. This contradiction shows that the rewrite has no non-trivial positive integer solutions and proves Fermat’s Last Theorem.

A NOTE ON JEŚMANOWICZ’ CONJECTURE CONCERNING NONPRIMITIVE PYTHAGOREAN TRIPLES

2020 ◽  
pp. 1-11
Author(s):  
YASUTSUGU FUJITA ◽  
MAOHUA LE
Keyword(s):  
Positive Integer ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Baker’S Method ◽  
Pythagorean Triples ◽  
Integer Solutions ◽  
Positive Integers

Abstract Jeśmanowicz conjectured that $(x,y,z)=(2,2,2)$ is the only positive integer solution of the equation $(*)\; ((\kern1.5pt f^2-g^2)n)^x+(2fgn)^y=((\kern1.5pt f^2+g^2)n)^x$ , where n is a positive integer and f, g are positive integers such that $f>g$ , $\gcd (\kern1.5pt f,g)=1$ and $f \not \equiv g\pmod 2$ . Using Baker’s method, we prove that: (i) if $n>1$ , $f \ge 98$ and $g=1$ , then $(*)$ has no positive integer solutions $(x,y,z)$ with $x>z>y$ ; and (ii) if $n>1$ , $f=2^rs^2$ and $g=1$ , where r, s are positive integers satisfying $(**)\; 2 \nmid s$ and $s<2^{r/2}$ , then $(*)$ has no positive integer solutions $(x,y,z)$ with $y>z>x$ . Thus, Jeśmanowicz’ conjecture is true if $f=2^rs^2$ and $g=1$ , where r, s are positive integers satisfying $(**)$ .

scholarly journals Characterizations and Identities for Isosceles Triangular Numbers

2021 ◽  
Vol 14 (2) ◽  
pp. 380-395
Author(s):  
Jiramate Punpim ◽  
Somphong Jitman
Keyword(s):  
Positive Integer ◽  
Sufficient Conditions ◽  
Necessary And Sufficient Conditions ◽  
Triangular Numbers ◽  
Triangular Number ◽  
Recursive Formulas ◽  
Necessary And Sufficient ◽  
Positive Integers ◽  
Special Case

Triangular numbers have been of interest and continuously studied due to their beautiful representations, nice properties, and various links with other figurate numbers. For positive integers n and l, the nth l-isosceles triangular number is a generalization of triangular numbers defined to be the arithmetic sum of the formT(n, l) = 1 + (1 + l) + (1 + 2l) + · · · + (1 + (n − 1)l).In this paper, we focus on characterizations and identities for isosceles triangular numbers as well as their links with other figurate numbers. Recursive formulas for constructions of isosceles triangular numbers are given together with necessary and sufficient conditions for a positive integer to be a sum of isosceles triangular  numbers. Various identities for isosceles triangular numbers are established. Results on triangular numbers can be viewed as a special case.

scholarly journals On the Diophantine equations z^2 = f(x)^2 ± f(x)f(y) + f(y)^2

2021 ◽  
Vol 27 (2) ◽  
pp. 88-100
Author(s):  
Qiongzhi Tang ◽  
Keyword(s):  
Positive Integer ◽  
Diophantine Equations ◽  
Pell Equation ◽  
Intersection Angle ◽  
Integer Solutions ◽  
Two Sides

Using the theory of Pell equation, we study the non-trivial positive integer solutions of the Diophantine equations $z^2=f(x)^2\pm f(x)f(y)+f(y)^2$ for certain polynomials f(x), which mean to construct integral triangles with two sides given by the values of polynomials f(x) and f(y) with the intersection angle $120^\circ$ or $60^\circ$.

Sign in / Sign up

Export Citation Format

Share Document