On some ternary pure exponential diophantine equations with three consecutive positive integers bases

2019 ◽  
Vol 129 (2) ◽  
Author(s):  
Ruiqin Fu ◽  
Bo He ◽  
Hai Yang ◽  
Huilin Zhu
Keyword(s):  
Diophantine Equations ◽  
Exponential Diophantine Equations ◽  
Positive Integers

TERAI'S CONJECTURE ON EXPONENTIAL DIOPHANTINE EQUATIONS

2011 ◽  
Vol 07 (04) ◽  
pp. 981-999 ◽  
Author(s):  
TAKAFUMI MIYAZAKI
Keyword(s):  
Diophantine Equations ◽  
Exponential Diophantine Equations ◽  
Terai’S Conjecture ◽  
Positive Integers ◽  
Integral Solutions

Let a, b, c be relatively prime positive integers such that ap + bq = cr with fixed integers p, q, r ≥ 2. Terai conjectured that the equation ax + by = cz has no positive integral solutions other than (x, y, z) = (p, q, r) except for specific cases. Most known results on this conjecture concern the case where p = q = 2 and either r = 2 or odd r ≥3. In this paper, we consider the case where p = q = 2 and r > 2 is even, and partially verify Terai's conjecture.

ON EXPONENTIAL DIOPHANTINE EQUATIONS CONTAINING THE EULER QUOTIENT

2014 ◽  
Vol 91 (1) ◽  
pp. 11-18
Author(s):  
NOBUHIRO TERAI
Keyword(s):  
Diophantine Equation ◽  
Diophantine Equations ◽  
Euler’S Totient Function ◽  
Exponential Diophantine Equations ◽  
Euler’S Theorem ◽  
Positive Integers

AbstractLet $a$ and $m$ be relatively prime positive integers with $a>1$ and $m>2$. Let ${\it\phi}(m)$ be Euler’s totient function. The quotient $E_{m}(a)=(a^{{\it\phi}(m)}-1)/m$ is called the Euler quotient of $m$ with base $a$. By Euler’s theorem, $E_{m}(a)$ is an integer. In this paper, we consider the Diophantine equation $E_{m}(a)=x^{l}$ in integers $x>1,l>1$. We conjecture that this equation has exactly five solutions $(a,m,x,l)$ except for $(l,m)=(2,3),(2,6)$, and show that if the equation has solutions, then $m=p^{s}$ or $m=2p^{s}$ with $p$ an odd prime and $s\geq 1$.

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).

Register machine proof of the theorem on exponential diophantine representation of enumerable sets

1984 ◽  
Vol 49 (3) ◽  
pp. 818-829 ◽  
Author(s):  
J. P. Jones ◽  
Y. V. Matijasevič
Keyword(s):  
Natural Number ◽  
Polynomial Time ◽  
Simple Proof ◽  
Diophantine Equations ◽  
Polynomial Equation ◽  
Exponential Diophantine Equations ◽  
Recursively Enumerable ◽  
Image Position ◽  
The Right ◽  
Exponential Relation

The purpose of the present paper is to give a new, simple proof of the theorem of M. Davis, H. Putnam and J. Robinson [1961], which states that every recursively enumerable relation A(a1, …, an) is exponential diophantine, i.e. can be represented in the formwhere a1 …, an, x1, …, xm range over natural numbers and R and S are functions built up from these variables and natural number constants by the operations of addition, A + B, multiplication, AB, and exponentiation, AB. We refer to the variables a1,…,an as parameters and the variables x1 …, xm as unknowns.Historically, the Davis, Putnam and Robinson theorem was one of the important steps in the eventual solution of Hilbert's tenth problem by the second author [1970], who proved that the exponential relation, a = bc, is diophantine, and hence that the right side of (1) can be replaced by a polynomial equation. But this part will not be reproved here. Readers wishing to read about the proof of that are directed to the papers of Y. Matijasevič [1971a], M. Davis [1973], Y. Matijasevič and J. Robinson [1975] or C. Smoryński [1972]. We concern ourselves here for the most part only with exponential diophantine equations until §5 where we mention a few consequences for the class NP of sets computable in nondeterministic polynomial time.

scholarly journals A Multi-Frey Approach to Some Multi-Parameter Families of Diophantine Equations

2008 ◽  
Vol 60 (3) ◽  
pp. 491-519 ◽  
Author(s):  
Yann Bugeaud ◽  
Maurice Mignotte ◽  
Samir Siksek
Keyword(s):  
Lower Bounds ◽  
Diophantine Equations ◽  
Galois Representations ◽  
Arbitrary Degree ◽  
Famous Theorem ◽  
Linear Forms ◽  
Positive Integers ◽  
Level Lowering

AbstractWe solve several multi-parameter families of binomial Thue equations of arbitrary degree; for example, we solve the equation5uxn − 2r3s yn = ±1,in non-zero integers x, y and positive integers u, r, s and n ≥ 3. Our approach uses several Frey curves simultaneously, Galois representations and level-lowering, new lower bounds for linear forms in 3 logarithms due to Mignotte and a famous theorem of Bennett on binomial Thue equations.

scholarly journals Some Diophantine Problems Related to k-Fibonacci Numbers

Mathematics ◽  
2020 ◽  
Vol 8 (7) ◽  
pp. 1047
Author(s):  
Pavel Trojovský ◽  
Štěpán Hubálovský
Keyword(s):  
Recurrence Relation ◽  
Diophantine Equations ◽  
Initial Conditions ◽  
Fibonacci Numbers ◽  
Fibonacci Sequence ◽  
Recursive Relation ◽  
Initial Values ◽  
Lucas Sequence ◽  
Diophantine Problems ◽  
Positive Integers

Let k ≥ 1 be an integer and denote ( F k , n ) n as the k-Fibonacci sequence whose terms satisfy the recurrence relation F k , n = k F k , n − 1 + F k , n − 2 , with initial conditions F k , 0 = 0 and F k , 1 = 1 . In the same way, the k-Lucas sequence ( L k , n ) n is defined by satisfying the same recursive relation with initial values L k , 0 = 2 and L k , 1 = k . The sequences ( F k , n ) n ≥ 0 and ( L k , n ) n ≥ 0 were introduced by Falcon and Plaza, who derived many of their properties. In particular, they proved that F k , n 2 + F k , n + 1 2 = F k , 2 n + 1 and F k , n + 1 2 − F k , n − 1 2 = k F k , 2 n , for all k ≥ 1 and n ≥ 0 . In this paper, we shall prove that if k > 1 and F k , n s + F k , n + 1 s ∈ ( F k , m ) m ≥ 1 for infinitely many positive integers n, then s = 2 . Similarly, that if F k , n + 1 s − F k , n − 1 s ∈ ( k F k , m ) m ≥ 1 holds for infinitely many positive integers n, then s = 1 or s = 2 . This generalizes a Marques and Togbé result related to the case k = 1 . Furthermore, we shall solve the Diophantine equations F k , n = L k , m , F k , n = F n , k and L k , n = L n , k .

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