scholarly journals THE EXPONENTIAL DIOPHANTINE EQUATION nx + (n + 1)y = (n + 2)z REVISITED

2009 ◽  
Vol 51 (3) ◽  
pp. 659-667 ◽  
Author(s):  
BO HE ◽  
ALAIN TOGBÉ

AbstractLet n be a positive integer. In this paper, we consider the diophantine equation We prove that this equation has only the positive integer solutions (n, x, y, z) = (1, t, 1, 1), (1, t, 3, 2), (3, 2, 2, 2). Therefore we extend the work done by Leszczyński (Wiadom. Mat., vol. 3, 1959, pp. 37–39) and Makowski (Wiadom. Mat., vol. 9, 1967, pp. 221–224).

2010 ◽  
Vol 81 (2) ◽  
pp. 177-185 ◽  
Author(s):  
BO HE ◽  
ALAIN TOGBÉ

AbstractLet a, b, c, x and y be positive integers. In this paper we sharpen a result of Le by showing that the Diophantine equation has at most two positive integer solutions (m,n) satisfying min (m,n)>1.


1970 ◽  
Vol 13 (2) ◽  
pp. 255-259 ◽  
Author(s):  
R. A. Macleod ◽  
I. Barrodale

Using the theory of algebraic numbers, Mordell [1] has shown that the Diophantine equation1possesses only two solutions in positive integers; these are given by n = 2, m = 1, and n = 14, m = 5. We are interested in positive integer solutions to the generalized equation2and in this paper we prove for several choices of k and l that (2) has no solutions, in other cases the only solutions are given, and numerical evidence for all values of k and l for which max (k, l) ≤ 15 is also exhibited.


2015 ◽  
Vol 11 (04) ◽  
pp. 1107-1114 ◽  
Author(s):  
Hai Yang ◽  
Ruiqin Fu

Let D1, D2, D, k, λ be fixed integers such that D1 ≥ 1, D2 ≥ 1, gcd (D1, D2) = 1, D = D1D2 is not a square, ∣k∣ > 1, gcd (D, k) = 1 and λ = 1 or 4 according as 2 ∤ k or not. In this paper, we prove that every solution class S(l) of the equation D1x2-D2y2 = λkz, gcd (x, y) = 1, z > 0, has a unique positive integer solution [Formula: see text] satisfying [Formula: see text] and [Formula: see text], where z runs over all integer solutions (x,y,z) of S(l),(u1,v1) is the fundamental solution of Pell's equation u2 - Dv2 = 1. This result corrects and improves some previous results given by M. H. Le.


2001 ◽  
Vol 64 (1) ◽  
pp. 99-105 ◽  
Author(s):  
Maohua Le

Let p be an odd prime with p > 3. In this paper we give all positive integer solutions (x, y, m, n) of the equation x2 + p2m = yn, gcd (x, y) = 1, n > 2 satisfying 2 | n of 2 ∤ n and p ≢ (−1)(p−1)/2(mod 4n.


2010 ◽  
Vol 107 (2) ◽  
pp. 161
Author(s):  
Bo He ◽  
Alain Togbé ◽  
Shichun Yang

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 $(x,y)$.


2021 ◽  
Vol 27 (3) ◽  
pp. 123-129
Author(s):  
Yasutsugu Fujita ◽  
◽  
Maohua Le ◽  

For any positive integer t, let ord_2 t denote the order of 2 in the factorization of t. Let a,\,b be two distinct fixed positive integers with \min\{a,b\}>1. In this paper, using some elementary number theory methods, the existence of positive integer solutions (x,n) of the polynomial-exponential Diophantine equation (*) (a^n-1)(b^n-1)=x^2 with n>2 is discussed. We prove that if \{a,b\}\ne \{13,239\} and ord_2(a^2-1)\ne ord_2(b^2-1), then (*) has no solutions (x,n) with 2\mid n. Thus it can be seen that if \{a,b\}\equiv \{3,7\},\{3,15\},\{7,11\},\{7,15\} or \{11,15\} \pmod{16}, where \{a,b\} \equiv \{a_0,b_0\} \pmod{16} means either a \equiv a_0 \pmod{16} and b \equiv b_0\pmod{16} or a\equiv b_0 \pmod{16} and b\equiv a_0 \pmod{16}, then (*) has no solutions (x,n).


1956 ◽  
Vol 3 (1) ◽  
pp. 55-56
Author(s):  
John Hunter

In the equationdis any positive integer which is not a perfect square. For convenience we shall consider only those solutions of (1) for which x and yare both positive. All the others can be obtained from these. In fact, it is well known that if (x0, y0) is the minimum positive integer solution of (1), then all integer solutions (x, y) are given byand, in particular, all positive integer solutions are given by


2018 ◽  
Vol 14 (05) ◽  
pp. 1223-1228
Author(s):  
Hai Yang ◽  
Ruiqin Fu

Let [Formula: see text] be a positive integer which is not a square. Further, let [Formula: see text] be the least positive integer solution of the Pell equation [Formula: see text], and let [Formula: see text] denote the class number of binary quadratic primitive forms of discriminant [Formula: see text]. If [Formula: see text] satisfies [Formula: see text] and [Formula: see text], then [Formula: see text] is called an exceptional number. In this paper, under the assumption that there have no exceptional numbers, we prove that the equation [Formula: see text] has no positive integer solutions [Formula: see text] satisfy [Formula: see text] and [Formula: see text].


2014 ◽  
Vol 64 (5) ◽  
Author(s):  
Jianping Wang ◽  
Tingting Wang ◽  
Wenpeng Zhang

AbstractLet n be a positive integer. In this paper, using the results on the existence of primitive divisors of Lucas numbers and some properties of quadratic and exponential diophantine equations, we prove that if n ≡ 3 (mod 6), then the equation x 2 + (3n 2 + 1)y = (4n 2 + 1)z has only the positive integer solutions (x, y, z) = (n, 1, 1) and (8n 3 + 3n, 1, 3).


2020 ◽  
Vol 55 (2) ◽  
pp. 195-201
Author(s):  
Maohua Le ◽  
◽  
Gökhan Soydan ◽  

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).


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