On Equal Products of Consecutive Integers

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.

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.


Mathematics ◽  
2021 ◽  
Vol 9 (15) ◽  
pp. 1813
Author(s):  
S. Subburam ◽  
Lewis Nkenyereye ◽  
N. Anbazhagan ◽  
S. Amutha ◽  
M. Kameswari ◽  
...  

Consider the Diophantine equation yn=x+x(x+1)+⋯+x(x+1)⋯(x+k), where x, y, n, and k are integers. In 2016, a research article, entitled – ’power values of sums of products of consecutive integers’, primarily proved the inequality n= 19,736 to obtain all solutions (x,y,n) of the equation for the fixed positive integers k≤10. In this paper, we improve the bound as n≤ 10,000 for the same case k≤10, and for any fixed general positive integer k, we give an upper bound depending only on k for n.


2012 ◽  
Vol 08 (03) ◽  
pp. 813-821 ◽  
Author(s):  
ZHONGFENG ZHANG ◽  
PINGZHI YUAN

Let a, b, c be integers. In this paper, we prove the integer solutions of the equation axy + byz + czx = 0 satisfy max {|x|, |y|, |z|} ≤ 2 max {a, b, c} when a, b, c are odd positive integers, and when a = b = 1, c = -1, the positive integer solutions of the equation satisfy max {x, y, z} < exp ( exp ( exp (5))).


2013 ◽  
Vol 94 (1) ◽  
pp. 50-105 ◽  
Author(s):  
CHRISTIAN ELSHOLTZ ◽  
TERENCE TAO

AbstractFor any positive integer $n$, let $f(n)$ denote the number of solutions to the Diophantine equation $$\begin{eqnarray*}\frac{4}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\end{eqnarray*}$$ with $x, y, z$ positive integers. The Erdős–Straus conjecture asserts that $f(n)\gt 0$ for every $n\geq 2$. In this paper we obtain a number of upper and lower bounds for $f(n)$ or $f(p)$ for typical values of natural numbers $n$ and primes $p$. For instance, we establish that $$\begin{eqnarray*}N\hspace{0.167em} {\mathop{\log }\nolimits }^{2} N\ll \displaystyle \sum _{p\leq N}f(p)\ll N\hspace{0.167em} {\mathop{\log }\nolimits }^{2} N\log \log N.\end{eqnarray*}$$ These upper and lower bounds show that a typical prime has a small number of solutions to the Erdős–Straus Diophantine equation; small, when compared with other additive problems, like Waring’s problem.


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).


2006 ◽  
Vol 02 (02) ◽  
pp. 195-206 ◽  
Author(s):  
MICHAEL A. BENNETT ◽  
ALAIN TOGBÉ ◽  
P. G. WALSH

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).


2018 ◽  
Vol 8 (1) ◽  
pp. 109-114
Author(s):  
Apoloniusz Tyszka

Abstract We define a computable function f from positive integers to positive integers. We formulate a hypothesis which states that if a system S of equations of the forms xi· xj = xk and xi + 1 = xi has only finitely many solutions in non-negative integers x1, . . . , xi, then the solutions of S are bounded from above by f (2n). We prove the following: (1) the hypothesis implies that there exists an algorithm which takes as input a Diophantine equation, returns an integer, and this integer is greater than the heights of integer (non-negative integer, positive integer, rational) solutions, if the solution set is finite; (2) the hypothesis implies that the question of whether or not a given Diophantine equation has only finitely many rational solutions is decidable by a single query to an oracle that decides whether or not a given Diophantine equation has a rational solution; (3) the hypothesis implies that the question of whether or not a given Diophantine equation has only finitely many integer solutions is decidable by a single query to an oracle that decides whether or not a given Diophantine equation has an integer solution; (4) the hypothesis implies that if a set M ⊆ N has a finite-fold Diophantine representation, thenMis computable.


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


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