ON A PROBLEM OF RICHARD GUY

Author(s):  
NGUYEN XUAN THO

Abstract In the 1993 Western Number Theory Conference, Richard Guy proposed Problem 93:31, which asks for integers n representable by ${(x+y+z)^3}/{xyz}$ , where $x,\,y,\,z$ are integers, preferably with positive integer solutions. We show that the representation $n={(x+y+z)^3}/{xyz}$ is impossible in positive integers $x,\,y,\,z$ if $n=4^{k}(a^2+b^2)$ , where $k,\,a,\,b\in \mathbb {Z}^{+}$ are such that $k\geq 3$ and $2\nmid a+b$ .

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).


2018 ◽  
Vol 11 (04) ◽  
pp. 1850056 ◽  
Author(s):  
Zahid Raza ◽  
Hafsa Masood Malik

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].


2005 ◽  
Vol 01 (04) ◽  
pp. 563-581 ◽  
Author(s):  
A. KNOPFMACHER ◽  
M. E. MAYS

The general field of additive number theory considers questions concerning representations of a given positive integer n as a sum of other integers. In particular, partitions treat the sums as unordered combinatorial objects, and compositions treat the sums as ordered. Sometimes the sums are restricted, so that, for example, the summands are distinct, or relatively prime, or all congruent to ±1 modulo 5. In this paper we review work on analogous problems concerning representations of n as a product of positive integers. We survey techniques for enumerating product representations both in the unrestricted case and in the case when the factors are required to be distinct, and both when the product representations are considered as ordered objects and when they are unordered. We offer some new identities and observations for these and related counting functions and derive some new recursive algorithms to generate lists of factorizations with restrictions of various types.


1985 ◽  
Vol 27 ◽  
pp. 5-10
Author(s):  
S. A. Burr ◽  
P. Erdös

Let A be a sequence of positive integers. Define P(A) to be the set of all integers representable as a sum of distinct terms of A. Note that if A contains a repeated value, we are free to use it as many times as it occurs in A. We call A complete if every sufficiently large positive integer is in P(A), and entirely complete if every positive integer is in P(A). Completeness properties have received considerable, if somewhat sporadic, attention over the years. See Chapter 6 of [3] for a survey.


1969 ◽  
Vol 10 (2) ◽  
pp. 126-129 ◽  
Author(s):  
Ian Anderson

Let s = s(a1, a2,...., ar) denote the number of integer solutions of the equationsubject to the conditionsthe ai being given positive integers, and square brackets denoting the integral part. Clearly s (a1,..., ar) is also the number s = s(m) of divisors of which contain exactly λ prime factors counted according to multiplicity, and is therefore, as is proved in [1], the cardinality of the largest possible set of divisors of m, no one of which divides another.


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.


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))).


2014 ◽  
Vol 2014 ◽  
pp. 1-3 ◽  
Author(s):  
Yahui Yu ◽  
Xiaoxue Li

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.


2021 ◽  
Vol 2021 ◽  
pp. 1-5
Author(s):  
Ze Gu

A proportionally modular numerical semigroup is the set S a , b , c of nonnegative integer solutions to a Diophantine inequality of the form a x   mod   b ≤ c x , where a , b , and c are positive integers. A formula for the multiplicity of S a , b , c , that is, m S a , b , c = k b / a for some positive integer k , is given by A. Moscariello. In this paper, we give a new proof of the formula and determine a better bound for k . Furthermore, we obtain k = 1 for various cases and a formula for the number of the triples a , b , c such that k ≠ 1 when the number a − c is fixed.


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).


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