scholarly journals Positive solutions of the diophantine equation

1982 ◽  
Vol 5 (2) ◽  
pp. 311-314 ◽  
Author(s):  
W. R. Utz

Integral solutions ofx3+λy+1−xyz=0are observed for all integralλ. Forλ=2the 13 solutions of the equation in positive integers are determined. Solutions of the equation in positive integers were previously determined for the caseλ=1.

2020 ◽  
Vol 16 (09) ◽  
pp. 2095-2111
Author(s):  
Szabolcs Tengely ◽  
Maciej Ulas

We consider equations of the form [Formula: see text], where [Formula: see text] is a polynomial with integral coefficients and [Formula: see text] is the [Formula: see text]th Fibonacci number that is, [Formula: see text] and [Formula: see text] for [Formula: see text] In particular, for each [Formula: see text], we prove the existence of a polynomial [Formula: see text] of degree [Formula: see text] such that the Diophantine equation [Formula: see text] has infinitely many solutions in positive integers [Formula: see text]. Moreover, we present results of our numerical search concerning the existence of even degree polynomials representing many Fibonacci numbers. We also determine all integral solutions [Formula: see text] of the Diophantine equations [Formula: see text] for [Formula: see text] and [Formula: see text].


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.


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].


2013 ◽  
Vol 89 (2) ◽  
pp. 316-321 ◽  
Author(s):  
MOU JIE DENG

AbstractLet $(a, b, c)$ be a primitive Pythagorean triple satisfying ${a}^{2} + {b}^{2} = {c}^{2} . $ In 1956, Jeśmanowicz conjectured that for any given positive integer $n$ the only solution of $\mathop{(an)}\nolimits ^{x} + \mathop{(bn)}\nolimits ^{y} = \mathop{(cn)}\nolimits ^{z} $ in positive integers is $x= y= z= 2. $ In this paper, for the primitive Pythagorean triple $(a, b, c)= (4{k}^{2} - 1, 4k, 4{k}^{2} + 1)$ with $k= {2}^{s} $ for some positive integer $s\geq 0$, we prove the conjecture when $n\gt 1$ and certain divisibility conditions are satisfied.


2020 ◽  
Vol 4 (2) ◽  
pp. 103
Author(s):  
Leomarich F Casinillo ◽  
Emily L Casinillo

A Pythagorean triple is a set of three positive integers a, b and c that satisfy the Diophantine equation a^2+b^2=c^2. The triple is said to be primitive if gcd(a, b, c)=1 and each pair of integers and  are relatively prime, otherwise known as non-primitive. In this paper, the generalized version of the formula that generates primitive and non-primitive Pythagorean triples that depends on two positive integers  k and n, that is, P_T=(a(k, n), b(k, n), c(k, n)) were constructed. Further, we determined the values of  k and n that generates primitive Pythagorean triples and give some important results.


2012 ◽  
Vol 55 (2) ◽  
pp. 435-440
Author(s):  
Konstantine Zelator

AbstractWe consider the diophantine equation x2 + y6 = ze, e ≥ 4. We show that, when e is a multiple of 4 or 6, this equation has no solutions in positive integers with x and y relatively prime. As a corollary, we show that there exists no primitive Pythagorean triangle one of whose leglengths is a perfect cube, while the hypotenuse length is an integer square.


1958 ◽  
Vol 23 (2) ◽  
pp. 183-187 ◽  
Author(s):  
Martin Davis ◽  
Hilary Putnam

Hilbert's tenth problem is to find an algorithm for determining whether or not a diophantine equation possesses solutions. A diophantine predicate (of positive integers) is defined to be one of the formwhere P is a polynomial with integral coefficients (positive, negative, or zero). Previous work has considered the variables as ranging over nonnegative integers; but we shall find it more useful here to restrict the range to positive integers, no essential change being thereby introduced. It is clear that the recursive unsolvability of Hilbert's tenth problem would follow if one could show that some non-recursive predicate were diophantine. In particular, it would suffice to show that every recursively enumerable predicate is diophantine. Actually, it would suffice to prove far less.


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