The Diophantine equation x2+3 = yn

1993 ◽  
Vol 35 (2) ◽  
pp. 203-206 ◽  
Author(s):  
J. H. E. Cohn
Keyword(s):  
Diophantine Equation ◽  
Special Cases ◽  
Recent Origin ◽  
Positive Integers

Many special cases of the equation x2+C= yn where x and y are positive integers and n≥3 have been considered over the years, but most results for general n are of fairly recent origin. The earliest reference seems to be an assertion by Fermat that he had shown that when C=2, n=3, the only solutions are given by x = 5, y = 3; a proof was published by Euler [1]. The first result for general n is due to Lebesgue [2] who proved that when C = 1 there are no solutions. Nagell [4] generalised Fermat's result and proved that for C = 2 the equation has no solution other than x = 5, y = 3, n = 3. He also showed [5] that for C = 4 the equation has no solution except x = 2, y = 2, n = 3 and x = 11, y = 5, n = 3, and claims in [6] to have dealt with the case C = 5. The case C = -1 was solved by Chao Ko, and an account appears in [3], pp. 302–304.

scholarly journals A note on the ternary Diophantine equation x 2 − y 2 m = z n

2021 ◽  
Vol 29 (2) ◽  
pp. 93-105
Author(s):  
Attila Bérczes ◽  
Maohua Le ◽  
István Pink ◽  
Gökhan Soydan
Keyword(s):  
Diophantine Equation ◽  
Diophantine Equations ◽  
Special Cases ◽  
Positive Integers

Abstract Let ℕ be the set of all positive integers. In this paper, using some known results on various types of Diophantine equations, we solve a couple of special cases of the ternary equation x2 − y2m = zn , x, y, z, m, n ∈ ℕ, gcd(x, y) = 1, m ≥ 2, n ≥ 3.

scholarly journals On the Ternary Exponential Diophantine Equation Equating a Perfect Power and Sum of Products of Consecutive Integers

Mathematics ◽  
2021 ◽  
Vol 9 (15) ◽  
pp. 1813
Author(s):  
S. Subburam ◽  
Lewis Nkenyereye ◽  
N. Anbazhagan ◽  
S. Amutha ◽  
M. Kameswari ◽  
...  
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Upper Bound ◽  
Exponential Diophantine Equation ◽  
Research Article ◽  
All Solutions ◽  
Sum Of Products ◽  
Consecutive Integers ◽  
Sums Of Products ◽  
Positive Integers

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.

scholarly journals On the 2-adic order of Stirling numbers of the second kind and their differences

2009 ◽  
Vol DMTCS Proceedings vol. AK,... (Proceedings) ◽  
Author(s):  
Tamás Lengyel
Keyword(s):  
Positive Integer ◽  
Stirling Numbers ◽  
Bell Polynomials ◽  
Binary Representation ◽  
Exact Order ◽  
Special Cases ◽  
International Audience ◽  
The Difference ◽  
Divisibility Properties ◽  
Positive Integers

International audience Let $n$ and $k$ be positive integers, $d(k)$ and $\nu_2(k)$ denote the number of ones in the binary representation of $k$ and the highest power of two dividing $k$, respectively. De Wannemacker recently proved for the Stirling numbers of the second kind that $\nu_2(S(2^n,k))=d(k)-1, 1\leq k \leq 2^n$. Here we prove that $\nu_2(S(c2^n,k))=d(k)-1, 1\leq k \leq 2^n$, for any positive integer $c$. We improve and extend this statement in some special cases. For the difference, we obtain lower bounds on $\nu_2(S(c2^{n+1}+u,k)-S(c2^n+u,k))$ for any nonnegative integer $u$, make a conjecture on the exact order and, for $u=0$, prove part of it when $k \leq 6$, or $k \geq 5$ and $d(k) \leq 2$. The proofs rely on congruential identities for power series and polynomials related to the Stirling numbers and Bell polynomials, and some divisibility properties.

scholarly journals A NOTE ON THE DIOPHANTINE EQUATION

2013 ◽  
Vol 89 (2) ◽  
pp. 316-321 ◽  
Author(s):  
MOU JIE DENG
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Pythagorean Triple ◽  
Positive Integers

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.

SOME NOTES ON A GENERALIZED VERSION OF PYTHAGOREAN TRIPLES

2020 ◽  
Vol 4 (2) ◽  
pp. 103
Author(s):  
Leomarich F Casinillo ◽  
Emily L Casinillo
Keyword(s):  
Diophantine Equation ◽  
Pythagorean Triples ◽  
Pythagorean Triple ◽  
Positive Integers

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.

A Note on the Diophantine Equation x2 + y6 = ze, e ≥ 4

2012 ◽  
Vol 55 (2) ◽  
pp. 435-440
Author(s):  
Konstantine Zelator
Keyword(s):  
Diophantine Equation ◽  
Positive Integers

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.

Reductions of Hilbert's tenth problem

1958 ◽  
Vol 23 (2) ◽  
pp. 183-187 ◽  
Author(s):  
Martin Davis ◽  
Hilary Putnam
Keyword(s):  
Diophantine Equation ◽  
Hilbert's Tenth Problem ◽  
Recursively Enumerable ◽  
Hilbert’S Tenth Problem ◽  
Essential Change ◽  
Recursive Predicate ◽  
Image Position ◽  
Positive Integers

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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