scholarly journals On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4)

2021 ◽  
Vol 6 (10) ◽  
pp. 10596-10601
Author(s):  
Yahui Yu ◽  
◽  
Jiayuan Hu ◽  

<abstract><p>Let $ k $ be a fixed positive integer with $ k &gt; 1 $. In 2014, N. Terai <sup>[<xref ref-type="bibr" rid="b6">6</xref>]</sup> conjectured that the equation $ x^2+(2k-1)^y = k^z $ has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. This is still an unsolved problem as yet. For any positive integer $ n $, let $ Q(n) $ denote the squarefree part of $ n $. In this paper, using some elementary methods, we prove that if $ k\equiv 3 $ (mod 4) and $ Q(k-1)\ge 2.11 $ log $ k $, then the equation has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. It can thus be seen that Terai's conjecture is true for almost all positive integers $ k $ with $ k\equiv 3 $(mod 4).</p></abstract>

2014 ◽  
Vol 90 (1) ◽  
pp. 9-19 ◽  
Author(s):  
TAKAFUMI MIYAZAKI ◽  
NOBUHIRO TERAI

AbstractLet $m$, $a$, $c$ be positive integers with $a\equiv 3, 5~({\rm mod} \hspace{0.334em} 8)$. We show that when $1+ c= {a}^{2} $, the exponential Diophantine equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(c{m}^{2} - 1)}\nolimits ^{y} = \mathop{(am)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$ under the condition $m\equiv \pm 1~({\rm mod} \hspace{0.334em} a)$, except for the case $(m, a, c)= (1, 3, 8)$, where there are only two solutions: $(x, y, z)= (1, 1, 2), ~(5, 2, 4). $ In particular, when $a= 3$, the equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(8{m}^{2} - 1)}\nolimits ^{y} = \mathop{(3m)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$, except if $m= 1$. The proof is based on elementary methods and Baker’s method.


2019 ◽  
Vol 15 (05) ◽  
pp. 1069-1074 ◽  
Author(s):  
Hai Yang ◽  
Ruiqin Fu

Let [Formula: see text] be a positive integer with [Formula: see text], and let [Formula: see text] be an odd prime. In this paper, by using certain properties of Pell’s equations and quartic diophantine equations with some elementary methods, we prove that the system of equations [Formula: see text] [Formula: see text] and [Formula: see text] has positive integer solutions [Formula: see text] if and only if [Formula: see text] and [Formula: see text] satisfy [Formula: see text] and [Formula: see text], where [Formula: see text], [Formula: see text] and [Formula: see text] are positive integers. Further, if the above condition is satisfied, then [Formula: see text] has only the positive integer solution [Formula: see text]. By the above result, we can obtain the following corollaries immediately. (i) If [Formula: see text] or [Formula: see text], then [Formula: see text] has no positive integer solutions [Formula: see text]. (ii) For [Formula: see text], [Formula: see text] has only the positive integer solutions [Formula: see text], [Formula: see text], [Formula: see text], [Formula: see text], [Formula: see text] and [Formula: see text].


2013 ◽  
Vol 90 (1) ◽  
pp. 20-27 ◽  
Author(s):  
NOBUHIRO TERAI

AbstractLet $q$ be an odd prime such that ${q}^{t} + 1= 2{c}^{s} $, where $c, t$ are positive integers and $s= 1, 2$. We show that the Diophantine equation ${x}^{2} + {q}^{m} = {c}^{n} $ has only the positive integer solution $(x, m, n)= ({c}^{s} - 1, t, 2s)$ under some conditions. The proof is based on elementary methods and a result concerning the Diophantine equation $({x}^{n} - 1)/ (x- 1)= {y}^{2} $ due to Ljunggren. We also verify that when $2\leq c\leq 30$ with $c\not = 12, 24$, the Diophantine equation ${x}^{2} + \mathop{(2c- 1)}\nolimits ^{m} = {c}^{n} $ has only the positive integer solution $(x, m, n)= (c- 1, 1, 2). $


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.


2016 ◽  
Vol 95 (1) ◽  
pp. 5-13 ◽  
Author(s):  
MOU-JIE DENG ◽  
DONG-MING HUANG

Let $a,b,c$ be a primitive Pythagorean triple and set $a=m^{2}-n^{2},b=2mn,c=m^{2}+n^{2}$, where $m$ and $n$ are positive integers with $m>n$, $\text{gcd}(m,n)=1$ and $m\not \equiv n~(\text{mod}~2)$. In 1956, Jeśmanowicz conjectured that the only positive integer solution to the Diophantine equation $(m^{2}-n^{2})^{x}+(2mn)^{y}=(m^{2}+n^{2})^{z}$ is $(x,y,z)=(2,2,2)$. We use biquadratic character theory to investigate the case with $(m,n)\equiv (2,3)~(\text{mod}~4)$. We show that Jeśmanowicz’ conjecture is true in this case if $m+n\not \equiv 1~(\text{mod}~16)$ or $y>1$. Finally, using these results together with Laurent’s refinement of Baker’s theorem, we show that Jeśmanowicz’ conjecture is true if $(m,n)\equiv (2,3)~(\text{mod}~4)$ and $n<100$.


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 $(x,y)$.


2020 ◽  
Vol 57 (2) ◽  
pp. 200-206
Author(s):  
Elif kizildere ◽  
Maohua le ◽  
Gökhan Soydan

AbstractLet l,m,r be fixed positive integers such that 2| l, 3lm, l > r and 3 | r. In this paper, using the BHV theorem on the existence of primitive divisors of Lehmer numbers, we prove that if min{rlm2 − 1,(l − r)lm2 + 1} > 30, then the equation (rlm2 − 1)x + ((l − r)lm2 + 1)y = (lm)z has only the positive integer solution (x,y,z) = (1,1,2).


2020 ◽  
Vol 16 (08) ◽  
pp. 1701-1708
Author(s):  
Xiao-Hui Yan

For fixed coprime positive integers [Formula: see text], [Formula: see text], [Formula: see text] with [Formula: see text] and [Formula: see text], there is a conjecture that the exponential Diophantine equation [Formula: see text] has only the positive integer solution [Formula: see text] for any positive integer [Formula: see text]. This is the analogue of Jésmanowicz conjecture. In this paper, we consider the equation [Formula: see text], where [Formula: see text] are coprime positive integers, and prove that the equation has no positive integer solution if [Formula: see text] and [Formula: see text].


2017 ◽  
Vol 96 (1) ◽  
pp. 30-35 ◽  
Author(s):  
MI-MI MA ◽  
YONG-GAO CHEN

In 1956, Jeśmanowicz conjectured that, for any positive integers $m$ and $n$ with $m>n$, $\gcd (m,n)=1$ and $2\nmid m+n$, the Diophantine equation $(m^{2}-n^{2})^{x}+(2mn)^{y}=(m^{2}+n^{2})^{z}$ has only the positive integer solution $(x,y,z)=(2,2,2)$. In this paper, we prove the conjecture if $4\nmid mn$ and $y\geq 2$.


2018 ◽  
Vol 98 (2) ◽  
pp. 188-195 ◽  
Author(s):  
MOU-JIE DENG ◽  
JIN GUO ◽  
AI-JUAN XU

Let $c\geq 2$ be a positive integer. Terai [‘A note on the Diophantine equation $x^{2}+q^{m}=c^{n}$’, Bull. Aust. Math. Soc.90 (2014), 20–27] conjectured that the exponential Diophantine equation $x^{2}+(2c-1)^{m}=c^{n}$ has only the positive integer solution $(x,m,n)=(c-1,1,2)$. He proved his conjecture under various conditions on $c$ and $2c-1$. In this paper, we prove Terai’s conjecture under a wider range of conditions on $c$ and $2c-1$. In particular, we show that the conjecture is true if $c\equiv 3\hspace{0.6em}({\rm mod}\hspace{0.2em}4)$ and $3\leq c\leq 499$.


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