A NOTE ON JEŚMANOWICZ’ CONJECTURE CONCERNING PRIMITIVE PYTHAGOREAN TRIPLES

2016 ◽  
Vol 95 (1) ◽  
pp. 5-13 ◽  
Author(s):  
MOU-JIE DENG ◽  
DONG-MING HUANG
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Character Theory ◽  
Positive Integer Solution ◽  
Pythagorean Triples ◽  
Pythagorean Triple ◽  
Positive Integers ◽  
Baker's Theorem

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 $n<100$.

scholarly journals A note on Jeśmanowicz’ conjecture for non-primitive Pythagorean triples

2021 ◽  
Vol 5 (1) ◽  
pp. 115-127
Author(s):  
Van Thien Nguyen ◽  
◽  
Viet Kh. Nguyen ◽  
Pham Hung Quy ◽  
◽  
...  
Keyword(s):  
Positive Integer ◽  
Simple Proof ◽  
Diophantine Equation ◽  
Necessary Conditions ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Pythagorean Triples ◽  
Pythagorean Triple

Let \((a, b, c)\) be a primitive Pythagorean triple parameterized as \(a=u^2-v^2, b=2uv, c=u^2+v^2\), where \(u>v>0\) are co-prime and not of the same parity. In 1956, L. Jesmanowicz conjectured that for any positive integer \(n\), the Diophantine equation \((an)^x+(bn)^y=(cn)^z\) has only the positive integer solution \((x,y,z)=(2,2,2)\). In this connection we call a positive integer solution \((x,y,z)\ne (2,2,2)\) with \(n>1\) exceptional. In 1999 M.-H. Le gave necessary conditions for the existence of exceptional solutions which were refined recently by H. Yang and R.-Q. Fu. In this paper we give a unified simple proof of the theorem of Le-Yang-Fu. Next we give necessary conditions for the existence of exceptional solutions in the case \(v=2,\ u\) is an odd prime. As an application we show the truth of the Jesmanowicz conjecture for all prime values \(u < 100\).

JEŚMANOWICZ’ CONJECTURE ON PYTHAGOREAN TRIPLES

2017 ◽  
Vol 96 (1) ◽  
pp. 30-35 ◽  
Author(s):  
MI-MI MA ◽  
YONG-GAO CHEN
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Pythagorean Triples ◽  
Positive Integers

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 $y\geq 2$.

ON THE EXPONENTIAL DIOPHANTINE EQUATION

2014 ◽  
Vol 90 (1) ◽  
pp. 9-19 ◽  
Author(s):  
TAKAFUMI MIYAZAKI ◽  
NOBUHIRO TERAI
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Baker’S Method ◽  
Elementary Methods ◽  
Positive Integers

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.

scholarly journals A note on the Diophantine equation $|a^x-b^y|=c$

2010 ◽  
Vol 107 (2) ◽  
pp. 161
Author(s):  
Bo He ◽  
Alain Togbé ◽  
Shichun Yang
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Integer Solutions ◽  
Positive Integers

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 $(x,y)$.

A note on the ternary purely exponential diophantine equation Ax + By = Cz with A + B = C2

2020 ◽  
Vol 57 (2) ◽  
pp. 200-206
Author(s):  
Elif kizildere ◽  
Maohua le ◽  
Gökhan Soydan
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Primitive Divisors ◽  
Lehmer Numbers ◽  
Positive Integers

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).

The Diophantine equation (m2 + n2)x + (2mn)y = (m + n)2z

2020 ◽  
Vol 16 (08) ◽  
pp. 1701-1708
Author(s):  
Xiao-Hui Yan
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Positive Integers ◽  
Coprime Positive Integers

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].

A NOTE ON THE DIOPHANTINE EQUATION

2013 ◽  
Vol 90 (1) ◽  
pp. 20-27 ◽  
Author(s):  
NOBUHIRO TERAI
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Elementary Methods ◽  
Positive Integers

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). $

A NOTE ON JEŚMANOWICZ’ CONJECTURE CONCERNING NONPRIMITIVE PYTHAGOREAN TRIPLES

2020 ◽  
pp. 1-11
Author(s):  
YASUTSUGU FUJITA ◽  
MAOHUA LE
Keyword(s):  
Positive Integer ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Baker’S Method ◽  
Pythagorean Triples ◽  
Integer Solutions ◽  
Positive Integers

Abstract Jeśmanowicz conjectured that $(x,y,z)=(2,2,2)$ is the only positive integer solution of the equation $(*)\; ((\kern1.5pt f^2-g^2)n)^x+(2fgn)^y=((\kern1.5pt f^2+g^2)n)^x$ , where n is a positive integer and f, g are positive integers such that $f>g$ , $\gcd (\kern1.5pt f,g)=1$ and $f \not \equiv g\pmod 2$ . Using Baker’s method, we prove that: (i) if $n>1$ , $f \ge 98$ and $g=1$ , then $(*)$ has no positive integer solutions $(x,y,z)$ with $x>z>y$ ; and (ii) if $n>1$ , $f=2^rs^2$ and $g=1$ , where r, s are positive integers satisfying $(**)\; 2 \nmid s$ and $s<2^{r/2}$ , then $(*)$ has no positive integer solutions $(x,y,z)$ with $y>z>x$ . Thus, Jeśmanowicz’ conjecture is true if $f=2^rs^2$ and $g=1$ , where r, s are positive integers satisfying $(**)$ .

scholarly journals On the Diophantine Equation x2 – kxy + ky2 + ly = 0, l = 2n

2017 ◽  
Vol 55 (1) ◽  
pp. 115-118
Author(s):  
Sukrawan Mavecha
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Positive Integer Solution

AbstractWe consider the Diophantine equation x2-kxy+ky2+ ly = 0 for l = 2nand determine for which values of the odd integer k, it has a positive integer solution x and y.

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.

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