SOME NOTES ON A GENERALIZED VERSION OF PYTHAGOREAN TRIPLES

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.

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 $n<100$.


2011 ◽  
Vol 90 (3) ◽  
pp. 355-370
Author(s):  
TAKAFUMI MIYAZAKI

AbstractLet (a,b,c) be a primitive Pythagorean triple such that b is even. In 1956, Jeśmanowicz conjectured that the equation ax+by=cz has the unique solution (x,y,z)=(2,2,2) in the positive integers. This is one of the most famous unsolved problems on Pythagorean triples. In this paper we propose a similar problem (which we call the shuffle variant of Jeśmanowicz’ problem). Our problem states that the equation cx+by=az with x,y and z positive integers has the unique solution (x,y,z)=(1,1,2) if c=b+1 and has no solutions if c>b+1 . We prove that the shuffle variant of the Jeśmanowicz problem is true if c≡1 mod b.


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.


2021 ◽  
Vol 5 (1) ◽  
pp. 115-127
Author(s):  
Van Thien Nguyen ◽  
◽  
Viet Kh. Nguyen ◽  
Pham Hung Quy ◽  
◽  
...  

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 \(u < 100\).


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 $y\geq 2$.


2005 ◽  
Vol 4 (1) ◽  
pp. 96-100
Author(s):  
A. K. Maran

We know already that the set Of positive integers, which are satisfying the Pythagoras equation Of three variables and four variables cre called Pythagorean triples and quadruples respectively. These cre Diophantine equation OF second power. The all unknowns in this Pythagorean equation have already Seen by mathematicians Euclid and Diophantine. Hcvwever the solution defined by Euclid are Diophantine is also again having unknowns. The only to solve the Diophantine equations wos and error method. Moreover, the trial and error method to obtain these values are not so practical and easy especially for time bound work, since the Diophantine equations are having more than unknown variables.


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.


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