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Paper Title
A Study on Integer Design of Solutions of Diophantine Equation ?(X^4+Y^4 )(?U^2+V^2 )=T^2 (C^2-D^2 )(Z^2-W^2 ) P^? With ?>0,?>0,?=2,3 and X
Authors
Dr T SRINIVAS
Keywords
Diophantine Equation, exponential, Pythagorean triplet, Integer design.
Abstract
This paper focused on a study to find integer design of solutions Diophantine Equation
?(X^4+Y^4 )(?U^2+V^2 )=T^2 (C^2-D^2 )(Z^2-W^2 ) P^? With ?>0,?>0 and X0,?>0 and X2 is parameterized by integers k and n, with variables defined as:
x = k^n,y=k^(n+1),z=k^(n+3),w=k^(n+2),p=k^n, U=2^(n+1), V=2^n , T=3(2)^n ,
?=(1+k^4 )(k^6-k^4 ) k^(?-2)n n^2 , and {?(C=(((1+k^4)/2)^2+1)n,D=(((1+k^4)/2)^2-1)n, if 1+k^4 is even@C=(((1+k^4 )^2+1)/2) )n,D=(((1+k^4 )^2-1)/2)n, if 1+k^4 is odd)}.
for ?=1 is x = k^n,y=k^(n+1),z=k^(n+3),w=k^(n+2),p=k^2n,U=2^(n+1),V=2^n ,T=3(2)^n ,
?=(1+k^4 )(k^6-k^4 ) n^2 and {?(C=(((1+k^4)/2)^2+1)n,D=(((1+k^4)/2)^2-1)n, if 1+k^4 is even@C=(((1+k^4 )^2+1)/2) )n,D=(((1+k^4 )^2-1)/2)n, if 1+k^4 is odd)}.
for ?=2 is x = k^n,y=k^(n+1),z=k^(n+3),w=k^(n+2),p=k^n,U=2^(n+1),V=2^n ,T=3(2)^n ,
?=(1+k^4 )(k^6-k^4 ) n^2 and {?(C=(((1+k^4)/2)^2+1)n,D=(((1+k^4)/2)^2-1)n, if 1+k^4 is even@C=(((1+k^4 )^2+1)/2) )n,D=(((1+k^4 )^2-1)/2)n, if 1+k^4 is odd)}.
Lemma 2: At ?=3, the Diophantine equation
?(X^4+Y^4 )(3U^2+V^2 )=T^2 (C^2-D^2 )(Z^2-W^2 ) P^? With ?>0,?>0 and X2 is parameterized by integers k and n, with variables defined as:
x = k^n,y=k^(n+1),z=k^(n+3),w=k^(n+2),p=k^n, U=3^n, V=3^n , T=2(3)^n ,
?=(1+k^4 )(k^6-k^4 ) k^(?-2)n n^2 , and {?(C=(((1+k^4)/2)^2+1)n,D=(((1+k^4)/2)^2-1)n, if 1+k^4 is even@C=(((1+k^4 )^2+1)/2) )n,D=(((1+k^4 )^2-1)/2)n, if 1+k^4 is odd)}.
for ?=1 is x = k^n,y=k^(n+1),z=k^(n+3),w=k^(n+2),p=k^2n,U=3^n, V=3^n , T=2(3)^n ,
?=(1+k^4 )(k^6-k^4 ) n^2 and {?(C=(((1+k^4)/2)^2+1)n,D=(((1+k^4)/2)^2-1)n, if 1+k^4 is even@C=(((1+k^4 )^2+1)/2) )n,D=(((1+k^4 )^2-1)/2)n, if 1+k^4 is odd)}.
for ?=2 is x = k^n,y=k^(n+1),z=k^(n+3),w=k^(n+2),p=k^n,U=3^n, V=3^n , T=2(3)^n ,
?=(1+k^4 )(k^6-k^4 ) n^2 and {?(C=(((1+k^4)/2)^2+1)n,D=(((1+k^4)/2)^2-1)n, if 1+k^4 is even@C=(((1+k^4 )^2+1)/2) )n,D=(((1+k^4 )^2-1)/2)n, if 1+k^4 is odd)}
IJCRT's Publication Details
Unique Identification Number - IJCRT2511047
Paper ID - 295939
Page Number(s) - a436-a445
Pubished in - Volume 13 | Issue 11 | November 2025
DOI (Digital Object Identifier) - https://doi.org/10.56975/ijcrt.v13i11.295939
Publisher Name - IJCRT | www.ijcrt.org | ISSN : 2320-2882
E-ISSN Number - 2320-2882
Cite this article
Dr T SRINIVAS, "A Study on Integer Design of Solutions of Diophantine Equation ?(X^4+Y^4 )(?U^2+V^2 )=T^2 (C^2-D^2 )(Z^2-W^2 ) P^? With ?>0,?>0,?=2,3 and X<Y<W<Z", International Journal of Creative Research Thoughts (IJCRT), ISSN:2320-2882, Volume.13, Issue 11, pp.a436-a445, November 2025, Available at :http://www.ijcrt.org/papers/IJCRT2511047.pdf
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