Generator of Pythagorean triples and Je$\acute{s}$manowicz conjecture

Nainrong Feng

Published 2020-01-23Version 1

Let $a,b,c$ be relatively prime positive integers such that $a^2+b^2=c^2, 2|b$. In this paper, we show that Pythagorean triples $(a, b,c)$ must satisfy $abc\equiv{0\; (\mod3\cdot{4}\cdot{5}})$ and $c\neq{0\; (\mod{3}})$, and we also prove that for $(a,b,c)\in\{(a,b,c)|a\equiv{0\;(\mod{3}}),b\equiv{0\;(\mod{4}}),c\equiv{0\; (\mod{5}})\}\bigcup\{(a,b,c)|b\equiv{0\;(\mod{12}}),c\equiv{0\;(\mod{5}})\}$, the only solution of $$a^x+b^y=c^z\qquad{z},y,z\in{N}$$ in positive integers is $(x, y, z) = (2, 2,2)$.