关于丢番图方程(1023n)^(x)+(64n)^(y)=(1025n)^(z)

On diophantine equation(1023n)^(x)+(64n)^(y)=(1025n)^(z)

设a,b,c是两两互素的正整数且满足商高数条件,即当a,b,c为本原商高数时,方程(an)^(x)+(bn)^(y)=(cn)^(z)仅有正整数解(x,y,z)=(2,2,2).而现有的丢番图方程形式并没有将b的具体形式与初等数论紧密结合,利用奇偶分析法、简单同余理论、将b取为26并与初等数论相结合,还运用了分类讨论、反证法的思想,具体为先采用反证法进行假设,根据所化简的等式选取合适的模数进行推算得出与假设相悖的结论,即证明了:若n为正整数,当(a,b,c)=(1023,64,1025)时,丢番图方程(1023n)x+(64n)y=(1025n)z仅有正整数解(x,y,z)=(2,2,2),以此验证Jesmanowicz猜想成立,这个证明结果使Jesmanowicz猜想更加充实.

Let a,b,c be positive integers such that any two among them were coprime and satisfied primitive pythagorean triples condition,namely,Let a,b,c be primitive pythagorean triples,the Diophantine equation(an)^(x)+(bn)^(y)=(cn)^(z)had no solutions in positive integer other than(x,y,z)=(2,2,2).However,the existing diophantine equation form did not closely combine the specific of b with elementary number theory.This paper not only used the parity analysis and simple congruence method to combine the six power of b with the elementary number theory.It also used the idea of classification discussion and reduction to absurdity.Specifically,the counter-evidence method was used to make the hypothesis,and the appropriate modulus was selected according to the simplified equation to calculate the conclusion that was contrary to the hypothesis.Namely,in order to verify Jesmanowicz’s conjecture,let n be a positive integer,if(a,b,c)=(1023,64,1025),the exponential diophantine equation(1023n)x+(64n)y=(1025n)z had only one positive integer solution(x,y,z)=(2,2,2),this proved that the result made the Jesmanowicz’s conjecture more substantial.