摘要
设a,b,c是互素的正整数,使得a^2+b^2=c^2.1956年Jesmanowicz猜测对于任意的正整数n,不定方程(an)~x+(bn)~y=(cn)~z只有唯一的解(x,y,z)=(2,2,2).根据已有引理,使用分类讨论和素数的唯一分解定理,证明a或者b为2^(r_1+1)n_1的情形下Jesmanowicz成立,其中r_1,n_1为任意正整数.
Let a,b,c be relatively prime positive integers such that a^2+b^2=c^2.In 1956,Jesmanowicz conjecture that for any positive integer n,the only solution of(an)~x+(bn)~y=(cn)~z in positive integers is(x,y,z)=(2,2,2).In this article,according to the existing lemmas,the unique decomposition theorem of prime numbers is discussed and used to prove Jesmanowicz conjecture which aor bis 2^(r_1+1)n_1 where r1,n1 is any positive integer.
出处
《淮海工学院学报(自然科学版)》
CAS
2017年第3期1-3,共3页
Journal of Huaihai Institute of Technology:Natural Sciences Edition