摘要
设(a,b,c)是一组满足a^2+b^2=c^2,gcd(a,b)=1,2|b的本原商高数,运用初等数论方法讨论方程(an)~x+(bn)~y=(cn)~z正整数解(x,y,z,n),证明了:当(a,b,c)=(143,24,145)时,方程仅有正整数解(x,y,z,n)=(2,2,2,m),其中m是任意正整数,上述结果说明此时Jesmanowicz猜想成立.
Let （a, b, c） be a primitive Pythagorean triple such that a^2＋b^2 = c^2, gcd （a, b） = 1 and21b. In this paper, using some elementary number theory methods, the positive integer solutions （x, y, z, n） of the equation （an）x ＋ （a, b, c） = （143, 24, 145）, then the equation has （2, 2, 2, m）, where m is an arbitrary positive conjecture for this case. bny = （cn）z are discussed. We prove that if only the positive integer solutions （x, y, z, n） = integer. The above result verifies Jesanowicz conjecture for this case.
出处
《数学的实践与认识》
北大核心
2017年第20期178-182,共5页
Mathematics in Practice and Theory
