摘要
设A={ak}∞k=1和B={bk}∞k=1是正整数数列.本文证明了:如果当k>1时,min(ak,bk)>1,则方程(a1an)m+(b1bm)n=(a1b1am+nbm+n)m+n无正整数解(m,n).根据上述结果,解决了有关数论函数δ(k)和Ψ(k)的一个方程.
Let A = {a k }∞k=1and B = {b k }∞k=1be two sequences of positive integer. In this paper, we prove that if min( ak , bk ) > 1for k > 1,then the equation (a1…an) m + (b1…bm)n = (a1b1…am+n bm+n)m+n has no positive integer solution ( m , n ). With the above-mentioned result,an equation concerning the arithmetic functions δ (k)andψ (k)is solved.
出处
《河套学院论坛》
2007年第4期12-13,共2页
HETAO COLLEGE FORUM