设a所有约数和为F(a)
首先对互质的两个数a,b,F(ab)=F(a)*F(b)
令a约数1=x1,x2.xm
b约数1=y1,y2,...yn
由a,b互质,a2->am与b2->bn无公共元素
那么
F(ab)=
sigma(1<=i<=m,1<=j<=n)xiyj
=(sigma(1<=i<=m)xi)*(sigma(1<=j<=n)yj)
=F(a)*F(b)
而
F(pi^qi)=1+pi+pi^2+...+pi^qi=(pi^(qi+1)-1)/(pi-1)
又pi^qi与pj^qj互质
所以
F(n)=∏(1<=i<=k)[pi^(qi+1)-1)/(pi-1)]
得证