两边展开,整理后得:
x=(a^2+1)/(a+1)=[(a+1)^2-2(a+1)+2]/(a+1)
=a-1+2/(a+1)
a为整数,且关于x的方程ax(x+1)-1=x(ax-1)+a^2有整数解,故2/(a+1)为整数,故a+1=2或1或-1或-2
故a=1或0或-2或-3
两边展开,整理后得:
x=(a^2+1)/(a+1)=[(a+1)^2-2(a+1)+2]/(a+1)
=a-1+2/(a+1)
a为整数,且关于x的方程ax(x+1)-1=x(ax-1)+a^2有整数解,故2/(a+1)为整数,故a+1=2或1或-1或-2
故a=1或0或-2或-3