a=1,2x+2=0,x=-1
a不等于1,一元二次方程.
x1+x2=(a2-3)/(a-1)=a+1-[2/(a-1)]
x1x2=(a2+a)/(a-1)=a+2+[2/(a-1)]
当a=3,2,0,-1时,x1+x2,x1x2为整数
又,[(a2-3)^2]-4(a-1)(a2+a)>=0,
a=0,-1
代入,验证,得
a=0,-1
综上,a=1,0,-1