ax^2+1为偶函数。f(x)为奇函数，所以bx+c为奇函数。c=0。(a+1)/b=2,b=(a+1)/2(4a+1)/2b=(4a+1)/(a+1)=4-(3/a+1)<3a+1<3a=1或0,因为b为整数故a=1,b=1综上,a=1,b=1,c=0
