f(x+1)+f(x-1)=2x^2-4x

f(x)=ax^2+bx+c
f(x+1)+f(x-1)
=a(x+1)^2+b(x+1)+c+a(x-1)^2+b(x-1)+c
=2ax^2+2a+2bx+2c
=2ax^2+2bx+(2a+2c)
=x^2-4x
所以2a=1,2b=-4,2a+2c=0
a=1/2,b=-2,c=-1/2
f(x)=(1/2)x^2-2x-1/2

易知x是2次多项式
设置f(x)=ax^2+bx+c
f(x+1)+f(x-1)=2ax^2+2a+2bx+2c
对照 a=1 b=-2 a+c=0 c=-1
因此f(x)=x^2-2x-1