设f（x）=ax²+bx+c
f（x+1）=a（x+1）²+b（x+1）+c
f（x-1）=a（x-1）²+b（x-1）+c
f(x+1)+f(x-1)=2ax²+2bx+2a+2c
于是2ax²+2bx+2a+2c=2x^2-4x
2a=2,2b=-4,2a+2c=0
得a=1.b=-2,c=-1
f（x）=x²-2x-1