令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)=a(x+1)²+b(x+1)+c+a(x-1)²+b(x-1)+c
=2ax²+2a+2bx+2c
=2ax²+2bx+2a+2c
=2x²+4x
所以
2a=2,2b=4,2a+2c=0
所以
a=1,b=2,c=-1
f(x)=x²+2x-1