恒等式f(x-y)=f(x)-y(2x-y+1),可化為f(x-y)-(x-y)²-(x-y)=f(x)-x²-x
設(shè)g(x)=f(x)-x²-x,則g(x-y)=g(x)對(duì)于任意實(shí)數(shù)x,y成立.
∴g(x)為常數(shù)函數(shù),
又g(0)=f(0)=1,
∴g(x)必定恒為1,即f(x)=x²+x+1.