设f（x ）是R上的函数,且满足f(0)=1,并且对任意实数x,y,都有f(x-y)=f(x)-y(2x-y+1),

令 y = x得：
f(x - x) = f(x) - x(2x - x + 1)
f(0) = f(x) - x(x + 1)
1 = f(x) - x(x + 1)
f(x) = x(x + 1) + 1
f(x) = x² + x + 1