在第4步取 d=yd=yd=y: Q(x,y)=Q(x+y,0)Q(x,y)=Q(x+y,0)Q(x,y)=Q(x+y,0)。齐次性给出 Q(x+y,0)=Q(1,0)(x+y)n−1Q(x+y,0)=Q(1,0)(x+y)^{n-1}Q(x+y,0)=Q(1,0)(x+y)n−1。由于 P(1,0)=Q(1,0)=1P(1,0)=Q(1,0)=1P(1,0)=Q(1,0)=1,得到 Q=(x+y)n−1Q=(x+y)^{n-1}Q=(x+y)n−1,从而 P=(x−2y)(x+y)n−1P=(x-2y)(x+y)^{n-1}P=(x−2y)(x+y)n−1。