设有k1,k2,k3,k4使k1(a1+a2)+k2(a2+a3)+k3(a3+a4)+k4(a4+a1)=0即(k1+k4)a1+(k1+k2)a2+(k2+k3)a3+(k3+k4)a4=0由题意a1,a2,a3,a4线性无关,则k1+k4=0k1+k2=0k2+k3=0k3+k4=0显然k1=k3=1,k2=k4=-1是其一组解,k1,k2,k3,k4...