已知数列an=(x^n-y^n)/(x-y)(n=1,2,3......)其中x,y是方程t^2-t-1=0的两个根。证明数列的项是正整数且任意相邻两项均互质。谢谢

解：因为xy为t^2-t-1=0的二根，所以x+y=1,xy=-1
当n=1时a1=(x^1-y^1)/(x-y)=1 为正整数
当n=2时a2=(x^2-y^2)/(x-y)
=(x-y)(x+y)/(x-y)=1　为正整数
当n=3时a3=(x^3-y^3)/(x-y)=(x-y)(x^2+xy+y^2)/(x-y)
=(x-y)((x+y)^2-xy)/(x-y)=2　为正整数
若n=k时ak=(x^k-y^k)/(x-y) 为正整数
a(k+1)=(x^(k+1)-y^(k+1))/(x-y) 为正整数且与ak互质
则a(k+1)=(x^(k+1)-y^(k+1))/(x-y)
=(x^(k+1)-y^(k+1))(x+y)/(x-y) 因为x+y=1
=(x^(k+2)-y^(k+2)+x^(k+1)y-xy^(k+1))/(x-y)
=(x^(k+2)-y^(k+2))/(x-y)+xy(x^k-y^k)/(x-y)
=a(k+2)-ak 因为xy=-1
所以a(k+2)=a(k+1)+ak 两正整数相加为一正整数
因为ak与a(k+1)互质，所以 a(k+2)=a(k+1)+ak与a(k+1)互质