证明
:(1)当n=1时,xn+yn=x+y能被x+y整除.(2)假设n=2k-1时命题成立,即(x2k-1+y2k-1)能被x+y整除,那么n=2k+1时有x2k+1+y2k+1=x2k-1·x2+y2k-1·y2
=x2k-1·x2+x2k-1·y2-x2k-1·y2+y2k-1·y2
=x2k-1(x2-y2)+(x2k-1+y2k-1)·y2
=x2k-1(x+y)(x-y)+(x2k-1+y2k-1)·y2.
∵x2k-1(x+y)(x-y)与(x2k-1+y2k-1)y2都能被x+y整除,∴n=2k+1时命题成立.
综合(1)(2)可知对一切正奇数n,xn+yn都能被x+y整除.