n=k+1时 x^2k+1 + y^2k+1
=x^[2k-1]*x2+y^[2k-1]*y^2
=x^[2k-1]*x2+y^[2k-1]*y^2+y^[2k-1]*x^2-y^[2k-1]*x^2
=[^[2k-1]*x2+^[2k-1]*x^2]+[y^[2k-1]*y^2-y^[2k-1]*x^2]
=x^2*[x^2k-1 + y^2k-1]+y^[2k-1]*[y^2-x^2]
由归纳假设x^2k-1 + y^2k-1能整除 x+y
y^2-x^2能整除 x+y
所以 x^2*[x^2k-1 + y^2k-1]+y^[2k-1]*[y^2-x^2] 能整除 x+y
即当n=k+1时时 命题成立
所以 x^2n-1 + y^2n-1 能被 x+y整除