1.n=1,显然能整除
2.假设n=k(k>1)能被整除,即x^(2k-1)+y^(2k-1)=A(x+y)
当n=k+1时,x^(2k+1)+y^(2k+1)
=x^(2k-1)*x^2+y^(2k-1)*y^2
=A(x+y)-y^(2k-1)*x^2+y^(2k-1)*y^2
=A(x+y)+y^(2k-1)(y^2-x^2)
=A(x+y)+y^(2k-1)(y+x)(y-x)
以上两项都能被(x+y)整除,得证
1.n=1,显然能整除
2.假设n=k(k>1)能被整除,即x^(2k-1)+y^(2k-1)=A(x+y)
当n=k+1时,x^(2k+1)+y^(2k+1)
=x^(2k-1)*x^2+y^(2k-1)*y^2
=A(x+y)-y^(2k-1)*x^2+y^(2k-1)*y^2
=A(x+y)+y^(2k-1)(y^2-x^2)
=A(x+y)+y^(2k-1)(y+x)(y-x)
以上两项都能被(x+y)整除,得证