x^n-y^n 　　当n=2时,x^2-y^2=(x+y)(x-y),所以(x^2-y^2)/(x+y)=x-y=f,f为整数. 　　设当n=2k之前都成立即x^2k-y^2k能被x+y整除,即有(x^2k-y^2k)/(x+y)=g,g为整数 　　当n=2(k+1)时, 　　x^2(k+1)-y^2(k+1) 　　=x^2*x^2k-y^2*y^2k 　　所以 　　[x^2(k+1)-y^2(k+1)]/(x+y) 　　=(x^2*x^2k)/(x+y)-(y^2*y^2k)/(x+y)说明:将(x+y)移进去, 　　=(x^2/(x+y)*x^2k)-(y^2*y^2k)/(x+y)说明:由(x^2-y^2)/(x+y)=f知道x^2/(x+y)=f-y^2/(x+y) 　　=[f-y^2/(x+y)]*x^2k-(y^2*y^2k)/(x+y) 　　=f*x^2k-y^2/(x+y)*x^2k-(y^2*y^2k)/(x+y) 　　=f*x^2k-y^2*[x^2k/(x+y)-y^2k/(x+y)]说明:n=2k时我们已经有(x^2k-y^2k)/(x+y)=g 　　=f*x^2k-y^2*g 　　f是整数,g是整数,x^2k,y^2都是整数, 　　所以[x^2(k+1)-y^2(k+1)]/(x+y)=h,h为整数. 　　即n=2(k+1)时也成立,所以对所有n=2k(k=1,2,3...)都成立. 　　考虑到太乱可能看不懂,我加了说明. 　　另外,事实上：有 　　x^n-y^n=(x+y)[x^(n-1)-x^(n-2)y+x^(n-3)y^2-…………+xy^(n-2)-y^(n-1)] 　　不过跟题目无关,因为题目要归纳法证明.