用数学归纳法证明：X的2n次方—y的2n次方能被X+Y整除（

证明：n=1时，x^2-y^2=(x+y)(x-y)，满足结论；假设n=k时也满足结论，即x^2k-y^2k能被x+y的整除，当n=k+1时，x^2(k+1)-y^2(k+1)=(x^2k)*(x^2)-(y^2k)*(y^2)+(x^2)*(y^2k)-(x^2)*(y^2k)=(x^2)(x^2k-y^2k)+(y^2k)(x+y)(x-y)，显然也能被x+y整除。综上所述， X的2n次方—y的2n次方能被X+Y整除。

证：
n=1时,x²-y²=(x+y)(x-y),包含因式x+y,能被x+y整除.
假设当n=k(k∈N+且k≥1)时,x^(2k)-y^(2k)能被x+y整除,则当n=2(k+1)时,
x^[2(k+1)]-y^[2(k+1)]
=[x^(2k+1)-y^(2k+1)](x+y)-yx^(2k+1)+xy^(2k+1)
=[x^(2k+1)-y^(2k+1)](x+y)-xy[x^(2k)-y^(2k)]
[x^(2k+1)-y^(2k+1)](x+y)中包含因式x+y,能被x+y整除；xy[x^(2k)-y^(2k)]中包含能被x+y整除的因式x^(2k)-y^(2k),能被x+y整除.即当n=k+1时,x^[2(k+1)]-y^[2(k+1)]能被x+y整除.
综上,得x^(2n)-y^(2n)能被x+y整除.