Question

当整数n>=0,求证:x^(n+2)+(x+1)^(2n+1)能被x^2+x+1整数

Answer 1

显然n=0时冲让：
x^(n+2)+(x+1)^(2n+1)=x^2+x+1
能被x^2+x+1整除.
如果假设n=k时：
x^(n+2)+(x+1)^(2n+1)=x^(k+2)+(x+1)^(2k+1) 能被x^2+x+1整除, 那么n=k+1时:
x^(n+2)+(x+1)^(2n+1)
=x^(k+3)+(x+1)^(2k+3)
=x^(k+3)+(x^2+2x+1)(x+1)^(2k+1)
=x(x^(k+2)+(x+1)^(2k+1))+(x^2+x+1)(x+1)^(2k+1)
也能被x^2+x+1整渣姿除.
所以当如判绝整数n>=0,x^(n+2)+(x+1)^(2n+1)能被x^2+x+1整除.