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：

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整除.

所以当整数n>=0,x^(n+2)+(x+1)^(2n+1)能被x^2+x+1整除.