记f(n)=a^(n+2)+(a+1)^(2n+1)
1)f(1)=a^3+(a+1)^3=(2a+1)(a^2-a^2-a+a^2+2a+1)=(2a+1)(a^2+2a+1)能被a^2+2a+1整除
2)假设n=k时成立,n=k+1时
f(k+1)-f(k)=a^(k+3)+(a+1)^(2k+3)-a^(k+2)-(a+1)^(2k+1)
=(a-1)a^(k+2)+(k^2+2a)(a+1)^(2k+1)
=(a-1)a^(k+2)+(a^2+a+1)*(a+1)^(2k+1)+(a-1)(a+1)^(2k+1)
=(a^2+a+1)*(a+1)^(2k+1)+(a-1)[a^(k+2)+(a+1)^(2k+1)]
=(a^2+a+1)*(a+1)^(2k+1)+(a-1)f(k)
所以f(k+1)-f(k)能被a^2+a+1整除,故f(k+1)也能被a^2+a+1整除
证毕!