试求出所有的整数n,使得n3-n+5/n2+1 是一个整数

（n3-n+5）/（n2+1 ）=[(n^3+n)-(2n-5)]/(n^2+1)=n-(2n-5)/(n^2+1)
所以 (2n-5)/(n^2+1) 必须为整数.
=> |2n-5|>n^2+1 或者2n-5=0.n为整数=>2n-5不等于0.
当n>5/2时,2n-5>n^2+1 =>n^2-2n+6(n+1)^2+5n^2+2n-4 -1-根号5