$n^5+1=n^5-n^2+n^2+1=n^2(n^3+1)+n^2-1$
để $n^5 +1$ chia hết cho $n^3+1 $hay $n^2(n^3+1)+n^2-1$ chia hết cho $n^3+1 => n^2-1$ chia hết cho $n^3+1$
$=> (n+1)(n-1)$ chia hết $(n+1)(n^2-n+1) $
$=> (n-1) $chia hết cho $(n^2-n+1)$
$=> n(n-1)$ chia hết cho $(n^2-n+1)$
$=> n^2-n$ chia hết cho $n^2-n+1$
$=> n^2-n+1-1$ chia hết cho$ n^2-n+1$
$=> 1$ chia hết cho $n^2-n+1$
$=> n^2-n +1 \in Ư(1)=(1;-1)$