$n^5+1⋮n^3+1$

$\Rightarrow n^6+n⋮n^3+1$

Viết về dạng phân số:

$\dfrac{n^6+n}{n^3+1}=\dfrac{n^6-1+n+1}{n^3+1}=\dfrac{n^6-1}{n^3+1}+\dfrac{n+1}{n^3+1}$

$=\dfrac{\left(n^3-1\right)\left(n^3+1\right)}{n^3+1}+\dfrac{n+1}{\left(n+1\right)\left(n^2-n+1\right)}$

$=n^3-1+\dfrac{1}{n^2-n+1}$

Để $n^5+1⋮n^3+1$ thì: $\dfrac{1}{n^2-n+1}\in Z\Leftrightarrow n^2-n+1\in\left\{1;-1\right\}$

$\circledast$Với: \(n^2-n+1=1\Leftrightarrow n^2-n=0\Leftrightarrow n\left(n-1\right)=0\Leftrightarrow\left[{}\begin{matrix}n=0\=1\end{matrix}\right.\)

$\circledast$ Với $n^2-n+1=-1\Leftrightarrow n^2-n+2=0$

$\Rightarrow n^2-n+\dfrac{1}{4}+\dfrac{7}{4}=0$

$\Rightarrow\left(n-\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\Leftrightarrow$ pt vô nghiệm

Vậy $n\in\left\{0;1\right\}$