Đáp án:
\[\dfrac{n+1}{n^2+1}\in Z\]
\[↔n+1\qquad\vdots\qquad n^2+1\]
\[↔(n-1)(n+1)\qquad\vdots\qquad n^2+1\]
\[↔n^2-1\qquad\vdots\qquad n^2+1\]
\[↔n^2+1-2\qquad\vdots\qquad n^2+1\]
Vì $n^2+1\qquad\vdots\qquad n^2+1$
$→2\qquad\vdots\qquad n^2+1$
Vì $n^2+1≥1$
$\to n^2+1\in U_{\{2\}}=\{1;2\}$
$\to n\in \{0;±1\}$