tìm n thuộc Z biết :5n^4-2n^2+3n-7 chia hết cho (n-1)

Ta thấy $\displaystyle ( n\ -\ 1)( n\ +\ 1) \ =\ n^{2} \ -\ n\ +\ n\ -\ 1\ =\ n^{2} \ -\ 1$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
5n^{4} \ -\ 2n^{2} \ +\ 3n\ -\ 7\\
=\ 5n^{4} \ -\ 5n^{2} \ +\ 3n^{2} \ -\ 3n\ +\ 6n\ -\ 7\\
=\ 5n^{2}\left( n^{2} \ -\ 1\right) \ +\ 3n( n\ -\ 1) \ +\ 6n\ -\ 7\\
=\ 5n^{2}( n\ -\ 1)( n\ +\ 1) \ +\ 3n( n\ -\ 1) \ +\ 6n\ -\ 7\\
=\ ( n\ -\ 1)\left[ 5n^{2}( n\ +\ 1) \ +\ 3n\right] \ +\ 6n\ -\ 7
\end{array}$
Dễ thấy $\displaystyle ( n\ -\ 1) \ \vdots \ ( n\ -\ 1)$ ⟹ $\displaystyle ( n\ -\ 1)\left[ 5n^{2}( n\ +\ 1) \ +\ 3n\right] \ \vdots \ ( n\ -\ 1)$
Vậy để $\displaystyle 5n^{4} \ -\ 2n^{2} \ +\ 3n\ -\ 7\ \vdots \ ( n\ -\ 1)$ thì $\displaystyle ( 6n\ +\ 7) \ \vdots \ ( n\ -\ 1)$ tức là $\displaystyle \frac{6n\ +\ 7}{n\ -\ 1} \ \in \ Z$
$\displaystyle \frac{6n\ +\ 7}{n\ -\ 1} \ =\frac{6n\ -\ 6\ +\ 13}{n\ -\ 1} \ =\ 6\ +\ \frac{13}{n\ -\ 1}$
Để $\displaystyle \frac{6n\ +\ 7}{n\ -\ 1} \ \in \ Z$ thì $\displaystyle \frac{13}{n\ -\ 1} \ \in \ Z$
⟹ $\displaystyle n\ -\ 1\ \in \ Ư( 13)$
⟹ $\displaystyle n\ -\ 1\ \in \ \{-13;\ -1;\ 1;\ 13\}$
⟹ $\displaystyle n\ \in \ \{-12;\ 0;\ 2;\ 14\}$