Đáp án:
$n \in \{−293,−146,−97,−48,−41,−20,−13,−6,−5,−2,−1,0,2,3,4,7,8,15,22,43,50,99,148,295\}.$
Giải thích các bước giải:
$(103n^2+121n+70) \ \vdots \ (n-1)\\ \Rightarrow \dfrac{103n^2+121n+70}{n-1} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{103n^2-103n+224n-224+294}{n-1} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{103n(n-1)+224(n-1)+294}{n-1} \in \mathbb{Z}\\ \Leftrightarrow 103n+224 + \dfrac{294}{n-1} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{294}{n-1} \in \mathbb{Z}(\text{Do } n \in \mathbb{Z})\\ n \in \mathbb{Z}, \dfrac{294}{n-1} \Rightarrow (n-1) \in Ư(294)\\ \Leftrightarrow (n-1) \in \{−294,−147,−98,−49,−42,−21,−14,−7,−6,−3,−2,−1,1,2,3,6,7,14,21,42,49,98,147,294\}\\ \Leftrightarrow n \in \{−293,−146,−97,−48,−41,−20,−13,−6,−5,−2,−1,0,2,3,4,7,8,15,22,43,50,99,148,295\}.$
