[tex]\it x\in\mathbb{Z},\ \ \dfrac{4}{x-3}\in\mathbb{Z} \Rightarrow x-3\in D_4 \Rightarrow x-3\in\{\pm1,\ \pm2,\ \pm4\} \Rightarrow \\ \\ \\ \Rightarrow x-3\in\{-4,\ -2,\ -1,\ 1,\ 2,\ 4\}|_{+3} \Rightarrow x\in\{-1,\ 1,\ 2,\ 4,\ 5,\ 7\}=A\\ \\ \\ A\cap P=\{2,\ 5,\ 7\}[/tex]
