1 problem found
Let \(\mathrm{g}(x)=ax+b.\) Show that, if \(\mathrm{g}(0)\) and \(\mathrm{g}(1)\) are integers, then \(\mathrm{g}(n)\) is an integer for all integers \(n\). Let \(\mathrm{f}(x)=Ax^{2}+Bx+C.\) Show that, if \(\mathrm{f}(-1),\mathrm{f}(0)\) and \(\mathrm{f}(1)\) are integers, then \(\mathrm{f}(n)\) is an integer for all integers \(n\). Show also that, if \(\alpha\) is any real number and \(\mathrm{f}(\alpha-1),\) \(\mathrm{f}(\alpha)\) and \(\mathrm{f}(\alpha+1)\) are integers, then \(\mathrm{f}(\alpha+n)\) is an integer for all integers \(n\).
Solution: If \(g(0) \in \mathbb{Z} \Rightarrow b \in \mathbb{Z}\). If \(g(1) \in \mathbb{Z} \Rightarrow a+b \in \mathbb{Z} \Rightarrow a \in \mathbb{Z}\), therefore \(a \cdot n + b \in \mathbb{Z}\), in particular \(g(n) \in \mathbb{Z}\) for all integers \(n\). \(f(0) \in \mathbb{Z} \Rightarrow C \in \mathbb{Z}\), \(f(1) \in \mathbb{Z} = A+ B + C \in \mathbb{Z} \Rightarrow A+ B \in \mathbb{Z}\) \(f(-1) \in \mathbb{Z} = A- B + C \in \mathbb{Z} \Rightarrow A- B \in \mathbb{Z}\) \(\Rightarrow 2A, 2B \in \mathbb{Z}\) \begin{align*} f(n) &= An^2 + Bn + C \\ &= An^2-An + An+Bn + C \\ &= 2A \frac{n(n-1)}2 + (A+B)n + C \\ &\in \mathbb{Z} \end{align*} Consider \(g(x) = f(x + \alpha)\), therefore \(g(0), g(1), g(-1) \in \mathbb{Z} \Rightarrow g(n) \in \mathbb{Z} \Rightarrow f(n+\alpha) \in \mathbb{Z}\)