Problem source

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 source

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 $g(0), g(1), g(-1) \in \mathbb{Z} \Rightarrow g(n) \in \mathbb{Z} \Rightarrow f(n+\alpha) \in \mathbb{Z}$