I am working through Computing the Continuous Discretely and they give the definition of a reflexive polytope as $$ P=\{x \in \mathbb{R}^d : Ax \leq 1\}$$ where all entries from $A$ are integers.
It's left as an exercise to the reader to show, that $$ P^{\mathrm{o}} \cap \mathbb{Z}^d = \{0\} \text{ and } (t+1)P^{\mathrm{o}} \cap \mathbb{Z}^d=tP \cap \mathbb{Z}^d$$ where t is a natural number, follows from the given definition.
I would appreciate any help.
Obeserve that $P$ is full-dimensional (otherwise it would not make sense to talk about the interior of $P$ in $\mathbb{R}^d$) as it contains a small ball around the origin.
Since $A$ is integral, we have $Az\leq 0$ for $z\in P^{\mathrm{o}}\cap\mathbb{Z}^d$. Note that $C=\lbrace x\in\mathbb{R}^d : Ax\leq 0\rbrace$ defines a polyhedral cone. The boundedness of $P$ combined with $C\subseteq P$ implies that the cone is bounded itself. This can only be the case if $C=\lbrace 0\rbrace$. Thus, $z = 0$.
For the second claim, we observe that $(t+1)P^{\mathrm{o}} = \lbrace x\in \mathbb{R}^d : Ax < (t+1)1\rbrace$. Again, we have $Az\leq t1$ for $z\in (t+1)P^{\mathrm{o}}\cap \mathbb{Z}^d$ which shows $z\in tP\cap\mathbb{Z}^d$. The reverse inclusion follows from $tP\subseteq (t+1)P^{\mathrm{o}}$.