Rational matrices whose powers have bounded denominators Let $d\ge 1$ be a natural number. Let $A$ be a square $d\times d$ matrix with rational entries :  $A \in M_{d}(\mathbb{Q})$.  The following statements about the matrix $A$ are equivalent

*

*The sequence of powers $(A^n)_{n\ge 1}$  has bounded denominators.


*The characteristic equation of $A$ has integral coefficients.


*$A$ is conjugated ( over $\mathbb{Q}$) to a matrix with integral entries.
Notes:
I've discovered    stumbled upon this property while playing with linear recurrences -- it may well be a classical thing.

*

*& 3. can be generalized, but then the test 2. is still elusive


*$\rightarrow$ 3.  is canonical forms (done right).


*$\rightarrow$ 1.$\wedge$ 2 is easy
Below is my proof, ignore it if you want to solve the problem on your own.
Thank you for your interest!
$\bf{Added:}$ The part 2. $\rightarrow$ 3. appeared before on this site.
$\bf{Added:}$  Indebted to Ewan Delanoy for the case $A$ a companion matrix
 A: Here is first 1. $\rightarrow$ 3.
A more general fact. Consider $\mathfrak{S}$ a semigroup of matrices from $M_d(\mathbb{Q})$  ( closed under multiplication) with bounded denominators ( there exists $0\ne D \in \mathbb{N}$ such that $D \cdot S \in M_d(\mathbb{Z})$ for all $S \in \mathfrak{S}$).  Then there exists $T \in GL(n, \mathbb{Q})$ such that
$T \cdot \mathfrak{S} \cdot T^{-1} \subset M_d(\mathbb{Z})$.
Indeed, consider the set of integral vectors
$$L \colon = \{ v\ \in \mathbb{Z}^d \ | \ \mathfrak{S} \cdot v \subset \mathbb{Z^d} \}$$
It is easy to see that $L$ is a subgroup, and moreover $D \cdot \mathbb{Z}^d \subset L \subset \mathbb{Z}^d$.  We conclude ( invariant factors for e.g.) that $L$ is a lattice. Moreover, $\mathfrak{S} L \subset L$. Indeed, we have
$$\mathfrak{S} ( \mathfrak{S} L) \subset \mathfrak{S} L \subset \mathbb{Z}^d$$
Now consider a basis of $L$. In this basis all the entries of the elements of $\mathfrak{S}$ are integral.
(this is just like the proof of the standard fact: a rational representation of a finite group is equivalent to an integral representation -- see W. Burnside, Theory of groups of finite  order)
Let's prove 1. $\rightarrow$ 2.  It is enough to show the following: consider $P(x)= x^d + c_1 x^{d-1} + \cdots + c_d$ the characteristic polynomial of $A$.  Assume that not all of the coefficients $c_k$ are integral. Let $p$ be a prime number such that for some $k$ we have for the $p$-adic valuation we have $v_p(c_k) < 0$. Below is the proof that uses some algebraic number theory.
Consider $K$ a number field containing all of the roots of $P(x)$. Let $v$ an extension to $K$ of the $p$-adic valuation of $\mathbb{Q}$. Now, from the Vieta relations we have $v(x_i) < 0$ for some $i$. Consider $x_1$,  $\ldots$, $x_l$ of smallest $v$. Let now $P_n(x)$ be the characteristic polynomial of $A^n$. It has the roots $x_i^n$, $1\le i \le d$. Write
$$P_n(x) = x^d + c_{n,1} x^{d-1} + \cdots + x_{n,d} $$
Now from Vieta relation it is easy to see that
$$v(c_{n,l}) = n \cdot v_(c_l)$$
and this will approach $-\infty$ as $n\to \infty$. This implies that the denominators of $A^n$ approach infinity as $n\to \infty$.
I will only say that 2. $\rightarrow$ 3. uses canonical forms (over $\mathbb{Q}$) and the fact that all of the factor of a monic integral polynomial are also integral.
$\bf{Added:}$  For 2. $\rightarrow$ 3.  Consider a matrix $A$ in $M_d(K)$ with an annihilating polynomial $P(x)$. Then $A$ is similar to a block diagonal matrix formed of companion matrices of factors of $P(x)$.
The statement would work for any $R$ integrally closed domain instead of $\mathbb{Z}$.
