Rational matrix having roots of every degree As the result of another question, now deleted, I am interested in the following problem. 
Problem. Let $A\in M_n(\mathbb Q)$ be an invertible matrix with the property that the equation $X^k=A$ has solutions (in $M_n(\mathbb Q)$) for any $k\ge 1$. Prove all eigenvalues of $A$ are equal to $1$. (This is the proposed problem 398 at page 36 of this journal.)
Remarks. 1. The question is the $\mathbb Q$-version of a well known contest problem for matrices in $M_n(\mathbb Z)$. (In that case the conclusion is stronger, and we have $A=I_n$.)
2. The converse also holds.
I do not know a proof of the main problem, nor do I know of a proof of the remarks. Remarks on any of these would be appreciated.  
 A: From the following, you can deduce the required result. See also REMARK at the end of the post.
Prop. 1. Let $A\in M_n(\mathbb{Q})$ s.t. there is an increasing sequence of integers $(k_i)_i$ s.t. for every $i$, there exists $B_i\in M_n(\mathbb{Q})$ satisfying $B_i^{k_i}=A$. Then $A$ is similar to $diag(0_p,L_{n-p})$ s.t. there is $r$ satisfying $L^r=I+N$ where $N$ is nilpotent.
Def: Let $u$ be a non-zero algebraic number with minimal polynomial $a_0\Pi_{k=1}^n(X-u_k)$. Its Mahler measure (cf. http://en.wikipedia.org/wiki/Mahler_measure ) is $M(u)=|a_0|\Pi_{k=1}^n\max\{1,|u_k|\}$ ; its height is $H(u)=M(u)^{1/n}$. Then $M(u)\geq 1$ and $M(u)=1$ iff $u$ is a root of unity. Moreover $H(u^k)=H(u)^k$. 
Lemma (due to Dobrowski): If $u$ is not a root of unity, then $M(u)\geq 1+\dfrac{1}{(20\log(n))^3}$.
Prop. 2. Let $\omega$ be a non-zero algebraic number of degree $\leq n$ s.t. there is an increasing sequence of integers $(k_i)_i$ s.t. for every $i$, there is an algebraic $\alpha_i$ of $degree \leq n$ satisfying $\alpha_i^{k_i}=\omega$. Then $\omega$ is a root of unity.
Proof . For every $i$, $H(\omega)^{1/k_i}=H(\alpha_i)$. Then $M(\alpha_i)\leq H(\omega)^{n/k_i}<$ (for $i$ great enough) $1+\dfrac{1}{(20\log(n))^3}$. According to Lemma, such $\alpha_i$ is a root of unity and $\omega$ too. $\square$
Proof of Prop 1. Part 1. Let $\lambda$ be a non-zero eigenvalue of $A$ ; it is algebraic of degree $\leq n$. For every $i$, there is an eigenvalue $\mu$ of $B_i$ s.t. $\mu^{k_i}=\lambda$ and $\mu$ is algebraic of degree $\leq n$. According to Prop 2, $\lambda$ is a root of unity. 
Part 2. Then $A$ is similar to $diag(M,U)$ where $M$ is nilpotent and $U^r=I+N$ for some $r$. There is $k\geq n$ s.t. $B^k=A$. Since $AB=BA$, $B=diag(P,V)$ ; since $P^k=M$, $M=0$. $\square$
REMARK 1. There is a complete answer to the required question in Posts 4 and 7 (by grobber) in http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=909877333&t=42444
REMARK 2. There is also a complete answer (in french) in the French review: p.180-184,2,Janvier 2007, RMS.
