What can we say about $m_A(x)$ with respect to $m_{A^2}(x)$ when $A$ is diagonalizable? Suppose $A$ is a real $n\times n$ matrix and diagonalizable over $R$.
Is one of the following propositions is true? 
(1) $m_{A^2}(x)$ divides $m_A(x)$
(2) $m_A(x)$ divides $m_{A^2}(x)$
I think that both of the propositions above are false because we don't know if the polynomial  $m_{A^2}(x)$ annihilates A. 
Also, I think that if A was complex matrix or if the exponent of A was different (i.e different natural number) the result would be the same (both of the propositions were false).
Am I correct?
 A: For a simple counterexample consider $A= 2 I_n$. The minimal polynomial is $X-2$, while $A^2= 4I_n$ and the minimal polynomial is $X-4$. 
More generally, let $\lambda_1, \dots, \lambda_k$ denote the (distinct) eigenvalues of $A$. 
Since $A$ is diagonalisable, the eigenvalues of $A^2$ are $\lambda_1^2, \dots, \lambda_k^2$ (not necessarily distinct anymore). 
Thus the roots of the polynomials do not necessarily coincide, and in general there is also no containement relation (either way).  Showing again that in general neither of the divisibility conditions holds.  
(As an aside note that containement of these sets is all that is relevant for divisibility since the minimal polynomials of diagonalizable matrices only have simple roots.)
In fact these sets hardly ever coincide or are contained in each other. Indeed, one can derive that one has  $m_A(x)$ divides $m_{A^2}(x)$ if an only if $A= A^2$. 
And, the converse divisibility relation can be characterized by saying that eigenvalues of $A$ are contained in $\{0,1,-1\}$. 
You are also correct in assuming that this would not change much for other powers or considering complex numbers, except for the fact that you would get the last divisibilty under the considition that the nonzero eigenvalues are appropriate roots of unity.
