if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$ Prove or disprove:

if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$.

I'm pretty confident this is not true, but I've tried and tried to find a counter example without success. If someone contradicts this, I'd appreciate if you can outline your thought process when constructing the matrix $A$.
Also, another question in the same batch, asks:

Let $p(t)=t(t-0.25)(t-1)$ be the characteristic polynomial of $A^2$, is $A$ diagonalizable?

I'm thinking this is suppose to answer the previous question, assuming the answer here is false.
Here I know $A^2$ is diagonalizable, but I haven't made any substantial progress other than that.
I know I can go from the eigenvalues of $A$ to the eigenvalues of $A^k$, but not the other way around. 
 A: Try $A=\begin{pmatrix} 0&0&1 \\ 0&0&0 \\ 0&0&0\end{pmatrix}$. The idea is that the null matrix is diagonalizable, but there exist matrices which satisfy $A^2=0$ and $A$ is not diagonalizable.
For the second question, note that the eigenvalues of $A^2$ are distinct, and therefore the eigenvalues of $A$ are distinct, which implies that $A$ is diagonalizable.
If $P_A$ denotes the minimal polynomial of $A$ then $P_{A^2}(X^2)$ is an annulating polynomial for $A$. This means that $P_A|P_{A^2}(X^2)$. If $A^2$ is diagonalizable then $P_{A^2}$ has simple roots. If zero is not  root of $P_{A^2}$ then all roots of $P_{A^2}(X^2)$ are simple so all roots of $P_A$ are simple and $A$ is diagonizable. (here we need to be sure that the roots of $P_A$ are all in the field of elements of $A$; if we work on $\Bbb{C}$ then we are fine)
So if zero is not an eigenvalue of $A$ then $A^2$ diagonizable implies $A$ diagonizable.
A: The answer to the second problem is yes.  In general, the (in general complex) eigenvalues of $A^2$ are the squares of the eigenvalues of $A$.  In this case $A^2$ has the eigenvalues $0$, $\frac{1}{4}$, and $1$.  Therefore $0$ is an eigenvalue of $A$, either $\frac{1}{2}$ or $-\frac{1}{2}$ is an eigenvalue of $A$, and either $1$ or $-1$ is an eigenvalue of $A$.  This means that $A$ has $3$ distinct real eigenvalues, which implies it is diagonalizable.  The key is that the eigenvalues of $A^2$ are distinct and nonnegative, which implies that the eigenvalues of $A$ are distinct and real.
Edit: I accidentally overlapped somewhat with Beni Bogosel's earlier answer.  (More was added after I started writing.)
A: Try $A=\begin{pmatrix} 0&-1&0 \\ 1&0&0 \\ 0&0&1\end{pmatrix}$.
I found it by considering the complex matrix $\begin{pmatrix} 1&0&0 \\ 0&i&0 \\ 0&0&-i\end{pmatrix}$ and finding its characteristic polynomial, then using rational canonical form over $\mathbb{R}$.
A: To the second question: yes. If all eigenvalues of $A^2$ are pairwise distinct, then so are the eigenvalues of $A$, because if $Av=\lambda v$ then $A^2 v = \lambda ^2 v$, so $\ker(A-\lambda I) \subset \ker(A^2 - \lambda^2 I)$.
A matrix with pairwise distinct eigenvalues is  diagonalizable. That's a simple and well known theorem.
