Find all real matrices $A$ such that $A^2 = \mathrm{diag}(1,1,2,3,5,8,13)$ Let $A \in \mathcal{M}_{7 \times 7} (\mathbb{R})$ such that 
$$A^2= \begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&2&0&0&0&0 \\ 0&0&0&3&0&0&0 \\ 0&0&0&0&5&0&0 \\ 0&0&0&0&0&8&0 \\ 0&0&0&0&0&0&13\end{pmatrix} $$
How many matrices, which satisfy this condition could you find?
My friend told to me, that the correct answer is infinity. But I can find only one:
$$A^2= \begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&\sqrt{2}&0&0&0&0 \\ 0&0&0&\sqrt{3}&0&0&0 \\ 0&0&0&0&\sqrt{5}&0&0 \\ 0&0&0&0&0&\sqrt{8}&0 \\ 0&0&0&0&0&0&\sqrt{13} \end{pmatrix} $$
I will grateful for hints, how find any other examples. I suppose that $A$ should depend from parametr which don't govern on product $A \cdot A$.
 A: Hint: $$\begin{pmatrix} \cos \varphi & \sin \varphi & 0 & \ldots & 0 \\ \sin \varphi & -\cos \varphi & 0 & \ldots & 0 \\ 0 &  0& 0& \ldots &  0\\ \vdots &  \vdots& \vdots& \ddots &  \vdots\\ 0 &  0& 0& \ldots &  0 \end{pmatrix}^2 = ?$$
A: We have $$\begin{pmatrix} \cos \varphi & \sin \varphi & 0 & \ldots & 0 \\ \sin \varphi & -\cos \varphi & 0 & \ldots & 0 \\ 0 &  0& 0& \ldots &  0\\ \vdots &  \vdots& \vdots& \ddots &  \vdots\\ 0 &  0& 0& \ldots &  0 \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 & 0 & \ldots & 0 \\ 0 & 1 & 0 & \ldots & 0 \\ 0 &  0& 0& \ldots &  0\\ \vdots &  \vdots& \vdots& \ddots &  \vdots\\ 0 &  0& 0& \ldots &  0 \end{pmatrix}$$
Additionaly I noticed that 
$$\begin{pmatrix} \cos \varphi & \sin \varphi & 0 \\ \sin \varphi & -\cos \varphi & 0 \\ 0 &  0& x \end{pmatrix} ^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 &  0& x^2 \end{pmatrix} $$ 
so I suppose that we can take 
$$A= \begin{pmatrix} \cos \varphi&\sin \varphi &0&0&0&0&0\\\sin \varphi&-\cos \varphi&0&0&0&0&0\\0&0&\sqrt{2}&0&0&0&0 \\ 0&0&0&\sqrt{3}&0&0&0 \\ 0&0&0&0&\sqrt{5}&0&0 \\ 0&0&0&0&0&\sqrt{8}&0 \\ 0&0&0&0&0&0&\sqrt{13} \end{pmatrix} $$
Of course, $\varphi$ is any so we have infinity matrices. 
