find $a,b,c$ such that $e^{2\pi A}=I$ 
Given $A$ is real valued matrix of the form $\begin{pmatrix}a&b\\c&0\end{pmatrix}$ find all $a,b,c$ such that $e^{2\pi A}=I$

My attempt is the following, Calculating the eigenvalues I get $\frac{a\pm\sqrt{a^2+4bc}}{2}$ and I said that $a=0$ otherwise we won't get the desired result and $bc<0$ to get the right form of diagonlize matrix of $A$ and afterwards that $bc\in\mathbb{Z}$ but not sure if I've done this right
 A: The eigenvalues should be in the form $\{ki|k\in\Bbb{Z}\}$. $a$ can't be imaginary, so it must be 0. So the eigenvalues are $\pm\sqrt{bc}$. So $\sqrt{bc} \in \{ki|k\in\Bbb{Z}\}$. You are right that it works for $\sqrt{bc} < 0$ but it works also for $\sqrt{bc}=0$. Therefore $bc \in \{-k^2|k \in \Bbb{Z}\}$.
There is just one exception for $bc=0$. If only $b$ or only $c$ is zero, there will be a 2×2 Jordan block, so the exponential of this matrix wouldn't be the identity matrix. Only if both $b$ and $c$ are zero, it works.
A: Since you've already gotten to
$$
A=\begin{bmatrix}0&b\\c&0\end{bmatrix}\tag1
$$
where $bc=-\lambda^2\lt0$, let's compute...
$$
\begin{align}
A^2
&=\begin{bmatrix}bc&0\\0&bc\end{bmatrix}\tag{2a}\\[6pt]
&=-\lambda^2I\tag{2b}
\end{align}
$$
Thus, $A^{2n}=(-1)^n\lambda^{2n}I$ and $A^{2n+1}=(-1)^n\lambda^{2n}A$. Therefore,
$$
\begin{align}
e^{2\pi A}
&=\sum_{n=0}^\infty\left(\frac{(2\pi A)^{2n}}{(2n)!}+\frac{(2\pi A)^{2n+1}}{(2n+1)!}\right)\tag{3a}\\
&=\sum_{n=0}^\infty\left(\frac{(-1)^n(2\pi\lambda)^{2n}}{(2n)!}I+\frac{(-1)^n(2\pi\lambda)^{2n+1}}{\lambda(2n+1)!}A\right)\tag{3b}\\
&=\cos(2\pi\lambda)I+\frac1\lambda\sin(2\pi\lambda)A\tag{3c}
\end{align}
$$
So $bc=-\lambda^2$ where $\lambda\in\mathbb{Z}$, except for $\lambda=0$ because of the division by $0$ (which actually gives $I+2\pi A$, which works if $A=0$).
A: You already got $a=0$ and $bc=-\lambda^2$. Suppose $bc\neq0$. Note that
$$ x(t)=e^{tA}x_0 $$
is the solution to $$x'=Ax $$
or
$$ x_1'=bx_2,x_2'=cx_1,x_1(0)=x^0_1. \tag 1$$
From (1), one has
$$ x_1=c_1\cos(\lambda t)+c_2\sin(\lambda t),x_2=\frac{1}{b}\bigg[-c_1\lambda\sin(\lambda t)+c_2\lambda\cos(\lambda t)\bigg] $$
and hence
$$ x(t)=\binom{x_1}{x_2}=\left(\begin{matrix}\cos(\lambda t)&\sin(\lambda t)\\
-\frac{1}{b}\lambda\sin(\lambda t)&\frac{\lambda}{b}\cos(\lambda t)\end{matrix}\right)\binom{c_1}{c_2}.$$
So
$$ e^{At}=\left(\begin{matrix}\cos(\lambda t)&\sin(\lambda t)\\
-\frac{1}{b}\lambda\sin(\lambda t)&\frac{\lambda}{b}\cos(\lambda t)\end{matrix}\right). $$
Since $e^{2\pi A}=I$, one has
$$ \sin(2\pi\lambda)=0, \cos(2\pi\lambda)=1,b=\lambda $$
which implies
$$ \lambda=k, b=k, c=-k, k\in N. $$
