Let $\alpha$ be a solution of a monic quadratic polynomial with integer entries and $|\alpha|=1$.Then prove that $\alpha^{12}=1$. ($a$) Let $\alpha$ be a solution of a monic quadratic polynomial with integer entries and $|\alpha|=1$.Then prove that $\alpha^{12}=1$.
($b$)Let $A \in M_2(\mathbb{Z})$ such that $A^n=I$ for some $n$ then show that $A^{12}=I$

completely stuck on it.can I get some help?
 A: Since $\alpha$ is a zero of the polynomial we ca write the polynomial as
$$P(x)=(x-\alpha)(x-\beta)$$
for some $\beta\in\mathbb C$. Now either $\alpha$ is real, then $\alpha=\pm 1$ and we are done, or it is complex $\alpha=e^{i\phi}$ for some $\phi$. Then $\beta=e^{-i\phi}$, because the complex zeros of a real polynomial come in conjugate pairs. Now the coefficients of $P$ are integers, thus $\alpha+\beta=2\cos(\phi)=p$ for some $p\in \mathbb Z$.
Thus $$\phi=\arccos(\frac p2)\in\{\frac \pi3,\frac{2\pi}{3},\pi,0,\frac\pi2\}+2\pi\mathbb Z$$
In any case $$\alpha^{12}=e^{12i\phi}=1$$
For the second question: The characteristic polynomial of $A$ satisfies the conditions on the polynomial in part a). The (complex) eigenvalues are solutions of this polynomial and since $A^n=I$, the matrix $A$ is diagonalisable over $\mathbb C$  (This is clear if the eigenvalues are distinct and easy to see if they coincide). In particular the $n^{th}$ power of the eigenvalues is 1, which implies $|\alpha|=1$. Finally $A$ is (up to base-change) a diagonal matrix where the entries saitsfy $\alpha^{12}=\beta^{12}=1$.
A: 1) If $\,\alpha\in\Bbb R\;$ then $\,\alpha=\pm1\implies \alpha^{12}=1\,$ and we're done, otherwise:
2) $\;\alpha\in\Bbb C-\Bbb R\implies\;$ also $\,\bar\alpha\,$ is a root of the same monic integer polynomial $\,x^2+bx+c\in\Bbb Z[x]\,$ .
But then
$$\alpha+\bar\alpha=2\text{Re}(\alpha)=-b\;,\;\;\alpha\bar\alpha=c=|\alpha|^2=1\implies$$
Writing $\,\alpha=e^{ix}\;,\;x\in\Bbb R\,$ , we get:
$$-1\le\cos x=-\frac b2\le1\implies -2\le -b\le 2\le \implies-2\le b\le 2\implies$$
$$\implies b\in\{\,-2,\,-1,\,0,\,1,\,2\,\}\implies\cos x\in\left\{\;0\,,\frac12\,,\,1\;\right\} \implies $$
$$\implies x\in\pm\left\{\;0\,,\,\frac\pi3\,,\,\frac\pi2\,,\,\frac{2\pi}3\;\right\}\ldots\ldots$$
Well, now finish the proof.
