Prove $X=0$, if $\det(I+pX)=1$ and $(I+pX)^n=I$, Question:

Let the matrix $X=(a_{ij})_{2\times 2},a_{ij}\in Z,$, and $p> 2,p\neq 4, p\in \mathbb{Z}$, 
$(1):$
  such that 

$\det(I+pX)=1$. 

$(2):$Suppose there exists positive integer $n$ such $(I+pX)^n=I$,
Show that: $X=0$.

My try: let
$$X=\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}
$$
where $a,b,c,d\in Z$
since $X\in M_{2}(Z)$,and $\det(I+pX)=1$, 
then we have
$$\begin{vmatrix}
pa+1&pb\\
pc&pd+1
\end{vmatrix}=1
$$
then
$$(pa+1)(pd+1)-p^2bc=1$$
$$p^2ad+pa+pd+1=p^2bc+1 \Longrightarrow pad+a+d-pbc=0$$
$$\Longrightarrow p(ad-bc)+a+d=0$$
and since there exist $n$ such that

$$(I+pX)^n=I$$
  $$\Longrightarrow
\begin{bmatrix}
pa+1&pb\\
pc&pd+1
\end{bmatrix}^n=\begin{bmatrix}
1&0\\
0&1\\
\end{bmatrix}$$
  and Then I can't prove $a=b=c=d=0$.

maybe this problem use other methods?
Thank you  Salman post his solution,But this is wrong,because 
$$I + p\binom{n}{2}X + \ldots + p^{n-1}\binom{n}{n-1}X^{n-1} + p^n X^n = I.$$
we can't have this 
Taking determinants of our equation, we get:
$$1 + \binom{n}{2}(a + d) + \ldots + \binom{n}{n-1}(a+d)^{n-1} + (a + d)^n = 1.$$
because 
$$det(A_{1}+A_{2}+\cdots+A_{n})\neq det(A_{1})+det(A_{2})+\cdots+det(A_{n})$$
Thank you very much!
 A: Let $A = I + pX$. Then $A$ is an integer matrix satisfying $A^n = I$, i.e. an integer matrix with finite order. This necessarily forces the minimal polynomial to be a product of distinct cyclotomic polynomials.
Given that $A$ is $2\times 2$, one can easily narrow the minimal polynomial down to one of the following:
$$\mu_A(x) = \begin{cases}\Phi_1(x) = x-1
\\ \Phi_2(x) = x+1
\\ \Phi_1(x)\Phi_2(x) = x^2 - 1
\\ \Phi_3(x) = x^2 + x + 1
\\ \Phi_4(x) = x^2 + 1
\\ \Phi_6(x) = x^2 - x + 1\end{cases}$$
The remaining is rather tedious, but straight forward.


*

*The first case with $\mu_A = \Phi_1$ trivially forces $X=0$. 

*The second case with $\mu_A = \Phi_2$ gives
$$A+I = 2I + pX = 0\implies 2I=-pX$$
This requires $p \le 2$ which doesn't work out. 

*If we have $\mu_A = \Phi_1\Phi_2$ then the determinant is $-1$. No good.

*If we have $\mu_A = \Phi_3$ then we get
$$(I + pX)^2 + A + I = 2I + 2pX + p^2X^2 + A = 3A + p^2X^2=0$$
This gives us
$$3A = -p^2X^2 \implies 9 = p^4(\det X)^2$$
This is impossible since $\det X$ is an integer and $p > 2$. 

*Continuing, we have $\mu_A = \Phi_4$. Then we have
$$(I + pX)^2 + I = 2I + 2pX + p^2X^2 = 2A + p^2X^2 = 0$$
Again taking determinants, we end up with $4 = p^4 (\det X)^2$. Again, this is impossible. 

*Finally, with $\mu_A = \Phi_6$ we have
$$(I+pX)^2 - A + I = 2I + 2pX + p^2X^2 - A = A + p^2X^2=0$$
which comes down to the determinant equation $1 = p^4(\det X)^2$. Again, impossible.
This shows that any matrix $A$ satisfying the hypotheses of the problem necessarily has minimal polynomial $x - 1$, i.e. $A = I$. It appears that the restriction $p\neq 4$ isn't necessary, although $p>2$ is necessary to prevent the case $p=2$ and $X = -I$.
A: Hint: The eigenvalue of a matrix in the group $SL_2(\mathbb{Z})$ of finite order must be one of the following: $\pm 1$, $\pm i$, $\pm 1/2\pm\sqrt{3}/2 i$.
