If $A$ is a $2\times2$ integer matrix with $\det(A)=1$ and $|\text{tr}(A)|>2$, then $A^n\neq I$. 
If $A \in \Bbb Z^{2\times 2}$ with $\det(A)=1$ and $\left| \text{tr} (A)\right|>2$, then $A^n\neq I$ for all $n\in \mathbb{N}$.


I tried to prove this by induction and by contradiction, but I'm quite a bit lost here because it seems that it all reduces to obtain a contradiction about the numbers. I don't see a way around that. Any suggestions?
 A: The condition that $\det(A)=1$ is redundant. Let $\lambda_1$ and $\lambda_2$ be the two eigenvalues of $A$ over $\mathbb C$. Since $2<|\operatorname{tr}(A)|=|\lambda_1+\lambda_2|\le|\lambda_1|+|\lambda_2|$, we have $|\lambda_i|>1$ for some $i$. For every $n\in\mathbb N$, as $\lambda_i^n$ is an eigenvalue of $A^n$ and $|\lambda_i^n|=|\lambda_i|^n>1$, $A^n$ cannot possibly be equal to $I$.
A: Elementary proof without eigenvalues
$$
A^n_{11} + A^n_{22} = A^{n-1}_{11}A_{11} + A^{n-1}_{12}A_{21} + A^{n-1}_{21}A_{12} + A^{n-1}_{22}A_{22}\\ = (A_{11}^{n-1} + A_{22}^{n-1})(A_{11} + A_{22}) - (A_{11}^{n-2} + A_{22}^{n-2}) (A_{11}A_{22} - A_{12}A_{21})
$$
which can be verified easily by expansion. This is basically : $tr(A^n) = tr(A^{n-1})tr(A) - tr(A^{n-2})\det(A)$.
We now argue by induction , that $|tr(A^n)| > |tr(A^{n-1})|$ for all $n > 0$. Clearly, for $n=1$ it is given as $|tr(A)| > |tr(A^0)| = tr(I) = 2$.  Now, in general we have $$|tr(A^n)| = |tr(A^{n-1})tr(A) - tr(A^{n-2})| \geq 2|tr(A^{n-1})| - |tr(A^{n-2})| \geq |tr(A^{n-1})|$$
which tells us that $|tr(A^n)| > 2$ for all $n$. There's no way then that $A^n = I$ for any $n$, for $I$ is a matrix with trace exactly $2$.

EDIT : The identity concerning the trace and determinant comes from the well known analogue for the eigenvalues. In particular, with multiplicity if $\alpha,\beta$ are the eigenvalues, then $$
\alpha^n + \beta^n = (\alpha^{n-1} + \beta^{n-1})(\beta + \alpha) - (\alpha^{n-2} + \beta^{n-2})\alpha \beta
$$
can be easily verified. Once the link between the eigenvalues of the power of a matrix, and the link between the trace , determinant and eigenvalues is made clear, this reduces to the identity I verified by expansion.
A: Here is a different take that works over $\mathbb Q$.
Suppose $A^n=I$. Then the possible irreducible factors for the minimal polynomial of $A$ are the cyclotomic polynomials of degree at most $2$:
$$
x-1,\quad x+1,\quad x^2+x+1,\quad x^2+1,\quad x^2-x+1
$$
Therefore, the possible characteristic polynomials are
$$
(x-1)^2,\quad (x+1)^2,\quad (x-1)(x+1),\quad x^2+x+1,\quad x^2+1,\quad x^2-x+1
$$
The absolute values of the coefficients of the $x$ term in these polynomials are all at most $2$ and so $|\text{tr}(A)|\le 2$.
A: If $A^n=I$ for some $n$, $A$ could only have $-1,1$ as eigenvalues. Moreover, since $\det(A) = \lambda_1 \lambda_2 = 1$, the eigenvalues would be both $1$ or both $-1$. This is however impossible, since $|\mbox{tr}(A)| = |\lambda_1 +\lambda_2| >2$.
