I have encountered the following linear algebra/number theory question in my work (low-dimensional topology), so I thought I should ask the experts.

Let $A \in SL(n,\mathbb{Z})$ be a matrix , $n \geq 3$. One can easily show that there is some $j \in \mathbb{N}$ such that $Tr(A^j) \geq 3$ (proof given at the end). My question is that how long does it take for this to happen?

**Question 1** Given $n \geq 3$, is there a number $k \in \mathbb{N}$ (depending only on n) such that for all $A \in SL(n,\mathbb{Z})$ there is some $1 \leq i \leq k$ such that $Tr(A^i) \geq 3$?

**Question 2**: If the answer to the above question is yes, then how $k$ depends on $n$?

Proof of the claimed fact: Assume the contrary that $Tr(A^i) \leq 2$ for all natural numbers $i$. Let $\lambda_1 , ... , \lambda_n$ be the eigenvalues of $A$, therefore $T_i := Tr(A^i) = \lambda_1 ^i + ... + \lambda_n ^i$. Since $T_i$ are bounded above and $\lambda_1 ... \lambda_n =1$ then we should have $|\lambda_i| = 1$ for each $1 \leq i \leq n$. This implies that all $\lambda_1 , ..., \lambda_n$ are roots of unity (because all of their Galois conjugates are on the unit circle) and therefore for some $j \in \mathbb{N}$ we have $\lambda_1 ^j = ... = \lambda_n ^j = 1$. This implies $T_j \geq n \geq 3$. Contradiction.

PS: Replace the assumption $A \in SL(n,\mathbb{Z})$ by $A \in Sp(2n,\mathbb{Z})$, if it makes things easier.

3more comments