the following is a question from a qualifying exam for abstract algebra:
If $A$ is an $n$ by $n$ matrix with entries in $\mathbb{Z}$ and odd determinant then show for some positive number $k$, all entries of the matrix $A^k - I$ are even. Hint: work over the finite field $\mathbb{Z}/(2)$
My first effort was to try to use the characteristic polynomial somehow, but I unfortunately could not get what I wanted, so I attempted a different approach.
Working over $\mathbb{Z}/(2)$, since det($A$) is odd, det($A) \equiv 1 \mod2$.
So we can consider $A$ as an element of $GL(n, \mathbb{Z}/(2))$, the general linear group over the finite field $\mathbb{Z}/(2)$.
$GL(n,F_q)$ has order $\displaystyle \Pi_{i=1}^n (q^n - q^{i-1})$, so in our case
$$|GL(n, \mathbb{Z}/(2))| = \Pi_{i=1}^n (2^n - 2^{i-1})$$
The most important part is that $GL(n, \mathbb{Z}/(2))$ has finite order, and for $A \in GL(n, \mathbb{Z}/(2))$, $|A|$ divides $|GL(n, \mathbb{Z}/(2))|$, so $A$ has finite order.
Thus there exists some positive $k$ so that $A^k = I$.
So $A^k - I = 0$ $(\mod 2$), meaning that $A^k -I$ has all even entries.
Any and all feedback on this proof would be very much appreciated!
Your proof works perfectly well! It would be slightly quicker, however, to simply take $k = |GL(n,\Bbb Z_2)|$, noting only that this group is finite (there is no need to explicitly state its order).