TITLE: Matrix with even integers entries doesn't have odd eigenvalue
QUESTION [5 upvotes]: Let $A \in M_n(\mathbb{Z})$ with even entries. Prove that $A$ doesn't have odd eigenvalue.

REPLY [6 votes]: We would like to show, in other words, that $A - kI$ is invertible for any odd integer $k$.  To see that this is the case, we compute $\det(A - k I)$ modulo $2$.  Notably, the determinant is a polynomial on the entries of a matrix, which means that the determinant defines a function over $M_n(\Bbb F_2)$.  That is, it holds that
$$
A \equiv B \pmod 2 \implies \det(A) \equiv \det(B) \pmod 2.
$$
Because the entries of $A$ are even, we have $A - k I \equiv -kI \equiv I \pmod 2$.  Thus, 
$$
\det(A - kI) \equiv \det(I) \equiv 1 \pmod 2.
$$
That is, $\det(A - kI)$ is odd and thus non-zero.  So, $(A - k I)$ has a non-zero determinant and is thus invertible.

Another proof: let $B = A/2$.  We note that by Gauss's lemma, any eigenvalue of $B$ is either an integer or irrational.  
Now if $\lambda$ is an eigenvalue of $A$, then $\lambda = 2 \mu$ for some eigenvalue $\mu$ of $B$.  If $\mu$ is irrational, then $\lambda = 2 \mu$ is also irrational and hence not an integer.  If $\mu$ is an integer, then $\lambda = 2\mu$ is even.  In either case, $\lambda$ cannot be an odd integer.