Determinant of $A+A^T$ is an odd integer if $\text{det}(A-A^T)=1$. Let $A\in\text{Mat}(2n\times 2n;\mathbb{Z})$ be an integer matrix such that $\text{det}(A-A^T)=1$. I want to show that $\text{det}(A+A^T)$ is an odd integer. Murasugi claims in his book "Knot Theory and its Applications" that this is trivial and it probably follows immeditiately from some determinant property of skew-symmetric resp. symmetric matrices. But I just can not proof it. Any help would be greatly appreciated!
 A: The point is that, so long as we enforce zeros on the diagonal, then skew-symmetric and symmetric matrices coincide in characteristic 2.
More formally, consider the ring homomorphism $\phi:\mathbb Z\longrightarrow R\cong \mathbb F_2$
then define $\Phi$ to apply $\phi$ component-wise to the matrix $B\in \mathbb Z^{m\times m}$.  Then
$\phi\Big(\det\big(B\big)\Big)=\det\Big(\Phi\big(B\big)\Big)$
and with $B:=A+A^T$ we have
$\phi\Big(\det\big(A+A^T\big)\Big)=\det\Big(\Phi\big(A+A^T\big)\Big)=\det\Big(\Phi\big(A-A^T\big)\Big)=\phi\Big(\det\big(A-A^T\big)\Big)=\phi(1)=1$
so $\det\big(A+A^T\big)\%2 =1$, i.e. the determinant is odd.
A: Given a matrix $A_{2(n+1)\times 2(n+1)}\in \mathrm{Mat}(2(n+1)\times 2(n+1),\mathbb{Z})$ there is matrices
$$
\begin{array}{c c c}
A_{2n \times 2n}\in \mathrm{Mat}(2n\times 2n,\mathbb{Z}), 
& \quad &
B_{2n \times 2}\in \mathrm{Mat}(2n\times 2,\mathbb{Z}),
\\ && \\
C_{2 \times 2n}\in \mathrm{Mat}(2 \times 2n,\mathbb{Z}),
& \quad  &
D_{2\times 2} \in  \mathrm{Mat}(2 \times 2,\mathbb{Z})
\end{array}
$$
such that
$$
A_{2(n+1)\times 2(n+1)}
=
\left\lgroup
\begin{array}{cc}
A_{2n \times 2n}  & B_{2n\times 2}
\\
 C_{2 \times 2n } & D_{2 \times 2 }
\end{array}
\right \rgroup.
$$
Now you just use induction and the formula
$$
\det\begin{pmatrix}A & B \\ C & D\end{pmatrix} = \det(A) \det\left(D - C A^{-1} B\right).
$$
to proof that
$$
\det\big( A_{_{2(n+1)\times 2(n+1)}} + A_{_{2(n+1)\times 2(n+1)}}^T \big)
\\
=\det\begin{pmatrix}
A_{_{2n\times 2n}} + A_{_{2n\times 2n}}^{T} 
& 
B_{_{2n\times 2}}^{}+C_{_{2\times 2n}}^{T}
\\
C_{_{2\times 2n}}^{}+B_{_{2n\times 2}}^{T}
&
2 D_{_{2\times 2}}
\end{pmatrix}
\\
=\det
\begin{pmatrix}
A_{_{2n\times 2n}} + A_{_{2n\times 2n}}^{T} 
\end{pmatrix}
\det\Big(
2D_{_{2\times 2}} - (C_{_{2\times 2n}}^{}+B_{_{2n\times 2}}^{T})(A_{_{2n\times 2n}} + A_{_{2n\times 2n}}^{T})^{-1}(B_{_{2n\times 2}}^{}+C_{_{2\times 2n}}^{T}) 
\Big)
$$
is equal to $1$.
