Let $A$ is a symmetric matrix with all diagonal element zero.
How to show that the dot product of $x$ with $Ax$ is divisible by $2$ ?
I can only verify for matrix of order (size) $2, 3, 4 $ but not sure about general $n×n$ matrix.
Edit: Here, $x$ is any vector in $\Bbb{Z}^n$.
If $A$ is real symmetric matrix with all diagonal elements $0$, then prove that $x^{T}Ax$ is divisible by $2$. ( Here $x$ is any vector in $\mathbb{Z}^{n}$, and $x^T$ denotes transpose of $x$ )
$2$nd edit: If $A$ is integer entry symmetric matrix with all diagonal elements $0$, then prove that $x^{T}Ax$ is divisible by $2$.
Let $A=(a_{ij})$ be a symmetric matrix with $a_{ii}=0$.
Then by symmetry $a_{ij}=a_{ji}$
Let $Q_A:\Bbb{R}^n\to \Bbb{R}$ is the quadratic form associated with $A$.
$\begin{align}Q(x) =x^TAx&=\sum_{i,j}a_{ij}x_ix_j\\&=2\sum_{i\neq j, i>j}a_{ij}x_ix_j\end{align}$
If $A\in M_n(\Bbb{Z})$ and $x\in \Bbb{Z}^n$ , then clearly $2\mid Q(x) $
Let $A=\begin{pmatrix}0&{\frac{1}{2}}\\ {\frac{1}{2}}&0\end{pmatrix}$
Then $\begin{align}Q(x) =x^TAx&= \frac{1}{2}x_1x_2+\frac{1}{2}x_2x_1\\&=x_1x_2\end{align}$
Clearly $2\mid Q(x) $ for all $x\in \Bbb{R^2}\setminus \{0\}$ is false.
$A=\begin{pmatrix}0&1\\ {1}&0\end{pmatrix}$
Then $Q(x) =2x_1x_2$
Let $x=\begin{pmatrix}\frac{1}{2}\\\frac{1}{2}\end{pmatrix}$
Then $Q(x) =\frac{1}{2}$ and $2\nmid Q(x) $.
Hence $2\mid Q(x) $ if $a_{ij},x_i\in \Bbb{Z} $