Let $n$ be a positive odd integer and let $A$ be a symmetric $n\times n$ matrix of integer entries such that $a_{ii}=0,i=1,2.....n$. Show that the determinant of $A$ is even.
I tried using definition of determinant. Also I can't use induction since the problem is for odd integers. Please give some thoughts on how to solve it.
The basic trick is to note that if $\sigma\in S_n$, the complete permutation group on $n$ elements, and $\sigma^2=1$, then there is a fixed point of $\sigma$. That's only true if $n$ is odd.
So this means that in the definition:
$$\det A = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) a_{1\sigma(1)}\cdots a_{n\sigma(n)}$$
the terms with $\sigma^2=1$ contribute zero, since $a_{ii}=0$, and the term from other $\sigma$ can be paired with th term from $\sigma^{-1}$. Those terms are equal by the symmetry of the matrix and the sign of a permutation being equal to the sign of its inverse.
(Technically, you don't need to know the signs are the same, since if they are different, they still contribute an even number together, $0$.)