Question is to prove that
Determinant of a symmetric matrix of odd degree with integer entries and zeros in the diagonal is even..
For $A=(a_{ij})$ with $a_{ii}=0$ and $a_{ij}=a_{ji}$..
Determinant is given by $$\sum_{\sigma}sgn(\sigma)\prod a_{i\sigma(i)}$$
Now, we can neglect $\sigma$ that fixes at least one element in $\{1,2\dots,n\}$ as we would then have $a_{ii}$ in that product which is zero..
Let $S$ be the collection of $\sigma$ that is remaining...
Let $\sigma\in S$ then $sgn(\sigma)=sgn(\sigma^{-1})$
$$\prod a_{i\sigma(i)}=\prod a_{\sigma(i)i}=\prod a_{j\sigma^{-1}(j)}$$
For pair $\sigma,\sigma^{-1}$ the product turns out to be same..
As $n$ is odd, there can be no element of order $2$ in $S$ so, $\sigma\neq \sigma^{-1}$
Suppose $\sigma$ is an element of order $2$.. We assume that $\sigma$ does not fix any element..
Start with $1$.. $\sigma(1)\neq 1$ we can suppose $\sigma(1)=2$.. As $\sigma^2=Id$ we should have $\sigma(2)=1$.. So, $\sigma=(1~2)\tau$
Start with $3$ $\sigma(3)\neq 3$ we can suppose $\sigma(3)=4$.. As $\sigma^2=Id$ we should have $\sigma(4)=3$.. So, $\sigma=(1~2)(3~4)\tau'$
Repeating this, we end up with $\sigma=(1~2)\cdots(2m-1~2m)(2m+1)$ a contradiction as we have assumed that $\sigma$ does not fix any element but here we have no option..
So, if $n$ is odd and $\sigma$ does not fix any element then $\sigma^2\neq Id$.. So, $\sigma\neq \sigma^{-1}$
So, sum has same number twice.. Thus determinant is even...
Help me to clear this...
If we change the sign of any member of matrix the deterninant doesnt change modulo 2. So replace all $a_{ij}, i<j$ by $-a_{ij}$ and receive skew-simmetric matrix of odd order which determinant is even of course - because it becomes zero.