Silver matrix properties of order n

Question

A square matrix of order n is a silver matrix if any i-th row and the i-th column contain all the elements from 1 to 2 − 1. Show that there is no silver matrix of order 1997. I can see by trial that it works for even n cases upto 4 but can't prove it. Please help me.

Answer 1

Here's an elaboration of the parity argument.

There are $2n-1$ different numbers in the array, and only $n$ entries on the diagonal, so if $n>1$ there must be some number $x$ that doesn't occur on the diagonal. (Note that if $n=1$ the $1\times1$ matrix whose only entry is $1$ is a silver matrix.)

For $i=1,2,\dots,n$, $x$ occurs once in the $i$th row and $i$th column. That is, there is a $1\leq j\leq n$ with $j\neq i$ such that exactly one of $a_{ij}=x$ and $a_{ji}=x$ is true. Let us assign $j$ to $i$. The same procedure assigns $i$ to $j$, for there cannot be a $k\neq i$ such that $a_{jk}=x$ or $a_{kj}=x$.

This procedure groups the numbers $1,2,\dots,n$ into pairs, so $n$ is even.