IMO 1997 Problem 4

An $ n \times n$ matrix whose entries come from the set $ S = \{1, 2, \ldots , 2n - 1\}$ is called a silver matrix if, for each $ i = 1, 2, \ldots , n$, the $ i$-th row and the $ i$-th column together contain all elements of $ S$. Show that:

(a)  there is no silver matrix for $ n =1997$;

(b)  silver matrices exist for infinitely many values of $ n$.

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(a) Every number in $[2n-1]$ appears exactly $n$ times when we count the $i$-th column and row for every $i\in[n]$. There are at least $2n-1-n=n-1$ numbers not on the main diagonal, and each of them appears exactly $n/2$ times in the matrix. Hence when $n$ is odd no silver matrix exists.

(b) Given an $n\times n$ silver matrix $A_n$, we build an $2n\times2n$ silver matrix $$A_{2n}=\begin{pmatrix}A_n & B \\ C & A_n\end{pmatrix},$$where $B$ and $C$ are $n\times n$ latin squares with entries from $\{2n,2n+1,\dots,3n-1\}$ and $\{3n,3n+1,\dots,4n-1\}$, respectively. Hence, silver matrix exists for every $n$ that is a power of two.