An old and interesting problem in combinatorics from Russia Mathematics Olympiad

Can the numbers from $1$ to $81$ be written on a $9 \times 9$ board, so that the sum of the numbers in each $3\times 3$ square is the same?

I believe I have not made much progress and am missing the key insight here. Any advice?

Solution 1:

I believe if you take the following matrix:

$$\left[\begin{array}{ccccccccc}0&3&6&0&3&6&0&3&6\\1&4&7&1&4&7&1&4&7\\2&5&8&2&5&8&2&5&8\\3&6&0&3&6&0&3&6&0\\4&7&1&4&7&1&4&7&1\\5&8&2&5&8&2&5&8&2\\6&0&3&6&0&3&6&0&3\\7&1&4&7&1&4&7&1&4\\8&2&5&8&2&5&8&2&5\end{array}\right]$$

it already satisfies the conditions, and so does the transpose $A^T$:

$$A^T=\left[\begin{array}{ccccccccc}0&1&2&3&4&5&6&7&8\\etc.\end{array}\right]$$

and so does an affine combination $B=A+9A^T+1$, where "$1$" is the matrix filled with "all ones". One should verify that the resulting matrix $B$ has all the different values from $1$ to $81$. It begins something like this:

$$B=\left[\begin{array}{ccccccccc}1&13&25&28&40&52&55&67&79\\etc.\end{array}\right]$$