This question is from RMO (Regional Mathematical Olympiad), the second round of selection to the IMO in India.
The 64 squares of an 8×8 chessboard are filled with positive integers in such a way that each integer is the average of the integers on the neighbouring squares. Two squares are neighbours if they share a common edge or a common vertex. Thus a square can have 8, 5 or 3 neighbours depending on its position. Show that in fact all the 64 entries are equal.
On
Choose the largest integer. Because this is the average of the integers on the neighboring squares, and it is the largest, we must have all the neighboring squares having the same number. It is possible to reach any one square from every other square by traveling from square to adjacent square, so we can conclude every square must have this largest integer; i.e. that they are all equal.
There must be a minimal values as they are integers. If the numbers weren't all equal there would be one square with minimal value that was adjecent to a number that is not minimal. This would mean that the mean of the neighbouring squares would be larger than the value at the square.