1. The problem statement, all variables and given/known data A square is divided into 81 smaller squares by lines parallel to its sides. The numbers 1, 2, ..., 81 are entered in an arbitrary fashion, one in each square. Show that, however the numbers are entered, it is possible to find two small squares with an edge in common whose entries differ by more than 5. 2. Relevant equations 3. The attempt at a solution (Proof by contradiction) Assume the numbers 1 to 81 have been entered so that any two adjacent squares have entries by at most 5. Some of the non-adjacent squares have a common adjacent square. Therefore, no squares diagonally opposite from each would have a difference of more than 10. However, all squares have at least two adjacent squares, so the difference of diagonally opposite squares cannot be more than 9 i.e. 1/5 6/11 will not work, but 1/5/9 6/10 will. That's as far as I've gotten. Do i just continue doing that? Or is there another way?