evert wrote:An 8- or 9-number proof looks like a nice challenge.

This should not be too difficult to do I think...

The basic idea would be to first list all valid patterns of 9 cells up to equivalence. Although there are a huge number of choices for 9-out-of-81 cells, I think the requirement that no "chute" has 2 empty rows or columns will reduce this dramatically. Then take only one representative for each equivalence class, and use that as the underlying pattern.

Then partition those 9 cells in all possible ways into 8 or 9 parts - this is trivial in that either all 9 numbers appear, one per cell, or only 8 numbers appear, in which case there are 36 choices for two cells sharing a number, and the remaining 7 cells collect one number each.

Then it will be trivial to show that each of these configurations has either zero or more than one solution.

The only "unknown" here is that I have not tried to estimate how many valid patterns there are, but I can't imagine it is too big...

(Now I wait for Guenter to prove me wrong by showing it is too large...)

Gordon