I'm curious to how other people solved Q5. I'm pretty sure that the correct approach involved a proof by contradiction in 2 cases: Case 1 = 2 corners on the same side have the same colour and Case 2 = 2 corners on the same diagonal have the same colour
I was fairly short on time for this question so I only got the first case done (in a kind of rushed but valid way)
I think, thinking about the problem now the easiest way to reach a contradiction is to first observe that we must have 4 squares of each colour since we can apply the pigeon-hole principal on the 4 non-overlapping 2x2 squares. Then in each case you can show if the 2 corners have the same colour then the grid will have 5 of one colour.
In the paper in the first case i made the observation that each square must have a different colour to each adjacent and diagonal square. Then I WLOG chose the colour of all the the squares on the side where the corners were the same colour e.g R B G R and then deduced that the 2 squares below the "R" and "G" must both be Y but this is a contradiction since they are adjacent. Then i ran out of time for case 2...