Self Referential Number Square III Solution

Let the label the squares as following:

First we will determine the possible values of the green, blue and red squares. Consider the following table:

Of course, Green and Blue must be different, as must Blue and Red. This rules out rows 1, 2 and 3. Thus green is either 4 or 5, blue either 2 or 3 and red either 2 or 4. In particular, this means that E5 is not 2 and one of either B2 and C3 is not 2. Therefore there are at most three 2's on the down-right diagonal. Given that the clue implies there is an even number, this means there are two 2's, and thus it follows that D4 is 1. Hence D1 is 4 and A1 is 2. It follows that A2 is 1. The grid now looks as follows:

E5 must be equal to 1, as there is nowhere else it could go in Row 5. Therefore E1 is 5. Now, we must decide if B1 is 1 or 3. Let us first suppose it is 1, and so C1 = 3, and then C3 = 1. Since the right-down diagonal has two 2's, it follows that B2 = 2. This means that the first set of numbers for the coloured squares are correct, and thus B4 = 4, B5 = 3 and C5 = 2. The state of play looks as follows:

Obviously then we can use the Latin Square property to give B3 = 5, A5 = 4, D5 = 5, A4 = 5, A3 = 3, E4 = 2. And then there is no value that can go in C4! (Alternatively other contradictions can be obtained without any extra insight.)

Therefore the assumption that B1 = 1 was incorrect, and thus B1 = 3. It follows that B3 = 1 and also that the second set of values are correct for the coloured square: B4 = 5, B5 = 2, C5 = 4. Hence the board looks as follows:

Using the Latin Square property we get B2 = 4. Due to there being two 2s in the down-right diagonal, we must have C3 = 2. Using Latin Square again, we get C2 = 5, C4 = 3, E4 = 2, A4 = 4, D2 = 2, E2 = 3, E3 = 4.

As this point there are only four squares remaining blank, as shown below:

Recall the clue that states the sum of one diagonal is one off a multiple of the sum of the other diagonal. The sum of the bottom-right diagonal is 10. The sum of the top-right diagonal is 14+n, where n is the value of A5, and can either be 3 or 5. If n = 3, then we have the values of 10 and 17, which does not fulfil the property. However, if n = 5 then the values are 10 and 19, which do have this property.

Thus A5 = 5 and the solution follows.