Self-Referential Square IV Solution

As there is a 'path' between A and B, we must have A = 1 and B = 5.

Consider the clues on the bottom row. We have two numbers which sum to a third number. As they are all on the same row, they must all be distinct and between 1 and 5 exclusive. Hence the two numbers must be 2 and 3, and so the sum is 5. Hence D = 5. By the Latin square property, C = 4.

The clue which says that the cell at Row D, Column C is 2 means that the order of the squares in the box can be detemined to be on the order 3 then 2.

Note that E cannot be four, for if it were then only one of the squares (N,N) is not N. But in that case E is not, and we already know that (3,3) cannot be 3. Therefore E is not four. But there must be a 4 either at E or immediately left or right of it in order to make a path from A to B. Thus, given that there is already a 4 in the first column, it must be in the third column. Therefore the cell left to E is in fact 4.

Consider the box in the second row. As each number in it is odd, it can sum to either 1+3 = 4, 1+5 = 6 or 3+5 = 8. It is equal in value to the box in the last column. However, this cannot possibly equal 8, for the five is already used in that column. In addition, it could not equal 4 if the other box does, for that can only be made by 1+3, and by the Latin Square property they cannot both be made of those sums. Therefore the box on row 2 equals 6 and has the items 1 and 5.

The vertical box therefore also equals 6. It cannot also be 1 and 5, so it must be 2 and 4. However, the two in the 4th row must be near A in order to make a path, and so we conclude the ordering is as follows:

Thus Row 1, Column 5 is 3. The box on row one equals 10, and therefore the other numbers must be 1 and 4. The order is obvious due to the 4 on Row 2, and also the square in column one can be filled in using the Latin Square property:

Filling in the second row using the Latin Square property is then easy. In particular, E = 2. It follows that the cell in the first column and second row must have value 2.

Now consider the path from A to B. One of the first three cells on the forth row must be 2. The first and second cannot now, and hence the cell in the third column is. Therefore the centre square is a 5.

Consider the first column. There must be a five in it, but this can only occur in row 4. Hence, by the Latin square property we have the cell in the third row of the same column as a 1. What is more, the fourth column can now be filled in.

The rest follows easily, and the grid ends up as follows: