Question:
A $5×5$ square is divided into 25 unit squares. One of the numbers $1,2,3,4,5$ is inserted into each of the unit squares in such away that each row, each column and each of the two diagonals contains each of the five numbers once and only once. The sum of the numbers in the four squares immediately below the diagonal from top left to bottom right is called the score. Show that it is impossible for the score to be 20. What is the highest possible score?
What have I done so far:
The only possible way to reach 20 in the row immediately below the diagonal is by adding four 5s, i.e., $5+5+5+5=20$. It follows that all the squares in those particular four squares immediately below the diagonal must be filled with 5. We show that this is impossible given the statement of the problem.
Proof
Assume for the sake of contradiction that such a score is possible. We call two squares adjacent if they share any common edge. It it given in the statement of the problem that the diagonals contain all the given integers exactly once. Hence it follows that the diagonal above our problem set of squares must contain exactly one five. It also follows that at least two squares containing 5s each must be adjacent to each other in a row or a column. Hence it is impossible to contain four 5s in the given set of squares. Going back, we conclude that it is impossible for the score to be 20.
The part where I am stuck is finding the maximum possible score. I am guessing that all the integers in the four squares have to be distinct but I cannot seem to prove it.
Thank you.