There are many ways of approaching this problem. Below is one of them:
We start by looking at the fourth row. Since we have five cells and five letters none of which can be repeated, we should use each letter exactly once. There is no B on the fourth row, so we should have one. Since B is in the middle of the square, and no letter should be repeated on the diagonals, none of the green squares in the figure on the right should be B. Therefore, the only square on the
fourth row that should be B is the last one. So we complete it with B:
On the same row we should also have A and E. Since we do have an E on the fourth column, the fourth square on fourth row should be completed with A, and the second one with E. So the diagram becomes:
We should also have an E on the colored diagonal. But we cannot have it on any of the two red squares since we already have one on the last row and on the second column. So E should be in the pink box.
Now we look at the other diagonal. We should have an A on that diagonal. We cannot have it in the red squares as we already have an A on each of the last two columns, therefore the A should be in the pink cell:
Now on the first column, in the remaining two cells should be C and B. We cannot have B in the third row as already have a B there, therefore B should be in the second cell of the first row. We then have C in the third one:
On the last row, we should have a B. It cannot be on the third or last column as we already have a B on both of them, therefore B should be in the second cell of the last row. Similarly, C cannot be on the third column, so it should be on the last one. The remaining cell on the last row should then be completed with D:
We can now complete the required cell. We look at the diagonal it is on. On that diagonal, we already have E, B, A, and C. Therefore, we should complete the cell with D:
If we continue to complete the remaining squares in the same way, we get:
This problem is taken from the UKMT Mathematical Challenges.