Magiked Square

Looking at where to place the 4 in the diagonal going from the lower left to the upper right box, the 4 in the lower right box negates the possibility of putting a 4 in the lower left or upper right box. Since the 1 and 2 already take up two of the three remaining positions, there is only one place left to put the 4, namely in the blue box. To be complete, we now need to see if such a 5 x 5 grid can be completed. Indeed, it can, uniquely:

$\Large 1... 2... 3... 4... 5$

$\Large 5... 3... 4... 1... 2$

$\Large 4... 5... 2... 3... 1$

$\Large 2... 4... 1... 5... 3$

$\Large 3... 1... 5... 2... 4$

The answer is therefore $\boxed{\Large 4}$ .

Consider the main right diagonal , 1 , 2 are already there so the blue box may contain 4 or 3 or 5.

From the figure , we see that 4 is there in the rightmost corner at the bottom so 4 cannot be placed at the two ends of the diagonal(considering the rule to place it in a box) . And so the only place left for 4 is the blue box.

No need to solve the grid.

It is simply that if anything but 4 goes in the square, the 4 will be duplicated at either corner, and has no place else to go.

All the same here is a solved grid.