Can a 4 by 4 magic square
be completed with the numbers 1 through 16 for entries?

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My Solutions.

I first need to determine my target sum. The sum of all the values 1 through 16 is 136. Dividing this result gives 34, which is my target sum for each row, column, and diagonal.

I then make an array of the numbers 1 through 16:

Examining the sums, I find that only the sums of the diagonals are 34:

I try to balance the results by inverting the second and fourth rows:

This action changed the sums of the diagonals, but it yields a sum of 34 for all columns:

I need to balance my rows. I notice that the sums of the rows progressively increase from the first to the fourth. All of the lower values are in the first two rows, so I try balancing by inverting the second and third columns:

The result is now that everything but the diagonals add up to 34:

So I need to focus on the diagonals:

&

The sum of the first diagonal is 38 (over by 4) and the sum of the second is 30 (under by 4). I look for values that I can substitute into both diagonals but that will not affect the results of the rows and columns.

In the second row, I have 11 and 8 - a difference of 3 - and 8 is currently in the first diagonal. If I can switch the 8 and the 11, I can reduce the sum of the first diagonal from 38 to 35.

In the third row, I have 9 and 6 - difference of 3 - and 6 is currently in the second diagonal. If I can switch the 6 and the 9, I can increase the sum of the second diagonal from 30 to 33.

The sum of 11 and 6 (which are both in the first column) is 17, and the sum of 8 and 9 (which are both in the third column) is also 17. I can then replace the 11 and 6 into the third column and the 8 and 9 into the first column without affect the sums of those columns:

So I now have:

So I need to reduce the sum of the first diagonal by 1 and increase the sum of the second by 1.

By examining the array, I see that I can exchange the 10 and 7 from the second column with the 11 and 6 in the first:

And the result is a 4x4 magic square: