Problem. Magic Square (click here to see the problem statement)

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My Solution (Basic Square).

I can use only the whole numbers 1 through 9. I start by making a 3x3 array of those numbers:

I need to get an idea of how the sum are interacting, so I add up each row, column, and diagonal. I find that the middle row, the middle column, and both diagonals sum to 15:

The sum of the first row is 6 and the sum of the third row is 24. I need to balance the results, so I try switching the 2 and the 8 in the middle cloumn:

I try to balance the first and third columns more by switching the 4 and 6 in the midle row:

As 5 is in the middle square and I am trying to achieve a sum of 15 in all cases, I need to maintain pairs on the "outer ring" that each add to 10: (1,9) (2,8) (3,7) (4,6) and that are in opposite positions on the outer ring. I try rotating the outer ring counter-clockwise:

I luck out on my very first rotation. I try again:

then

then

then

And the next rotation will bring the ring full-circle.

So it appears that my solutions are:

I now need to consider possibilities other than rotations, so I can examine Reflections. For example, I can hold the middle column constant and switch the numbers in the the first and third columns:

If I define this array to be my base,

then a reflection about the middle column is:

Similarly, a reflection about the middle row
means that I hold the middle row constant
and switch the numbers in the first and third rows:

So my reflection is:

And reflections about each diagonal:

gives

gives

Last, I need to examine the possibility of placing numbers other than 5 in the middle square, such as 8

I can fairly quickly determine that this is not an option, because I will have 9 somewhere in the outer ring. As the sum of 8 & 9 is 17, I will have a row or column greater than 15. I can immediatley extend this to rule out having 9 in the middle square:

-&-

For 7 in the middle square, I need the sum of all pairs of remaining numbers to equal 8; however in my set
{1, 2, 3, 4, 5, 6, 8, 9},
I already have 8 and 9, so there will be at least one pair whose sum is breater than 8.

For 6 in the middle square, the sum of all remaining pairs must equal 9;
however my set contains 9, so there is at least one pair whose sum is greater than 9.

For 4, 3, 2, or 1 in the middle square, the sum of all pairs must equal 11, 12, 13, and 14, respectively;
however in each set we have 1, 2, 3, and 4 requiring the matches for each to be 10 (or greater),
none of which is included in the set.