## A Magic Difference Square

Place the numbers 1 through 9 into a 3x3 grid, such that in each of the 8 directions of 3 squares (horizontally, vertically and diagonally) the sum of the first and last numbers minus the center number gives the same result. How many different answers exist?

Source: Original, based on a puzzle in Michael Holt's Math Puzzles and Games, Vol. 1, Page 43.

Solutions were received from Denis Borris, Al Zimmermann, Philippe Fondanaiche, Joseph DiVincentis, Walkovers@aol.com, and Sandy Thompson.

Disregarding rotations and reflections, there is only one solution:

```   2 1 4
3 5 7
6 9 8```

Denis Borris sent this analysis and extension:
```Hmmm...melikes those: the kind you can do on the corner
of a Ronald MacDonald's napkin! Fill an XandO design:
A B C
D E F
G H I
Since A+I = B+H = C+G = D+F
Then 4(A + I) = 45 - E ; A + I = (45 - E) / 4
Also, 2E = A + I (by combining other equalities).

Now, after finishing your Big Mac, you have the
possibilities:  E   A+I
1    11
5    10
9     9
So combos are: 9,1  8,2  7,3  6,4 (since 2E = A + I)

Draw a couple more XandO designs, fiddle around to get,
around the middle of your 99 cents caramel sundae:
2 1 4
3 5 7
6 9 8
which is the ONLY solution; but which may be upsidedowned
and bedroomceilingmirrored into 7 other "appearances".

Similarly (but impossible on a napkin!), you can do this
with a 5by5, numbers 1 to 25:
A B C D E
F G H I J
K L M N O (hey: we have M in Middle!)
P Q R S T
U V W X Y
...(A+B-C+D+E) = (F+G-H+I+J) = and so on...

Don't feel like typing too much; arranging/combining
equalities will lead to: 4M = A + G + S + Y;
so we get a M = 13, and all 12 combos = 39.
Here's one solution:
25  6 17 21  4   (25+6-17+21+4 = 39)
7  1  9 22 18
15 11 13 14 12
19 23 10  2  5
3 20 16  8 24
```

Mail to Ken