Represent the numbers in the 9 boxes by letters A, B, ..., I.
A+B+C=D+E+F=G+H+I=-2; A+B+C+D+E+F+G+H+I=-6.
The central box E is involved in four sums if the magic square is a true one (rows, columns and diagonals all sum to -2):
A+E+I=B+E+H=C+E+G=-2; A+I=B+H=C+G=-(E+2);
I=-(A+E+2), H=-(B+E+2), G=-(C+E+2). These equations reduce the number of variables by 3.
Other squares are involved in at most 3 sums, some of which have already been noted. The two below haven't already been listed.
A+D+G=C+F+I=-2; A+D-(C+E+2)=C+F-(A+E+2)=-2.
-6=A+B+C+D+E+F-(C+E+2)-(B+E+2)-(A+E+2)=D-2E+F-6, so D+F=2E, or E is the average of D and F.
We can replace E by (D+F)/2. This reduces the number of variables to 5: A, B, C, D, F.
C=-(A+B+2); F=-(D+E+2)=-(D+(D+F)/2+2)=-(3D/2+F/2+2), 2F=-(3D+F+4), F=-(D+4/3). The remaining unknowns are A, B and D.
However, this implies fractional values because of the fraction 4/3. If D is an integer F will be a fraction; if F is an integer, D will be a fraction.
One solution (involving fractions) is:
| -4/3 |
2 |
-8/3 |
| -2 |
-2/3 |
2/3 |
| 4/3 |
-10/3 |
0 |
There are other solutions but they involve fractions, too. The above argument makes no assumptions about the relations between the variables (such as consecutive numbers, or no duplicates), but logically deduces that some of the numbers have to be fractions because of the fact that F=-(D+4/3).
Please note that if the given sum had been -3 instead of -2, there would have been an integer solution, for example:
Every magic square can be transformed by interchanging its rows and columns, tipping it (rotating) to lie on a different side or reflecting it, to produce other magic squares, and you can add, subtract, multiply or divide each number in the boxes by a constant to produce other magic squares, and there are different ways to create magic squares. But it's provable from the above logic that you can't produce a 3×3 magic square with fixed sum -2 for rows, columns and diagonals without introducing fractions.
