Number of magic squares with magic constant 0?

Question

How can we determine the number of magic squares with magic constant 0?

Answer 1

There are an infinite number of such squares

Take the example square below:

-3  2  1
 4  0 -4
-1 -2  3

To generate a new square, simply multiply each element of this square by any positive integer. As there are an infinite number of positive integers, there are an infinite number of possible squares.

Answer 2

If you add a number $k$ to every element of a magic square, you get another magic square whose magic constant is the original constant plus $3k$. In particular, if you set $k$ to $-\frac 13$ of the original constant, you get a magic square whose constant is $0$. Thus every magic square induces one whose constant is $0$.

Now, for a square with constant $0$, if we name the following elements, $$\begin{matrix}x & y & \cdot\\\cdot & z & \cdot\\\cdot & \cdot & \cdot\end{matrix}$$ we can start filling in the other elements from the requirement all columns, rows and diagonals sum to $0$.

$$\begin{matrix}x & y & -x-y\\\cdot & z & \cdot\\\color{red}{w} & -y-z & -x-z\end{matrix}$$ With $w$, we run into an impasse: to make the diagonal $0$, we must have $w = x + y - z$, and to make the bottom row $0$, we must have $w = x + y + 2z$. These can both be realized only if $z = 0$. With that change, we can complete the pattern: $$\begin{matrix}x & y & -x-y\\-2x -y & 0 & 2x+y\\x + y & -y & -x\end{matrix}$$ By the definition quoted by Ian MacDonald, any $x$ and $y$ would work. But I've always understood that the numbers in a magic square should also be distinct (Wikipedia agrees). If we add that condition then we have a list of restrictions on the values of $x$ and $y$: $$x \ne 0, y \ne 0, y \ne kx\text{ for }k \in \{1, -\frac 12, -1, -\frac 32, -2, -3\}$$

If we do not consider rotations/reflections as separate squares, then these restrictions allow only one magic square that consists of consecutive integers, the one that frodoskywalker has already produced: $$\begin{matrix}1 & 2 & -3\\-4 & 0 & 4\\3 & -2 & -1\end{matrix}$$

If you do consider rotations and reflections as separate squares, then there are 8 possible squares, depending on which of the 4 corners the $1$ is in, and whether the $2$ is clockwise or counter-clockwise from it.

If you allow non-consecutive integers, or non-integers, then there are infinitely many solutions, even if you don't allow multiples like frodoskywalker mentions. Just choose a non-zero $x$, then choose any value for $y$ other than the 7 that are restricted (and also relatively prime to $x$, if you want to avoid multiples of other squares).

Addendum: why there is only one magic square with consecutive integers.

Any pattern with $x < 0$ is a rotation of $180^\circ$ from one with positive $x$, so we can assume $x > 0$. Further, if $|y| < x$, then $-x < y$, so $x < 2x + y = |-2x - y|$. So by flipping the square if needed, we can also assume $|y| > x$.

Now, the sum of all the numbers in a magic square is the sum of the values of all three rows, and therefore is 3 times the magic number, which in this case is $0$. The only 9 consecutive integers that sum up to $0$ are those between $-4$ and $4$. So all the elements in our square need to be in this range. In particular $|2x + y| \le 4$. Consider the cases:

• $x = 1$: $y \ne -2, -3$ by the restrictions, and $2x + y < 4$ gives $y \le 2$. So $y = 2$ and $y = -4$ are the only choices, and both give the same square.
• $x = 2$: $y \ne -1, -2, -3, -4$ by the restrictions, and $2x + y < 4$ gives $y < 0$, so there is no solution.
• $x = 3$: Since $|y| > x$, the only possibilities are $-4$ and $4$, but $2\cdot 1 + 4 > 4$, so it must be that $y = -4$, which gives the same pattern again.
• $x = 4$: $|y| > x$ cannot be satisfied.

Which exhausts all possibilities.