Magic Square With Minimal Information

Claim: In a $3 \times 3$ magic square, the magic sum is equal to thrice the central square.

Proof: Let the numbers in the square be $a, b, c, d, e, f, g, h, i$ from left to right, top to bottom. Let the magic sum be S. We have $S = a + e + i, S = b + e + h, S = c + e + g$ . Also, we have $S = a + b + c$ and $S = g + h + i$ . Summing up the first 3 equations and subtracting the last 2, we get $S = 3 e$ .

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In this problem, since the magic sum is $S = 5 + 0 + 1 = 6$ , so
the center center square is $\frac{6}{3} = 2$ ,
the lower right square is $6 - 5 - 2 = -1$ ,
the lower center square is $6 - 1 - (-1) = 6$ ,
the upper center square is $6 - 6 - 2 = - 2$ .

Note: We can show that the upper right square is $6 - 5 - (-2) = 3$ and the center right square is $6 - 0 - 2 = 4$ .

Let the elements of the magic square be $a, b, c, d, e, f$ .

The sum of every row, column, and diagonal is 6. We can write these sums as a system of linear equations. Let's look at the row sums first:

$\begin{array}{ccc} 5 + a + b = 6 & & b = 1 - a \\ 0 + c + d = 6 & \implies & d = 6 - c\\ 1 + e + f = 6 & & f = 5 - e \\ \end{array}$

We can eliminate $b, d,$ and $f$ from the magic square by writing them in terms of $a, c,$ and $e$ .

The sum of second column is 6. We get $a + c + e = 6$ . Therefore, $e = 6 -a - c$ . We can eliminate $e$ from the magic square by writing it in terms of $a$ and $c$ .

The sum of the diagonals is also 6.

$\begin{array}{rcc} 5 + c + (a + c - 1) = 6 & \implies & a + 2c = 2\\ (1- a) + c + 1 = 6 & & -a + c = 4 \end{array}$

When we add the two equations, we get $3c = 6$ . Substituting back $c=2$ in the equation gives us $a = -2$ . When we substitute values of $a$ and $c$ in the magic square, we get

The number that goes in the yellow square is $-2$ .

This is a consistent solution since the sum of each row, column and diagonal is $6$ . This is also a unique solution as the solving the linear equation gave us this solution only.