Magic Squares

by Gilbert Keith

Say you have 9 numbers which form a magic square. If you’re not familiar with magic squares, they’re essentially an n x n grid with the property that the sum of every row and column, as well as both diagonals, is the same number. You know the numbers are in arithmetic progression; i.e. the 9 numbers {a1, a2, a3, … a9} are of the form {x0 + d, x0+2d, x0+3d…,x0+9d}.

Show that the central square must always be occupied by the 5th term in the sequence.

My solution: It’s a little wordy, but hopefully it’s clear.

We know that the numbers take the form {x0+d, x0+2d, x0+3d… x0+9d}. Since each cell in the magic square has an x0 in it, we can ignore it for now and add it back in the end.

So now we are dealt with allocating the numbers {d,2d,3d,…,9d} to a magic square.

We see that each number in the set is a multiple of d. We’re dealing with magic squares, here, where the sums of rows and columns, and diagonals must add up. Since multiplication is a distributive property, we can do the addition first and then do the multiplication.

i.e., we can divide the series by d first and multiply it back (prior to the addition of x0) in after we are done allocating the sequence of integers.

But now our sequence boils down to {1,2,3,4,5,6,7,8,9}

It has been show that there exists only one unique magic square which contains those numbers as their entries, and the central cell of that square contains 5. At this point, you can multiply each cell by d, and then add x0 to each cell.

Gautam Kandlikar

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