[Math] Proof of Magic Constant Formula

magic square

A magic square is a NxN square grid filled with distinct positive integers in the range 1,2…$N^2$.
Each cell contains a different integer.
The sum of the integers in each row, column and diagonal is equal and called magic constant.
The Formula for magic constant is given by

Magic Constant=$\frac{N*(N^2+1)}{2}$

How to derive this formula?

Best Answer

Yes. The sum in any row must be $\frac{1}{N}$ times the sum of all the entries in the matrix (because there are $N$ rows of equal sum, and the sum of all rows' sums is equal to the sum of all entries).

Note that the sum of all the entries in the matrix is $$1+2+\cdots+N^2 = \frac{N^2 \left(N^2+1\right)}{2}.$$ Hence the Magic Constant is $$\frac{1}{N}\times \frac{N^2 \left(N^2+1\right)}{2} =\frac{N\left(N^2+1\right)}{2} .$$

Related Solutions

[Math] The sum of integers in each row / column / major diagonal of a Magic Square of order $N$ is $\dfrac{N(N^2+1)}{2}$

Add all the numbers in the square. It's the sum from 1 to $n^2$, which is $n^2(n^2+1)/2$. This is equal to the sum of $n$ rows, so each row is $n(n^2+1)/2$.

Magic Square – Possible Sums of an n x n Magic Square

I don't know about the 4x4 case, but here are some things we can do for the 5x5 case.

Start with a typical 5x5 magic square (sum = 65):

\begin{array}{|c|c|c|c|c|} \hline 17 & 24 & 1 & 8 & 15\\ \hline 23 & 5 & 7 & 14 & 16 \\ \hline 4 & 6 & 13 & 20 & 22 \\ \hline 10 & 12 & 19 & 21 & 3 \\ \hline 11 & 18 & 25 & 2 & 9 \\ \hline \end{array}

Multiply all entries by 2 (sum = 130):

\begin{array}{|c|c|c|c|c|} \hline 34 & 48 & 2 & 16 & 30\\ \hline 46 & 10 & 14 & 28 & 32 \\ \hline 8 & 12 & 26 & 40 & 44 \\ \hline 20 & 24 & 38 & 42 & 6 \\ \hline 22 & 36 & 50 & 4 & 18 \\ \hline \end{array}

Add 1 to selected cells (sum = 131)

\begin{array}{|c|c|c|c|c|} \hline 34 & \color{red}{49} & 2 & 16 & 30\\ \hline \color{red}{47} & 10 & 14 & 28 & 32 \\ \hline 8 & 12 & \color{red}{27} & 40 & 44 \\ \hline 20 & 24 & 38 & 42 & \color{red}{7} \\ \hline 22 & 36 & 50 & \color{red}{5} & 18 \\ \hline \end{array}

OK, so this immediately shows that the sum need not be a multiple of n.

Obviously, I could have added 3 to those cells as well, so then the sum would be 133. I can't add 2, for then I would get duplicate entries. However, I can get a sum of 132 by adding 67 to these 5 cells in the original square. In fact, I could just stick to the original square and add any number greater than 25 to those 5 cells to get any sum of 90 or more. And by picking those 5 cells a little more carefully (or by choosing a different original 5x5 square), we can go even lower than that, though 65 is of course the minimum ... if we stick to positive numbers.

Assuming we are allowed to use negative numbers though, I can get any sum I want, since if the number I want is a multiple of 5, then add or subtract the appropriate number to each cell of the original magic square, and if what I want is not a multiple of 5, then multiply all entries by 5, and add or subtract the appropriate number to the 5 selected cells as we did above (we multiplied by 5 so as to ensure that we don't get duplicate entries when we add or subtract the number which is not a multiple of 5). E.g:

Wanted: sum = 10. OK, then take original square and subtract 11 from each entry:

\begin{array}{|c|c|c|c|c|} \hline 6 & 13 & -10 & -3 & 4\\ \hline 12 & -6 & -4 & 3 & 5 \\ \hline -7 & -5 & 2 & 9 & 11 \\ \hline -1 & 1 & 8 & 10 & -8 \\ \hline 0 & 7 & 14 & -9 & -2 \\ \hline \end{array}

Wanted: sum = 1. OK, then multiply all entries by 5 (sum = 325), and subtract 324 from the 5 selected cells:

\begin{array}{|c|c|c|c|c|} \hline 85 & -204 & 5 & 40 & 75\\ \hline -209 & 25 & 35 & 70 & 80 \\ \hline 20 & 30 & -259 & 100 & 110 \\ \hline 50 & 60 & 85 & 105 & -309 \\ \hline 55 & 90 & 125 & -314 & 45 \\ \hline \end{array}

OK, so this is just the 5x5 case, but it works for other n as well. For example, here is how you can pick 6 cells for a 6x6 to play with:

\begin{array}{|c|c|c|c|c|c|} \hline \color{red}{X} & X & X & X & X & X\\ \hline X & X & X & X & \color{red}{X} & X\\ \hline X & \color{red}{X} & X & X & X & X\\ \hline X & X & X & X & X & \color{red}{X}\\ \hline X & X & \color{red}{X} & X & X & X\\ \hline X & X & X & \color{red}{X} & X & X\\ \hline \end{array}

In fact, you can easily convince yourself that for any n>4 we can find these n cells for any nxn so that each row, column, and main diagonal has exactly one of those n cells.

Edit

OK, the 4x4 is also possible! (Thanks @RossMillikan !)

\begin{array}{|c|c|c|c|} \hline \color{red}{X} & X & X & X\\ \hline X & X & \color{red}{X} & X\\ \hline X & X & X & \color{red}{X}\\ \hline X & \color{red}{X} & X & X\\ \hline \end{array}

Best Answer

Related Solutions

[Math] The sum of integers in each row / column / major diagonal of a Magic Square of order $N$ is $\dfrac{N(N^2+1)}{2}$

Magic Square – Possible Sums of an n x n Magic Square

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