Magic Squares

This post was inspired by my daily number analysis of my diurnal age. I was struggling to find something of interest about 26102 when I discovered that a $4\times 4$ magic square could be created in which 26102 was the magic constant. This got me thinking about what numbers could be represented in this way as magic squares.

A $n\times n$ magic square has its $n^2$ cells filled with the numbers from $1\text{ to }{n}^2$ in such a way that the sums of all rows, columns and main diagonals are the same. This sum is called the magic constant:$$\frac{n(n^2+1)}{2}$$The $3\times 3$ magic square contains the numbers from 1 to 9 and all rows, columns and main diagonals add up to 15. Thus the magic constant is 15.$$\begin{array}{|ccc|}\hline 8& 1& 6\\ 3& 5& 7\\ 4& 9& 2\\ \hline\end{array}$$If we add 1 to each element of the above magic square, we get the next magic square and this has a magic constant of 18:$$\begin{array}{|ccc|}\hline 9& 2& 7\\ 4& 6& 8\\ 5& 10& 3\\ \hline\end{array}$$All subsequent $3\times 3$ magic squares will have magic constants that are multiples of 3. Thus the number 26097 = 3 x 8699 can be represented as:$$\begin{array}{|ccc|}\hline 8702& 8695& 8700\\ 8697& 8699& 8701\\ 58698& 8703& 8696\\ \hline\end{array}$$We could represent the various matrices in terms of a subscript for their size and a superscript for their magic sum. This is purely my own invention but it means we would have: M${}_3^{15}$, M${}_3^{18}$ and M${}_3^{26097}$ where M is any matrix that satisfies the condition of being magic and having a particular size and magic sum. The $4\times 4$ magic square containing the numbers 1 to 15 has a magic sum of 34:$$\begin{array}{|cccc|}\hline 16& 2& 3& 13\\ 5& 11& 10& 8\\ 9& 7& 6& 12\\ 4& 14& 15& 1\\ \hline\end{array}$$It would be designated as M${}_4^{34}$ in my system. The subsequent magic sums that are possible are 38, 42, 46, 50 and so on. To test whether a number can be the magic constant for a $4\times 4$ magic square, simply subtract 34 from the number and test whether the result is divisible by 4. In the case of 26102, it is and so we have:$$\begin{array}{|cccc|}\hline 6533& 6519& 6520& 6530\\ 6522& 6528& 6527& 6525\\ 6526& 6524& 6523& 6529\\ 6521& 6531& 6532& 6518\\ \hline\end{array}$$This magic square can be designated M${}_4^{26102}$. Figures 1 and 2 shows the magic squares together with their magic sums and associated planet or luminary.

Figure 1: source

Figure 2: source

The $5\times 5$ magic square has a constant of 65 and so any subsequent multiples of 5 will be representable as $5\times 5$ magic squares. The $6\times 6$ magic square has a constant of 111 which is 3 x 37. This means that 117, 123, 129 etc. can be represented as $6\times 6$ magic squares but these numbers are not multiples of 6. To determine whether a number can be the magic constant for a $6\times 6$ magic square, the 111 must first be subtracted and the result tested for divisibility by 6. This is like the $4\times 4$ case.

For the $7\times 7$ magic square, the constant is 175 = 7 x 25 and so any multiple of 7 can be represented by a $7\times 7$ magic square. For the $8\times 8$ magic square, the constant is 260 = 4 x 65 which is not divisible by 8 and the situation is as the same as with 4 and 6. For the $9\times 9$ magic square, the constant is 369 = 9 x 41 so all multiples of 9 above 369 qualify.

Figure 3 shows the magic constants for larger magic squares:

Figure 3: source

So generally, excluding 2, the prime factorisation of a number will quickly tell us what magic squares it can serve as a constant for. For example, 910 = 2 * 5 * 7 * 13 and so it can be represented by magic squares of side 5, 7 and 13. It can also be represented as a magic square with a composite number of sides but a little testing is required. For example, it can be represented as a $4\times 4$ magic square but not $8\times 8$. This site is great for such testing.

on May 14th 2021