A Magic Square of order $n$ is an arrangement of $n^2$ numbers, usually distinct integers, in a square, such that the $n$ numbers in all rows, all columns, and both diagonals sum to the same constant.
How to prove that a normal $3\times 3$ magic square where the integers from 1 to 9 are arranged in some way, must have $5$ in its middle cell?
I have tried taking $a,b,c,d,e,f,g,h,i$ and solving equations to calculate $e$ but there are so many equations that I could not manage to solve them.
On
The common sum must be $15$, because the sum of numbers from $1$ to $9$ is $45$. How can we write $15$ as sum of three distinct numbers between $1$ and $9$ (included)?
\begin{gather} 1+5+9 \\ 1+6+8 \\ 2+4+9 \\ 2+5+8 \\ 2+6+7 \\ 3+5+7 \\ 3+4+8 \\ 4+5+6 \end{gather}
We have just eight ways and we can try accommodating them in the square. The central place belongs to one row, one column and the two diagonals, so the number we put in it must appear four times in the above sums: the only one is $5$.
Similarly, in the four corners we have to place numbers that appear three times, that is: $2$, $4$, $6$ and $8$.
This also shows that basically only one $3\times3$ magic square is possible, up to symmetries of the square.
How to compose the magic square? First note how many rows, columns and diagonals each cell belongs to:
Now we know that $5$ must be in the center; choose arbitrarily an even number, say $2$ and place it in a corner. It could go in any corner, let's choose the upper left one; in the opposite corner we have to write $8$:
Now $4$ must go in one of the other corners and $6$ in the opposite one:
At this point, the other cells can be filled in a unique way:
We have four choices for placing $2$ and, for any choice we can choose two places for $4$. In total we have eight magic squares, but just one if we consider two of them identical after applying a symmetry of the square (there are eight of them).
If we subtract $5$ to each cell, we can better see the symmetry:
On
The proof are in the rules (or conditions) about how a magic square is built...
In this case a 3x3 grid, you need a sequence of nine integer and positive numbers to make a magic square. You must order that numbers by size... the number that goes in the center of the grid is always the middle number of that group you ordered.
For example, you have this group of nine numbers to make a $3\times 3$ MS: $15, 33, 12, 45, 18, 48, 27, 30$ and $42$. After you ordered them by size: $12,15,18,27,30,33,42,45,48$.
The one in the middle of this ordered group of nine numbers is $30$.
30 will be the number that goes to the center of the $3\times 3$ grid.
No need of any formulas, just follow the MS (Magic Square) rules.
On
Once 15 is established as the magic number the easiest way to see this is to reason as follows. If a row/column/diagonal is 8,6,1, we say that 8 is opposite 1.
If a number smaller than five is in the centre consider the number opposite the remaining smallest number. It will have to be greater than 10.
Similarly a number greater than five will not do.
The row, column, diagonal sum must be $15$, e.g. because three disjoint rows must add up to $1+\ldots +9=45$. The sum of all four lines through the middle is therefore $60$ and is also $1+\ldots +9=45$ plus three times the middle number.