How to prove that a $3 \times 3$ Magic Square must have $5$ in its middle cell? 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 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.
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.
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.