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.