Proof of Magic Constant Formula

Question

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?

score 2 · Accepted Answer · answered Dec 06 '19 at 19:13

2

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 $$\frac{1}{N}\times \frac{N^2 \left(N^2+1\right)}{2} =\frac{N\left(N^2+1\right)}{2} .$$

answered Dec 06 '19 at 19:13

Minus One-Twelfth

7,413

If you want a proof that $1+2+3+\cdots+n =\frac{n(n+1)}{2}$, see e.g. Proof for formula for sum of sequence $1+2+3+…+n$?. – Minus One-Twelfth Dec 06 '19 at 19:17
Thanks @Minus One-Twelfth – atin Dec 06 '19 at 19:24

Proof of Magic Constant Formula

1 Answers1