# Prove Magic Square

Suppose that $S$ is a square such that the sum of the entries in each row is some number $R$, and the sum of the entries in each column is some number $C$. Prove that $S$ is in fact a magic square, i.e., $R = C$.

What I thought?

The sum of a row is $(1+..+n-1)n+(1+..+n)$, and the same for any col, hence the square in magic. Can someone verify this and help me elaborate on this.

$\textbf{Hint:}$ Compute the sum of all the numbers in the square in two different ways.
I don't understand your attempt. What is $n$? And why does the sum have to equal that thing?
Let $a_{i,j}$ be the entry in the $i$th row and $j$th column.
So, in an $n$x$n$ square, the sum of each row is: $a_{i,1} + ... + a_{i,n} =R$
And the sum of each column is: $a_{1,j} + ... + a_{n,j}=C$
As Arthur said, the sum of all the entries is $$(a_{1,1} + ... + a_{1,n}) + (a_{2,1} + ... + a_{2,n})+ ...+(a_{n,1} + ... + a_{n,n}) = (R)+(R)+...+(R)=nR$$ $$||\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$ $$(a_{1,1} + ... + a_{n,1}) + (a_{1,2} + ... + a_{n,2})+ ...+(a_{1,n} + ... + a_{n,n}) = (C)+(C)+...+(C)=nC$$ Then divide both sides by $n$ $$nR=nC \quad\Rightarrow\quad R=C$$