0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$\textbf{Hint:}$ 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?

$\endgroup$
0
$\begingroup$

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 $n$ $$nR=nC \quad\Rightarrow\quad R=C$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.