Is there a $3\times 3$ magic square adding up to $7$?

Question

I suspect that there is no magic square with natural number entries (matrix where each row, column and long diagonal add up to the same number) which would add up to $7$. There is no restriction on which numbers have to appear in the slots, also the numbers that do appear are not resricted to single digit numbers, so 12 can appear in the magic square)

Is it correct? If yes, how can I prove it?

(I see there are quite a lot of questions about magic squares on the site, but I havent been able to find an exact duplicate.)

What follows is my attempt write a magic square adding up to 7.

2, 3,2 - 3,2,2 - 2,2,3 -

Answer 1

Suppose we have a $3 \times 3$ magic square with integer entries, and each row/column/main diagonal adds to $n$:

$$\begin{pmatrix} a &b & c \\ d& e & f \\ g &h &i \end{pmatrix}$$

Then

\begin{align} 3n& =(a+b+c)+(d+e+f)+(g+h+i) \\ &=e+(a+i)+(b+h)+(c+g)+(d+f) \\ &=e+(n-e)+(n-e)+(n-e)+(n-e) \\ &=4n-3e \end{align}

Thus $n=3e$ so $n$ must be a multiple of $3$. Unfortunately, $7$ is not a multiple of $3$.

Answer 2

I've just spent $5$ minutes trying all possible combinations for the central square from $7$ to $4$ and it's easy to see that none of these are possible central squares. I don't forsee there being much trouble in checking the elements $3$ to $0$ either, but the possible combinations of diagonal squares does get slightly larger for smaller values of the central - but not so large that $10$ minutes wouldn't be enough to check by hand.

Answer 3

Adam (and others if interested). Here is a general 3by3 magic square generator. Draw a 3by 3 grid and write down the following in each box: Top row: $x+y$ , $x-y-z$ , $x+z$, middle row: $x-y+z$ , $x$ , $x+y-z$ , bottom row: $x-z$ , $x+y+z$ , $x-y$ The sum of each row/column/diagonal is $3x$