Is there a proof of 3x3 Magic Square General Form? Below, I have included a picture of an article I read from Martin Gardner on additional structures found within 3x3 magic squares, concerning their row (and column) products’ sums.
He casually mentions that the form of 3x3 magic squares MUST fit the general form at the top-right, but I can’t find a proof anywhere. I may be looking in the wrong places.
Does anyone have a DOI link or journal reference for a general proof? The reference page in the article was also cut off, so I couldn’t find it there, if it is there.
Thank you / sorry for not being more helpful.

 A: This is actually a reasonable straightforward thing to work out algebraically: essentially, you are trying to solve $7$ linear systems which express that the sum of any row, column, or diagonal is equal to the sum of any other.
Let's prove this! Suppose that the following is a magic square:
$$\begin{pmatrix}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\A_{31} & A_{32} & A_{33}\end{pmatrix}$$
The main substance of the form that Martin Gardner notes is that the sum of every row, column, or diagonal is $3A_{22}$. To prove this, let $S$ be this sum. Note that
$$S=A_{11}+A_{22}+A_{33}=A_{21}+A_{22}+A_{23}=A_{31}+A_{22}+A_{13}.$$
We note that $A_{22}$ appears three times here and that the other coefficients are laid out in two rows. Notice that adding these three forms together gives
$$3S=3A_{22}+(A_{11}+A_{21}+A_{31})+(A_{13}+A_{23}+A_{33}).$$
However, the two parenthesized terms are columns, so sum to $S$ each, giving
$$S=3A_{22}$$
after cancellation.
The rest is less insightful - just a convenient choice of variables. First, $A_{22}$ is really important, so we'll give that a name $a$ and note that every row, column, and diagonal must sum to $3a$. We know that $A_{11}$ has to be something and we can certainly write any number in the form $a+b$. Likewise, $A_{13}$ has to be something, so let's call it $a+c$. We have done nothing in this step except choose some names.
Everything else follows easily now. For instance, $A_{33}$ must be $a-b$ because $A_{11}+A_{22}+A_{33}=3a$, but we already know two of the terms so can solve for the last one. Similarly, once we note that $A_{31}$ must be $a-c$, we then can solve for the four remaining values by looking at the outer rows and columns, as we know two of the terms in each. Finally, for mathematical completeness, we check that the equations whose truth we did not force - that the middle row and column sum to $3a$ - is indeed true. Thus, the given form is a magic square, but is also the only way to make a magic square because we deduced it from the mere statement "this is a magic square."
A: All of the rows, columns, and diagonals sum to $3a$, so it satisfies the conditions on a magic square. It would probably still be a magic square if you replaced $a$ with $2a$.
