I have a math proof that I think would be in the category of number-theory or algebra.
EDIT: The proof is incorrect as pointed out by Empy2 but still proves that the values must be all odd or all even.
The proof is that there doesn't exist a 3x3 magic square with all squares (see https://plus.maths.org/content/os/latestnews/may-aug10/magic/index)
Edit: Here is my proof:
So the basic idea is to solve variables
a^2 b^2 c^2
d^2 e^2 f^2
g^2 h^2 i^2
and we know that every row/column/diagonal must add to some value. Say s for sum. Now let's replace h^2 with s-b^2-e^2 since we know that the row must add to s.
a^2 b^2 c^2
d^2 e^2 f^2
g^2 s-b^2-e^2 i^2
Now let's replace the bottom right value i^2 with s-a^2-e^2 since that diagonal must add to s.
a^2 b^2 c^2
d^2 e^2 f^2
g^2 s-b^2-e^2 s-a^2-e^2
and now the same with the bottom left:
a^2 b^2 c^2
d^2 e^2 f^2
s-c^2-e^2 s-b^2-e^2 s-a^2-e^2
We can also replace d^2 with s-(s-c^2-e^2)-a^2 which can be simplified as c^2+e^2-a^2:
a^2 b^2 c^2
c^2+e^2-a^2 e^2 f^2
s-c^2-e^2 s-b^2-e^2 s-a^2-e^2
Let's also replace the middle-right f^2 with s-(s-a^2-e^2)-c^2 which is a^2+e^2-c^2.
a^2 b^2 c^2
c^2+e^2-a^2 e^2 a^2+e^2-c^2
s-c^2-e^2 s-b^2-e^2 s-a^2-e^2
We also know that the middle row must add to s therefore (c^2+e^2-a^2)+e^2+(a^2+e^2-c^2)=s. Which can be simplifyed as 3e^2=s. Therefore e^2=s/3 and we already know that s is the sum of the top row therefore a^2+b^2+c^2=s. Which means that e^2=(a^2+b^2+c^2)/3. Let's now replace that:
a^2 b^2 c^2
c^2+(a^2+b^2+c^2)/3-a^2 (a^2+b^2+c^2)/3 a^2+(a^2+b^2+c^2)/3-c^2
s-c^2-(a^2+b^2+c^2)/3 s-b^2-(a^2+b^2+c^2)/3 s-a^2-(a^2+b^2+c^2)/3
Let's also replace s with (a^2+b^2+c^2) and simplify:
a^2 b^2 c^2
c^2+(a^2+b^2+c^2)/3-a^2 (a^2+b^2+c^2)/3 a^2+(a^2+b^2+c^2)/3-c^2
a^2+b^2-(a^2+b^2+c^2)/3 a^2+c^2-(a^2+b^2+c^2)/3 b^2+c^2-(a^2+b^2+c^2)/3
subsection{Multiples of 4}
We know that every integer is one of these: 4m, 4m+1, 4m+2, 4m+3. However if we square all of them we get:
(4m)^2 = 16m^2 = 4(4m^2)
(4m+1)^2 = 16m^2+8m+1 = 4(4m^2+2m)+1
(4m+2)^2 = 16m^2+16m+4 = 4(4m^2+4m+1)
(4m+3)^2 = 16m^2+24m+9 = 4(4m^2+6m+2)+1
Which means that a square number can't be a multiple of 4 plus 2 or a multiple of 4 plus 3 and since all the values have to be squares none of them can be multiple of 4 plus 2 or a multiple of 4 plus 3. Now let's look at the center value of (a^2+b^2+c^2)/3. We know that all the squares in it can only be multiple of 4 or a multiple of 4 plus 1. Here is a table of the inputs and the outputs (0 is a multiple of 4, 1 is a multiple of 4 plus 1, and so on):
Output a^2 b^2 c^2
0 0 0 0
3 0 0 1
3 0 1 0
2 0 1 1
3 1 0 0
2 1 0 1
2 1 1 0
1 1 1 1
Which means we are left with these options:
Output a^2 b^2 c^2
0 0 0 0
1 1 1 1
However we can ignore a^2,b^2,c^2=4j,4k,4l because if that's the case then you could divide everything by 4 and they would still be squares. Therefore we can ignore it. This is the only option:
4j+1 4k+1 4l+1
4m+1 4n+1 4t+1
4p+1 4q+1 4r+1
Since they are actually of the form 4(m^2+m)+1 instead of 4j+1 (due to the fact that they are an odd squared) and the fact that j^2+j will always be even we can rewrite it like so:
8j+1 8k+1 8l+1
8m+1 8n+1 8t+1
8p+1 8q+1 8r+1
(PLEASE NOTE: the variables before and the variables after are different!) The sum of any row must be 12w+3 because the center value is a multiple of 4 plus 1. Now let's write all the possiblities for whether they're 16j+1 or 16j+9 (We can write anything say 6u+1 as either 12u+1 or 12u+6+1):
Sum a^2 b^2 c^2
16x+3 1 1 1
16x+11 1 1 9
16x+11 1 9 1
16x+19 1 9 9
16x+11 9 1 1
16x+19 9 1 9
16x+19 9 9 1
16x+11 9 9 9
Let's look at a^2,b^2,c^2=16j+1,16k+1,16l+9:
16j+1 16k+1 16l+9
? ? ?
? ? ?
We know that there can only be one 9 per row/colum/diagonal therefore:
16j+1 16k+1 16l+9
? ? 16t+1
? ? 16r+1
and we can also put 1s from the diagonal.
16j+1 16k+1 16l+9
? 16n+1 16t+1
16p+1 ? 16r+1
We also know that there must be one 9 per row/colum/diagonal:
16j+1 16k+1 16l+9
16m+9 16n+1 16t+1
16p+1 16q+9 16r+1
but wait! There arn't any 9s in the top-left to bottom-right diagonal. Therefore a^2,b^2,c^2=16j+1,16k+1,16l+9 is impossible. As well as a^2,b^2,c^2=16j+9,16k+1,16l+1 is impossible becuase of symetry. We can also ellimenate 2 more options by replacing 1 with 9 and 9 with 1. So we are down to these:
Sum a^2 b^2 c^2
16x+3 1 1 1
16x+11 1 9 1
16x+19 9 1 9
16x+11 9 9 9
Now let's look at a^2,b^2,c^2=16j+1,16k+9,16l+1:
16j+1 16k+9 16l+9
? ? ?
? ? ?
We know that there can only be one 9 per row/colum/diagonal therefore:
16j+1 16k+9 16l+1
? 16n+1 ?
? 16q+1 ?
and we can also put a 9 in the bottom right because there must be a 9 in the diagonal.
16j+1 16k+9 16l+1
? 16n+1 ?
? 16q+1 16r+9
and we can also put a 9 in the bottom left because there must be a 9 in the other diagonal.
16j+1 16k+9 16l+1
? 16n+1 ?
16p+1 16q+1 16r+9
but wait! There are two 9s in the bottom row therefore a^2,b^2,c^2=16j+1,16k+9,16l+1 is impossible. Here are the two options left:
Sum a^2 b^2 c^2
16x+3 1 1 1
16x+11 9 9 9
Looking back at the original table now looks like this:
16j+1 16k+1 16l+1
16m+1 16n+1 16t+1
16p+1 16q+1 16r+1
or
16j+9 16k+9 16l+9
16m+9 16n+9 16t+9
16p+9 16q+9 16r+9
However I now can prove that both of these are impossible! Imagine that we can split 16j+1 into 32j+1 or 32j+17 and we can keep repeating, but I can prove that all the numbers have to be the same option! First let's start with the scenaros:
a^2 b^2 c^2
n n n
n n m
n m n
n m m
m n n
m n m
m m n
m m m
where n and m are their values above some power of 2. (like: 128+17, n=17) Now let's get the basic idea: we want to show that for any different n and m both greater than 0 and smaller than the power of 2 (which we will now write as P) that it won't work unless all numbers are of the form 2^P+n or 2^P+m but not combos of them.
Let's start by proving n,n,m is impossible:
2^Pj+n 2^Pk+n 2^Pl+m
? ? ?
? ? ?
To simplify for looks let's replace them with just n or m not 2^Pj+n or 2^Pj+m:
n n m
? ? ?
? ? ?
We can now put n from the top right m (since there has to be one m per line or else they can't add to the same number. I have a proof of it but I don't think I need to show that here. It's already kinda long.) straight down, and diagonal from it:
n n m
? n n
n ? n
but wait! There are three n in the left diagonal therefore it can't work. We are now down to these options: (and since this doesn't work we can also remove the opposite and the backwards ones!)
a^2 b^2 c^2
n n n
n m n
m n m
m m m
Now let's prove that n,m,n is impossible.
n m n
? ? ?
? ? ?
Create a line down from it:
n m n
? n ?
? n ?
Add two m because of the diagonals must have a m:
n m n
? n ?
m n m
but wait! There are two m in the bottom row therefore it can't work! Meaning the all the top three values have to be like so:
a^2=2^Pj+z
b^2=2^Pk+z
c^2=2^Pl+z
where z is some constant amount. (We know that z must be constant because we showed that the only options are if it's the same number for all 3.) We also know that P can be as high as we want it to be as long as it's not infinity but a finite number. Let's imagine that 2^P is greater than a^2, b^2, or c^2. Then j,k, and l must all be 0. Which means they all equal z therefore they would be the same number and that is against the rules so there doesn't exist a solution with all squares.
Q.E.D.
Thanks in advance.
I think you missed the possibility of $$\left[\begin{array}[ccc] {} 16j+1&16k+9&16h+1\\16m+9&16n+9&16t+9\\16p+1&16q+9&16s+1\end{array}\right]$$