TITLE: How do I get my math proof checked?
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: I think you missed the possibility of $$\left[\begin{array}[ccc] {}
16j+1&16k+9&16h+1\\16m+9&16n+9&16t+9\\16p+1&16q+9&16s+1\end{array}\right]$$