Proof that magic squares of squares don't exist

Introduction to Magic squares
Magic squares are squares with numbers in them. The sum of all columns, rows and the two main diagonals must all equal the same number. and there are no duplicates. Magic squares of squares are squares with only square numbers in them. Again the sum of all columns, rows and both main diagonals must equal a constant, again, without duplicates. In the rest of the article, the constant will be notated as m and the magic square we'll look at has these variables in it. Notice that for b2,b4,b6 and b8, there are only two sums (all of which are squares) which involve each of them. This implies that m-(bn)^2 for even n can be written as the sum of two squares in at least two different ways. Similarly, b1,b3,b7 and b9 are involved in 3 sums of squares, implying that m-(bn)^2 for odd n not equal to 5, can be written as the sum of two squares in at least 3. By the same logic m-(bn)^2 for n=5 can be written as the sum of two squares in 4 different ways.

Finding a lower bound for m
As the equation for a circle is x^2+y^2=r^2 where r is the radius and that m-(bn)^2 can be written as the sum of two squares in some amount of ways, if we set the radius of the circle as sqrt(m-(bn)^2), then the equation becomes x^2+y^2=m-(bn)^2 due to the cancelling out of the square and the square root. As we are looking for integer squares, x and y must be integers too. So the question becomes, how many lattice points does said circle hit? $$4\sum \chi(f(n))$$ where f(n) is the nth factor of r^2. As we are looking for unique solutions of x and y, we can choose to only look at one quadrant of the grid (the other quadrants produce the same squares), allowing us to divide the number of lattice points by 4, giving us $$\sum\chi(f(n)) $$. Now we can split into two cases, a) where there is a solution for x=y and b) where there isn't. Let's start with the second case

The x≠y case
In that quadrant, there are still duplicates. One is x^2+y^2 and the other is y^2+x^2. So we'll need to cut the number of lattice points in half giving $$1/2 \sum\chi(f(n)) $$.

The x=y case
In that quadrant, we need to remove the x=y case, as that corresponds to duplicates in the magic square (e.g. m-(b5)^2=(b2)^2+(b8)^2, if the squares are equal, (b2)^2=(b8)^2, which is not allowed). After that, we can proceed as usual giving $$1/2(-1+\sum\chi(f(n))) $$.

Summary of cases
For the second case to give a whole number, the number of lattice points in the quadrant must be odd. The function also gives rise to a number that is less than one half of the number of lattice points, indicating the direction of rounding: $$\lfloor 1/2\sum\chi(f(n)) \rfloor $$ Let's name this function g(r) where r is the radius of the circle.

As $$\chi(2)=0$$ and $$\chi(3)=-1$$, neither can serve as factors to increase the number of sums of squares. The smallest number that works is $$\chi(5)=1$$, which is also true for every power of 5. If you want g(r) to be at least s, $$\sum\chi(f(n))$$ must be at least 2s. As in our scenario, it's purely factors of 5, if our number is equal to 5^c, then $$\sum\chi(f(n)) $$ is also equal to c. So c is at least 2s. must be As m-(bn)^2 for odd n not equal to 5 can be written in said form in 3 different ways, m-(b5)^2 can be written in 4 ways and m-(bn)^2 for even n can be written in 2 ways, the values of c are 6, 8 and 3 respectively. As all integer squares are positive, (b5)^2 can't be negative, and therefore, as m-(b5)^2 is at least 5^8, m is at least 5^8.

The proof
m-(b1)^2 and m-(b3)^2 both have a lower bound of 5^6. Also m-(b2)^2 has a lower bound of 5^4. As m has a lower bound of 5^8, b1 and b3 have a lower bound of 5^8-5^6. Similarly, b2 has a lower bound of 5^8-5^4. As m is the sum of b1, b2 and b3, the new lower bound becomes 2(5^8-5^6)+(5^8-5^4), which is noticeably bigger. In general, for a lower bound for m as d, the new lower bound is 2(d-5^6)+(d-5^4) which when iterated, diverges.