$3\times3$ Magic Square and hidden formula [Solved] I was doing the $3 \times3$ Magic Square of Squares problem -- found here: http://www.multimagie.com/English/SquaresOfSquaresSearch.htm -- and I figured out that if such Magic square exists with $9$ distinct numbers, then it must satisfy the following equation:
$$\sqrt{a} + \sqrt{b} - \sqrt{c} = \sqrt{d}$$
where the numbers $a,b,c$ and $d$ are distinct entries in the square.
Crucial notes:
1) $a+b$ must also be a square number
2) The numbers $a,b,c,d$ can be any distinct integers greater than 0.
My attempt was
$$\sqrt{36} + \sqrt{64} - \sqrt{75} = \sqrt{25}$$
which isn't quite equal and doesn't work because 75 isn't a perfect square.
In my amateur opinion there isn't a solution, but would like someone good in mathematics to prove or disprove above equation.
In advance, I'm very thankful for any answers.
 A: From what I understand in your post and the comments, you would like to find -- as JMoravitz said -- $a,b,c,d,e$ such that
$$a^2 + b^2 = c^2 + d^2=e^2$$
We necessarily have
$$a^2+b^2-d^2=c^2$$
And if I may explain how this relates to magic squares problems, as most people won't know. The existence of a 3x3 magic square of square numbers is currently unknown. There is a prize for finding one (Read here: http://www.multimagie.com/English/SquaresOfSquaresSearch.htm) But if a solution did exist it would have the following form
$$\begin{bmatrix}
a^2 & b^2 & c^2 \\
d^2 & e^2 & f^2 \\
g^2 & h^2 & i^2
\end{bmatrix}$$
Being a magic square means the sum of the entries of each column, row, and the 2 diagonals are equal. From this, one can show that
$$a^2 + i^2 = b^2 + h^2 = c^2 + g^2=2e^2$$
Not exactly what you wanted though -- I think. However from this we can derive other interesting equalities. For example
$$(ai)^2 + \Big(\frac{a^2-i^2}{2}\Big)^2=
(bh)^2 + \Big(\frac{b^2-h^2}{2}\Big)^2=
(cg)^2 + \Big(\frac{c^2-g^2}{2}\Big)^2 = e^4$$
Maybe this is what you were tackling?
Anyways, to solve your question -- and the magic square of squares problem -- it is very important to understand the two square identity
$$(a^2 + b^2)(c^2 + d^2) = (ac \mp bd)^2 + (ad \pm bc)^2$$
It's closely related to the Gaussian integers (integer complex numbers. read here: https://en.wikipedia.org/wiki/Complex_number#Multiplication_and_division)
To solve the first equation, lets break it up into to 2
$$a^2 + b^2 = e^2$$
$$c^2 + d^2 = e^2$$
Each equation is solved if we substitute 
$$a=2mn \quad b=m^2-n^2 \quad c = 2pq \quad d=p^2-q^2$$
$$e=m^2 + n^2 = p^2 + q^2$$
I mentioned the two square identity because it can be used to find this substitution. Now we must also find $m,n,p,$ and $q$ such that $m^2 + n^2=p^2+q^2$, which again can be done with the two square identity. The smallest example that gives a distinct solution without zeros is
$$(2^2 + 1^2)(3^2 + 2^2) \ \ =\ \ 8^2 + 1^2 \ \ =\ \  7^2 + 4^2 \ \ =\ \  65$$
So we can use $m=8$, $n=1$, $p=7$, and $q=4$ to get our first -- and the smallest -- solution
$$16^2 + 63^2 = 56^2 + 33^2 = 65^2$$
It would be fitting here to also explain why there are guaranteed to be infinitely many solutions. Any prime number that leaves a remainder of 1 when divided by 4, can be expressed as the sum of two squares (read more here: https://en.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares). For the previous solution I used the first two such primes: $5=2^2 + 1^2$ and $13 = 3^2 + 2^2$.
$\quad$ There are indeed infinitely many of this kind of prime just like there are infinitely many primes. The first few are $5,13,17,29,37,41,...$ (https://oeis.org/A002144). You can pick any two distinct products of these primes to make a new solutions. 
For example, starting with $m=7$, $n=4$, $p=5$, and $q=2$ will yield this solution
$$1836^2 + 427^2 = 1813^2 + 516^2 = 1885^2$$
That choice of $m,n,p,$ and $q$ works because $7^2+4^2=65=5\cdot 13$ and $5^2 + 2^2 = 29$. The numbers $5,13,$ and $29$ are primes of the form $4k+1$.
All solutions to $a^2 + b^2 = c^2 + d^2=e^2$ can either be generated with this method or by multiplying the $a,b,c,d$ and $e$ of some such solution by an arbitrary integer.
Hope this helps. Good luck getting that prize :)
