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Adam Dewbery's Proof of the importance of  x2 + y2 = 2 z2
(For the Magic Square of Squares)
A magic square has integer values in its boxes.  Every row,column, and
diagonal sums to the same value.

Therefore:-

a + b + c = k             (Equation 1)
d + e + f = k             (Equation 2)
g + h + i = k             (    "    3)
a + d + g = k                 etc.
b + e + h = k
c + f + i = k
c + e + g = k
a + e + i = k
a + b + c + d + e + f + g + h + i = 3k

Taking Eqs 1,6,7

a + b + c = c + f + i = c + e + g

therefore :-

a + b = f + i = e + g

similarly :-

b + c = d + g = e + i
h + i = a + d = c + e
g + h = c + f = a + e

From this we can see that

f + i = e + g
d + g = e + i

rearranging gives

g - i = f - e
g - i = e - d

therefore

f - e = e - d
2e = f + d

Similarly
2e = a + i
2e = b + h
2e = c + g

Adding these gives
9e = 3k
3e = k
e = k/3
 

Therefore the centre box in a magic square must be 1/3 of a row, column or
diagonal's total.
 

Now lets say that all the intergers in a magic square are made by
multiplying a rational number by the centre number.  The square looks like
this.

ae       be        ce
 

de        e        fe
 

ge       he        ie
 

We have already established that a row, column, diagonal adds up to 3 times
the centre number.

Therefore

ae + e + ie = 3e
e(a + 1 + i) = 3e
a + 1 + i = 3
a + i = 2
i = 2 - a

similarly

h = 2 - b
g = 2 - c
f = 2 - d

Now we know that a + i = 2.

ae needs to be a square.  As e has to be a square as it is in the centre,
then a must be a rational square, with its numerator and denominator being
factors of e.  This will give a square integer for box a.

Back to Ed Pegg.

Therefore, it must be a solution to a2 + b2 = 2 c2 in four different ways.  Finding just one solution is easy.  Pick two numbers m and n, and the following parameterization can be used:

a = m^2 + 2mn - n^2
b = m^2 - 2mn - n^2
c = m^2 + n^2
I looked for solutions to this equation, and found the answer in Beiler's Recreations in the Theory of Numbers.  First, realize that some numbers are the sum of squares.  5 = 22 + 12, 13 = 32 + 22, for example.  Let a and b be two such numbers.  It turns out that the product ab is a sum of squares in two different ways.  65 = 82 + 12 or 72 + 42.  This is due to the following algebraic identity:
(a2+b2)(c2+d2) = (ac+bd)2 + (ad-bc)2 = (ac+bd)2 + (ad-bc)2 .
Using the identity and the parameterization leads to a lot of possible middle squares.  Some of my dabbling is here.  What are the properties of the middle and side squares?  I formerly thought this problem was unsolvable, but the algebraic properties of squares provides enough tools to make the problem solvable.