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Adam Dewbery's Proof of the importance of x^{2} + y^{2} = 2 z^{2}
(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 a^{2} + b^{2} = 2 c^{2} in four different ways. Finding just one solution is easy. Pick two numbers m and n, and the following parameterization can be used: