Euler's 3x3 Magic Square of 7 Squares


Introduction
In the paper, "On squares of squares", by Andrew Bremner, Acta Arith. 88
(1999), Bremner writes about parameterizing special cases of a 3x3 magic
square of squares using Pythagorean triangles of equal area.

   "An equivalent parameterization had in fact been discovered
   much earlier by Euler.  This parameterization is equivalent
   to that at (11), so in fact the case can be made that Euler
   was essentially aware of this family (12) of squared squares".

I wonder if the case could be made that Euler solved some other magic square
of square problems.


An Historical Discovery
There is only one known 3x3 magic square containing 7 square entries.
But who discovered it first?  Was it Lee Sallows?  Was it Andrew Bremner?
I claim it was Leonhard Euler.  Here is my evidence.

We know that Euler wrote papers on the subject of magic squares. He produced
formulas for 4x4 and 5x5 magic and semi-magic squares of squares as well as
research on what is now known as orthogonal matrices, which is related to the
3x3 magic square of squares problem.  But did he ever work on 6-square or
7-square configurations of the 3x3 magic square of squares?

Opera Postuma I contains fragments from Euler's notebooks. Sections 82 and 83
are fragments about magic squares.  Just a little before it, in section 80,
there is a treatment and some solutions for the following "triple-equation".
     p2 + q2 = 2z2
     p2 + r2 = 2y2
     q2 + r2 = 2x2

Euler gives the following five solutions.
     p =  89,  97,  119,  23,   17
     q = 191, 553,  833, 289,  697
     r = 329, 833, 1081, 527, 1127

He also gives a treatment that is part of what is now known as binary
quadratic forms, including the composition of forms to produce new solutions
from old.

Here is a formulation for a 3x3 magic square of squares.
     c+b      c-(a+b)  c+a               A2  B2  C2
     c-(b-a)  c        c+(b-a)    ===>   D2  E2  F2
     c-a      c+(a+b)  c-b               G2  H2  I2

Here is a configuration where the 6 x-entries are required to be squares.
     - x -
     x x x
     x - x

It contains three 3-square arithmetic progressions given by
     c-(a+b), c-b, c-(b-a)   ===>  B2, I2, D2
     c-(a+b), c-a, c+(b-a)   ===>  B2, G2, F2
     c-(b-a), c,   c+(b-a)   ===>  D2, E2, F2

These progressions can also be formulated as
     B2 + D2 = 2I2
     B2 + F2 = 2G2
     D2 + F2 = 2E2
which is identical to the form of Euler's triple-equation.

This means we can use any of his solutions for this configuration,
producing a magic square with 6 square entries.


The fragments in Opera Postuma were published after Euler's death
by his followers who extracted them from Euler's notes.
So we don't really know what Euler had in mind when he made the notes.
But we can make educated guesses based on later knowledge
of the subject and also based on clues Euler left behind.


Euler's Clue
What is the criteria Euler used to order his solutions?  There doesn't seem
to be any.  The closest you can come to a pattern is that they are in
low to high order, based on r values, except that the 4th solution
is out of order.

Is Euler trying to call our attention to the 4th solution?

Let's try using it for our magic square configuration.
     p = B = 23,  q = D = 289, r = F = 527.
Computing the other three variables from the formula,
     z = I = 205, y = G = 373, x = E = 425.

Filling in the magic square yields
     ---    232  ---
     2892  4252  5272
     3732  ---   2052

The last three entries are determined from the magic sum requirements.
     5652    232  222121
     2892   4252    5272
     3732  360721   2052

Look what's in the upper left corner.
It's a 7th square!


Another Historical Discovery
Who was the first to find the smallest solution to this 6-square configuration?

Using Euler's 1st solution, we can fill in the magic square as follows.
    -13679  3292  -27959
      892   1492   1912
     2692  -63839  2412
It has a magic sum of 3x1492.

Here is Andrew Bremner's smallest magic sum solution for the same configuration.
     93961  1912   43801
      892   2412   3292
     2692   79681  1492
It has a larger magic sum of 3x2412.
Euler wins again.