Orthomagic Square of Squares

It's not known if there exists a 3x3 magic square of squares, i.e., 
a 3x3 arrangement of nine distinct integer squares such that the 
sum of each row, column, and main diagonal is the same.  A recent 
note discussed one approach to this problem, namely, to determine 
the form of all 3x3 arrangements of squares that satisfy the four 
sums involving the central number, and then see if any of those 
arrangements can be made to also satisfy the four outer sums.  In 
this way it was shown that no solution is possible if the central 
square is expressible as a sum of two squares in only four ways 
(which is the simplest non-trivial case).  It may be possible to 
extend that method to the general case, but I wonder if another
approach might be more effective.

Instead of looking at the 3x3 arrangements that satisfy the four sums
involving the central number, suppose we consider the arrangements 
that satisfy the six orthogonal sums, i.e.,the sums of the rows and
columns.  If these "orthomagic squares" of squares could be completely
characterized, it might be possible to show that they can never satisfy
the sums on the two main diagonals, thereby proving the impossibility
of a 3x3 magic square of squares.  (Of course, if this can't be shown,
this approach may help to construct an example.)

Remarkably, it turns out that most orthomagic squares of squares also
possess another property: the common sum of the rows and columns is a
square!  For example, the smallest orthomagic arrangement of distinct
squares is
                     4^2   23^2   52^2
 
                    32^2   44^2   17^2

                    47^2   28^2   16^2

and each rows and column of this arrangement sums to 3249 = 57^2.
The same is true for the next several OMSOS's.  In any case, this is
nice because we know the common sum of a completely magic arrangement
of squares must be of the form 3E^2 where E^2 is the central square.
Therefore, since a square can't be 3 times a square, we can 
immediately rule out all orthomagic arrangements whose common sum
is a square.

Of the twelve smallest OMSOS's, nine of them have a square common 
sum, so this just leaves three possibilities, and those can also 
be ruled out individually.  Interestingly, the smallest OMSOS that 
does NOT have a square common sum happens to be unique in another 
sense, namely, all the entries are squared primes:

                  11^2   23^2   71^2

                  61^2   41^2   17^2

                  43^2   59^2   19^2

The common sum of the rows and columns is 5691 = 3*7*271.  Obviously
we can permute the rows and columns of an OMSOS without affecting the
sums, but since 3*7*271 is not 3 times a square, we know this can't 
be permuted into a fully magic square.  Still, this is an interesting
square in its own right.  The next two "all-prime" OMSOS's (after the
one noted above) are based on the matricies

            13  67 149            17  71 149
            89 127  53            89 137  29
           137  79  43           139  61  67

It's also interesting that the next two "exceptional" OMSOS's (meaning
those whose common sum is NOT a square) also have common sums of the 
form 3*7*p where p is a prime congruent to 1 (mod 6).

Even though the OMSOS's with square common sums are immediately 
excluded from being completely magic, they are interesting in their
own right, and it's worthwhile to consider why the condition of equal
sums for the row and columns predisposes the common sum to be a 
square (when the elements themselves are squares).  First, notice
that they seem to occur in infinite families, and it's not too hard
to figure out parametric representations for some of them.  For 
example, there's an infinite family containing "(1)^2":


         (1)^2            (8+10k)^2      (4+16k+10k^2)^2

      (4+8k+6k^2)^2    (4+14k+8k^2)^2       (7+8k)^2

     (8+14k+8k^2)^2     (1+8k+6k^2)^2       (4+6k)^2


with the common sum (9+16k+10k^2)^2.  (Of course there are a few
values of k for which the elements of this array are not distinct,
so I exclude those from the set of orthomagic squares.  Note also 
that k can be positive or negative, because all the results are 
squared anyway.)  Similarly an infinite family containing (2)^2 
is given by


         (2)^2            (14+10k)^2      (5+14k+5k^2)^2

      (11+10k+3k^2)^2    (2+10k+4k^2)^2       (10+8k)^2

     (10+10k+4k^2)^2     (5+10k+3k^2)^2       (10+6k)^2


with the common sum (15+14k+5k^2)^2.  We can give a similar infinite
1-parameter family containing (n)^2 for any given n, so there ought 
to be a 2-parameter representation covering all of these.  Ideally 
we'd like to find a complete characterization of all OMSOS's, or at
least the possible common sums, to see if complete magicality can be
ruled out.

The class of orthomagic squares whose "common sum" is a square is 
closely related to quaternions, spatial rotation matricies, and 
representations of numbers as sums of FOUR squares.  This is an 
observation that essentially goes back to Euler (see Dickson's 
History).  Specifically, for any numbers a,b,c,d we can construct
a 3x3 matrix

 a^2 + b^2 - c^2 - d^2      2(bc - ad)            2(ac + bd)

     2(ad + bc)        a^2 - b^2 + c^2 - d^2      2(cd - ab)

     2(bd - ac)             2(ab + cd)        a^2 - b^2 - c^2 + d^2


Each row and column, regarded as a 3D vector, has the magnitude

                 L = a^2 + b^2 + c^2 + d^2

so obviously if we construct a 3x3 square whose components are the
squares of the components of the above matrix, it will be an 
"orthomagic square of squares" with the common sum L^2.  This accounts
for the frequent occurrance of OMSOS's with a square common sum.  For
example, with a=1, b=2, c=4, d=6 we have the basic matrix

                     -47   4  32
                      28 -23  44
                      16  52  17

and if we square each number this is the smallest orthomagic square 
of squares, with the common sum 3249 = 57^2.  The Determinant is 57^3.
Note that the three row vectors constitute an orthogonal triad, as do 
the three column vectors, and if we normalize each term by dividing it 
by the magnitude L=57 the above matrix is the rotation operator 
representing the space rotation relating the row triad to the column 
triad.

Obviously the product of two such base squares gives another base square.
For example, we have the product
    _             _   _             _       _                   _
   |  -47   4  32  | |  -51  18  46  |     |   3397  1190 -1850  |
   |   28 -23  44  | |   42 -19  54  |  =  |  -1250  3845   178  |
   |_  16  52  17 _| |_  26  66   3 _|     |_  1810   422  3595 _|

The second factor on the left side is given by setting a=1, b=3, c=5, and
d=6, it's determinant is 71^3, and the common sum of squares of its rows
and columns is 71^2. The matrix product is also a base square, i.e., the
squares of its elements form an orthogonal orthomagic square of squares,
with the common sum (57*71)^2 and the base square has determinant (57*71)^3.
It is produced by setting a=-61, b=-1, c=15, d=10, which can be inferred
from the four-square multiplication formula

  (A^2 + B^2 + C^2 + D^2)(a^2 + b^2 + c^2 + d^2) = w^2 + x^2 + y^2 + z^2

where
                      w = Aa + Bb + Cc + Dd
                      x = Ab - Ba + Cd - Dc
                      y = Ac - Bd - Ca + Db
                      z = Ad + Bc - Cb - Da

The above is an interesting example of how, when trying to work in three 
parameters (or dimensions), it often seems that we're led to a much more 
natural formulation by going to four.  I had been looking at sums of THREE 
squares, noting that the the most general solution (according to Dickson) 
of the equation  X^2 + Y^2 + Z^2  =  N^2  is of the form

               X = 2( a^2 + b^2 - c^2 )
               Y = 2( a^2 - b^2 + c^2 ) + 2a( b-3c)
               Z =  (-a^2 + b^2 + c^2 ) + 2b(2a-3c)
               N = 3( a^2 + b^2 + c^2 ) - 2c(2a+ b)

and in retrospect it's clear that there's a 4-parameter family hidden 
behind this, trying to get out.  Indeed, it was derived from Euler's 
four-square product formula, supressing one of the parameters.

Incidentally, the quaternionic formulas noted above are very similar 
to the generalized Heron's formula, relating the volume of a 
"perfect tetrahedron" in terms of the areas of its faces, as 
discussed in Heron's Formula For Tetrahedrons.

Anyway, the OMSOS's with square common sum seem to be well covered by 
the above parameterization, although I'm not quite sure it necessarily 
includes ALL OMSOS's with square sum.   Another interesting question 
is whether the OMSOS's that *don't* have a square sum are also given
by the 4-parameter matrix above, with non-integer values of a,b,c,d.
In other words, are there any sets of numbers a,b,c,d such that the
nine elements of the basic matrix are integers but the magnitude of 
the row and column vectors is not an integer?

There are 91 primitive orthomagic squares of squares with common sums
less than 30000.  Of those, 56 have a square common sum, whereas the 
remaining 35 do not.  Of the 35 non-square cases, none of them is of 
the form 3k^2, so they clearly can't give a complete magic square of
squares.  For more on the search for a magic square of squares, see 
the note Automedian Triangles and Magic Squares

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