Discordance Impedes Square Magic

As discussed in another note, the terms of a 3x3 magic square 
of squares can be re-arranged into a 3x3 bilinear array as shown 
below:
                  |  2v  |  2v  |

            --   a^2    b^2    c^2
            2u
            --   d^2    e^2    f^2
            2u
            --   g^2    h^2    i^2

The values of 2u and 2v are the vertical and horizontal "partials",
respectively.  Notice that the differences are even, because we have
the conditions

  2e^2  =  a^2 + i^2  =  b^2 + h^2  =  c^2 + g^2  =  f^2 + d^2

which show that if any of the terms a,b,..,i is even, then they
ALL must be even, so we can divide each of them by 2, and repeat
until all are odd.  (By a similar analysis we can show that all
the terms a,b,..,i must be congruent modulo 3, so all the squares
must be congruent modulo 36.)  Therefore, we have

     e^2 + 2u = b^2           e^2 - 2u = h^2                (1a,b)

     e^2 + 2v = f^2           e^2 - 2v = d^2                (2a,b)

     e^2 + 2(u+v) = c^2       e^2 - 2(u+v) = g^2            (3a,b)

     e^2 + 2(u-v) = a^2       e^2 - 2(u-v) = i^2            (4a,b)

Subtracting equation (1b) from (1a) gives

               b^2 - h^2      / b+h \  / b-h \
         u  =  ---------  =  ( ----- )( ----- )
                   4          \  2  /  \  2  /

Letting B and H denote the two integer factors in parentheses, we
see that
              u  =  BH          B^2 + H^2  =  e^2

Similarly from equations (2), (3), and (4) we deduce integers F, D,
C, G, A, and I such that

              v  =  FD          F^2 + D^2  =  e^2

            u+v  =  CG          C^2 + G^2  =  e^2

            u-v  =  AI          A^2 + I^2  =  e^2

These equations together imply that if a magic square of square exists
then there must be four rational right triangles on a common hypotenuse 
with areas A1, A2, A3, A4 such that A1+A2 = A3 and A1-A2 = A4.  This 
can also be expressed by the conditions

            A3 + A4 = 2 A1         A3 - A4 = 2 A2

Just to emphasize how surprising a magic square of squares would be,
it's worth noting that even if we drop the right hand requirement, 
it appears that there do not exist THREE rational right triangles
on a common hypotenuse with areas in arithmetic progression
(signified by the left hand condition).  This would correspond
with a set of integers a,b,..,f,H such that

      a^2 + b^2  =  c^2 + d^2  =  e^2 + f^2  =  H^2
                                                             (5)
                       ab + cd  =  2ef

No such integers have ever been found.  However, if we relax the
requirement by replacing H^2 with an arbitrary integer K, then there
exist some solutions, although even these are not terribly plentiful.
There are only 39 primitive solutions with K less than 52 million,
as listed below:

     K       diff       a       b        c      d        e      f
 --------  --------   ------  -----    -----  -----    -----  -----
    25345      5460        8    159       44    153       96    127
    32266      6720       15    179       55    171      125    129
    36490      6720        3    191       39    187       81    173
    99025     15960       48    311      104    297      183    256
    99125      9828       23    314       55    310       89    302
   325117     61200       69    566      186    539      379    426
   419050     87360       21    647      161    627      343    549
   743665    143220       96    857      276    817      569    648
  1006561    170940       81   1000      260    969      480    881
  1229045     83160      439   1018      551    962      758    809
  2047786    255360        5   1431      185   1419      375   1381
  3129802    235200      661   1641      851   1551     1179   1319
  4658425    228228      948   1939     1120   1845     1389   1652
  5602945    840840       16   2367      376   2337      768   2239
  7805890    779520      357   2771      651   2717      973   2619
  8181625    879648      844   2733     1235   2580     1908   2131
  9157850    993600      139   3023      473   2989      827   2911
  9551777   1096200      604   3031     1001   2924     1484   2711
  9699265    999180      672   3041     1036   2937     1473   2744
  9887266    549120     1179   2915     1421   2805     1725   2629
 14281930   2291520       33   3779      649   3723     1331   3537
 14751841   3127320       79   3840      920   3729     2000   3279
 18870865   3714480       96   4343      976   4233     2048   3831
 19734650   1330560     1741   4087     2183   3869     2921   3347
 21651370   2735040      891   4567     1551   4387     2389   3993
 22163530   1404480     1629   4417     2023   4251     2513   3981
 22884277   4395300      486   4759     1474   4551     2946   3769
 23386441    683760     1429   4620     1596   4565     1771   4500
 26809445   4241160      778   5119     1679   4898     2911   4282
 29713450   2735040        7   5451      511   5427     1029   5353
 30002050   2542848     1917   5131     2555   4845     3669   4067
 31676033   4781700     1057   5528     2023   5252     3488   4417
 33826325   4823172     1223   5686     2185   5390     3698   4489
 36133681   7332780      160   6009     1420   5841     3000   5209
 37425389   5242860     1358   5965     2365   5642     4067   4570
 37661026   2914560     1901   5835     2499   5605     3251   5205
 38878705   4084080     1063   6144     1776   5977     2592   5671
 42639466   7512960      875   6471     2135   6171     4021   5145
 50205361   6846840     1280   6969     2360   6681     3769   6000

For a conjecture related to this table, see the note entitled
No Four Rectangles in Line?

Suppose we place each of our three putative right triangles with their 
common hypotenuses on the x axis as shown below.  (Only the a,b triangle
edges are shown.)

          

We want the areas of our three rational right triangles to be in 
arithmetic progression, and since the triangles are on a common
base, it's clear that their heights must be in arithmetic progression.
Also, if the edges of the right triangle are rational, their areas
and therefore their heights are rational, so the y coordinates of their
upper vertices are rational.  We can also show that the x coordinates of
those vertices are rational because, referring to the above figure, we 
have

              4 u^2 v^2             (u^2 - v^2)^2
         m = -----------        n = -------------
             (u^2 + v^2)             (u^2 + v^2)

Therefore, if we re-scale the figure so that the radius of the
circumscribing circle is unity (which we can do by dividing each
length by half the length of the hypotenuse), the three upper vertices
have rational x,y coordinates on the unit circle.  Now, the usual way of
parameterizing rational points on a unit circle is with the mapping

                  1 - t^2             2t
              x = -------       y = -------                   (6)
                  1 + t^2           1 + t^2

(It follows that t = y/(1+x).)  So, letting r,s,t denote the rational 
parameters of our three points on the circle, the requirement for their
heights (i.e., their y coordinates) to be in arithmetic progression gives
the following necessary condition for a complete solution of (5)

                r           s           2t
             -------  +  -------  =  -------                  (7)
             1 + r^2     1 + s^2     1 + t^2

It's important to note that although this is necessary, it is NOT
sufficient, because while the rationality of the edges implies 
rationality of the x,y coordinates, the converse is not true.  We
can have rational solutions of (7) that don't correspond to triangles
with rational edge lengths, even though the vertices have rational
coordinates.  Here are the first 41 primitive sets of three 
rational points on the unit circle with heights in arithmetic
progression, which correspond to three rational right triangles 
on a common hypotenuse with areas in arithmetic progression:

   H    del y   x1     y1    x2    y2    x3    y3      rs      r/s
  ----  -----  ----  ----  ----  ----  ----  ----    ------  ------
   65    -4      25    60    33    56    39    52    0.3333  1.3333
  185    44     175    60   153   104   111   148    0.0833  0.3333
  205    39     200    45   187    84   164   123    0.0370  0.3333
  221    55     204    85   171   140   104   195    0.1200  0.3333
  425  -104     119   408   297   304   375   200    0.1875  3.0000
  725   161     696   203   627   364   500   525    0.0612  0.3333
  901   -15     424   795   451   780   476   765    0.3333  1.0800
 1105  -256     264  1073   744   817   952   561    0.2138  2.8739
 1105  -276    1020  1020   817   744  1001   468    0.1067  2.1600
 1469   185    1456   195  1419   380  1356   565    0.0133  0.3333
 1769  -260     319  1740   969  1480  1281  1220    0.3333  2.0833
 1885   -23     312  1859   427  1836   516  1813    0.6389  1.1206
 1885  -231     464  1827  1003  1596  1300  1365    0.3333  1.8148
 1885  -343    1643  1643  1365  1300  1624   957    0.1270  1.7076
 2405   -68     483  2356   741  2288   925  2220    0.5439  1.2237
 2425  -264     679  2328  1273  2064  1625  1800    0.3333  1.6875
 2465   376    2431   408  2337   784  2175  1160    0.0208  0.3333
 2465   729    2340   775  1953  1504  1044  2233    0.1026  0.2535
 3445  -484    1749  2968  2387  2484  2805  2000    0.1829  1.7857
 3485  -117     236  3477   925  3360  1276  3243    0.6365  1.3718
 3625  -129    1276  3393  1577  3264  1820  3135    0.3986  1.2024
 3965   649    3900   715  3723  1364  3416  2013    0.0248  0.3333
 4505 -1036     689  4452  2937  3416  3825  2380    0.2449  3.0000
 4745   976    4599  1168  4233  2144  3575  3120    0.0469  0.3333
 4745  -329    2144  4233  2697  3904  3120  3575    0.2793  1.3518
 5069  -455    1644  4795  2619  4340  3256  3885    0.3333  1.5306
 5185  -243     144  5183  1575  4940  2196  4697    0.6189  1.5284
 5525   511    5500   525  5427  1036  5304  1547    0.0068  0.3333
 5525  -324    1131  5408  2163  5084  2805  4760    0.4643  1.4219
 5525   961    5084  2163  4557  3124  3720  4085    0.0901  0.4614
 6205   679    5244  3317  4747  3996  4080  4675    0.1317  0.6374
 6565 -1551    1300  6435  4387  4884  5656  3333    0.2231  3.0000
 6641  1420    6409  1740  5841  3160  4809  4580    0.0533  0.3333
 6929  -920    1521  6760  3729  5840  4879  4920    0.3333  1.9200
 7085  2759    7076   357  6363  3116  3960  5875    0.0134  0.0474
 7565  1079    7476  1157  7227  2236  6800  3315    0.0178  0.3333
 7565 -1344    2236  7227  4756  5883  6052  4539    0.2458  2.2121
 8621 -2135    8155  8155  6171  6020  7696  3885    0.1157  2.0417
 9425   -49     776  9393  1233  9344  1560  9295    0.7791  1.0882
 9425  1587    8372  4329  7337  5916  5704  7503    0.1206  0.4905
 9881 -2360    2169  9640  6681  7280  8569  4920    0.2133  3.0000

In none of these cases are the edge lengths of the triangles rational, 
so they don't represent solutions of (5).  This raises the question 
of whether we can prove that a rational solution of (7) (which is a 
necessary condition for (5)) can NEVER give rational edge lengths.  
If we could prove this, then it would follow that no solution of (5) 
exists, and therefore no magic square of squares exists.  I don't 
have a proof, but it's interesting that we can rule out a large class 
of possible solutions by means of a certain "discordant form", as 
explained below.

Note that (7) is a quadratic in t, and if we solve for t we get an 
expression containing a square root of the quantity

          [(1+r^2)(1+s^2)]^2  -  [(r+s)(1+rs)]^2        (8a)

so a sufficient condition is for this to be a rational square.  On 
this basis equation (8a) implies a rational number q such that

      q^2  +  [(r+s)(1+rs)]^2  =  [(1+rs)^2 + (r-s)^2]^2

Dividing both sides by the right hand side, we can put this into a
form that closely resembles the usual parametric representation of
a circle         
                      _                 _
                     |     / r + s \     |2
                     |    ( ------- )    |
                     |     \1 + rs /     |
             Q^2  +  | ----------------- |  =   1             (9)
                     |       / r - s \2  |
                     |  1 + ( ------- )  |
                     |_      \1 + rs /  _|

where Q is an arbitrary rational number.  One obvious class of 
solutions to (9), in view of (6), is given by identifying the 
quantity (r-s)/(1+rs) with the rational parameter T, and then 
setting the quantity (r+s)/(1+rs) to either 1-T^2 or 2T, equating 
the large fraction in the above expression with x or y (respect-
ively) of equations (6).  

In the first case, we set the quantity (r+s)/(1+rs) to 1-T^2, 
leading to a quadratic equation in r and s which can be solved 
for either variable to give an expression involving the square 
root of a number of the form  5q^2 + 2q + 5  where q is rational, 
and we require this to be a rational square.  Thus there must be 
integers m,n,k (not all divisible by 3) such that

               5m^2 + 2mn + 5n^2  =  k^2

but this is impossible, as can be seen by inspection modulo 9.

In the second case, after identifying (r-s)/(1+rs) with the 
rational parameter T, we set quantity (r+s)/(1+rs) to 2T, which 
implies the condition r = 3s, or, by symmetry, r=s/3 or r=1/(3s) 
or r=3/s).  This obviously gives an infinite family of solutions, 
i.e., sets of three rational points on a unit circle with heights 
in arithmetic progression, and in fact we see that 22 of the 41 
sets listed above are of this form.  In addition, three more are 
of the form s=1/(3r), which is another infinite family of solutions, 
as shown in the note Bi-Rational Substitutions Giving Squares.

However, we can show that the EDGE LENGTHS corresponding to r,s are 
NEVER rational if r = 3s or 1/(3s).  To prove this, note that for 
any given vertex parameter t, corresponding to the vertex x,y, one 
of the edge lengths is given by 

                  E^2 = (1-x)^2 + y^2

Substituting the expressions from (6) into this equation E^2 as a 
function of the parameter t

                     4t^2
                    -------  =  E^2
                    1 + t^2            

Consequently, for the two parameters r and s, with r=3s, we require
both of the quantities  4s^2/(1+s^2)  and  36s^2/(1+9s^2)  to be
rational squares.  Inverting, and noting that the numerators are
already squares, we see that we must have 1+s^2 and 1+9s^2 both
equal to rational squares for some rational s.  Clearing fractions,
we need integers m,n such that 

                      m^2 + n^2  =  square

                      m^2 + 9n^2  =  square

This is a problem of Concordant Forms, and it's known that there
are no integers m,n that make both m^2 + n^2 and m^2 + kn^2 for many 
specific values of k, including 9.  Thus there can be no solution
to these equations.  An almost identical argument works for the case
r=1/(3s).

Therefore, none of the rational solutions of (7) given by setting
rs or r/s equal to 1/3 can possibly give three rational right 
triangles with areas in arithmetic progression.  This shows an 
interesting relation between concordant forms and the obstacles
to constructing a magic squares of squares.

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