Bi-Rational Substitutions Giving Squares

In the note Discordance Impedes Square Magic we considered the 
requirement to make the quantity

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

a rational square for rational values of r and s.  A sufficient
condition for this to be a square is if each of the quantities

     f(x,y)  =   [(1+r^2)(1+s^2)]  -  [(r+s)(1+rs)]        (1)

     F(x,y)  =   [(1+r^2)(1+s^2)]  +  [(r+s)(1+rs)]        (2)

is a square.  Any such solution gives a set of three rational points 
on the unit circle with heights in arithmetic progression.  The first 
several primitive solutions were listed in that note.  The actual 
values of r and s can be inferred from that table using the formulas

                     y1                y3
              r =  -------       s = ------
                   H + x1            H + x3

From the columns listing the values of rs and and r/s in that table
we can identify the solutions arising from the simple infinite 
families corresponding to r = 3s, 3/s, s/3, and 1/(3s).  Notice that 
if we substitute s = 3r into (1) and (2) with any rational r we have 
the square factorizations

             /              \ 2                 /              \ 2
  f(r,3r) = ( 3r^2 - 2r +  1 )       F(r,3r) = ( 3r^2 + 2r +  1 )
             \              /                   \              /

and similarly if we substitute s = 1/(3r) into (1) and (2) we have 
the square factorizations

              / 3r^2 - 2r + 1 \ 2                 / 3r^2 - 2r + 1 \ 2
 f(r,1/3r) = (  -------------  )     F(r,1/3r) = (  -------------  )
              \      3r       /                   \      3r       /

Thus, either of these conditions is sufficient to make both the
quantities f and F factor as squares, and we notice that of the 41 
solutions given in the article noted above, 22 of them are of this
form, so these two cases represent a large subset of all the 
solutions in this range.

In addition to those, the table contains 19 other solutions, whose r 
and s values don't fall into this simple "r=3s" family.  Following 
is a list of these "other" rational pairs (r,s).

              r               s
         ----------      ----------
          29    37         3    11
          12    25         2     9
          11    13        37    49
        1643  3528         3    11
          31    38         2     3
           5    31         7    11
           4     7         8    25
          57    61        47    69
           9    13        19    33
          51    83         5    11
          71    73         7    11
          13    16         4     7
          21   103        19    43
          31   107         5    11
           3   119        25    47
          73    99         1     3
          35    72         5    21
          93   101        11    13
           9    37        61   123

Notice that all four of the substitutions s=3r, s=1/(3r), s=3/r, 
and s=r/3 are linear fractional transformations (LFTs).  This
suggests that some of the other (r,s) solutions might be members of
other infinite families with algebraic factorizations given by some
other LFTs (with integer coefficients).  However, rather than trying
to find LFTs that make f and F both algebraic squares simultaneously, 
let's just try to make f an algebraic square.  In other words, we 
seek integers a,b,c,d such that

                /   ar+b \       / g(x) \ 2
              f( r, ----  )  =  ( ------ )
                \   cr+d /       \ cr+d /

It turns out that if T(r) is such an LFT, then by symmetry so are
the LFTs given by  1/T(r), invT(r), and 1/invT(r), so we only need
to consider one of each such set of four related solutions.  The
smallest LFTs with this property have the coefficients listed
below, along with the values of num/den = (a+d)^2 / (ad-bc).

           a     b     c     d       num   den
          ---   ---   ---   ---      ---   ---
          -1     0     0     1         0    -1
           3     0     0     1        16     3
           3    -3     3     5         8     3
           8    -3     3     0        64     9
           5    -8     8     7        16    11
          15    -8     8    -3       144    19
           7   -15    15     9         8     9
           8    -5     5    16        64    17

This apparently gives an infinite family of LFTs, each of which makes
equations (1) algebraically factorable into a square.  Also, it appears
that the LFTs that make the "conjugate" expression

     F(x,y)  =   [(1+r^2)(1+s^2)]^2  +  [(r+s)(1+rs)]^2        (2)

into an algebraic square are the same, EXCEPT that the b and c
coefficients are transposed.  As a result, one of these LFT's will
make BOTH (1) and (2) into algebraic squares if b=c, but this is
only the case for the first two LFTs in the above table, the first
of which is just the trivial solution r=-s.  Thus the only non-
trivial solution is (3r+0)/(0r+1) and its three associates, i.e.,
the set 3r, 1/(3r), r/3, 3/r.

Since the table of r,s values given above represents pairs that
make BOTH f and f into algebraic squares, and since they can't be
represented by any of the linear fractional transformations, this
shows that there are numerical solutions that make f a rational
square but that are not members of any infinite family based on
an LFT substitution.