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           (0)

a rational square for rational values of r and s.  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

A sufficient (but not necessary) condition for (0) to be a square is 
for each of the quantities

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

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

to be a square.  Several (but not all) of the solutions of (0) in 
that table also satisfy (1) and (2).  From the columns listing the 
values of rs and and r/s in that table we can identify the solutions 
arising from 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.   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.

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 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.

By the way, in addition to the solutions of (0) that have square
f and F factors as discussed above, the table also contains 19 other 
pairs r,s that make (0) a square but that do not make (1) or (2) a
square.  We can recognize these because the r and s values don't 
fall into any of the simple "r=3s" families.  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

For example, taking r = 29/37 and s = 3/11 we have

                                              / 193440 \2
    [(1+r^2)(1+s^2)]^2 - [(r+s)(1+rs)]^2  =  ( -------- )
                                              \ 165649 /

This makes (0) a square, but does not make either (1) or (2)
a square.