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.

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