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.

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