Let inf /===\ a a/24 | | a n h := eta(q ) = q | | (1 - q ). a | | n = 1 Then if r,s,t are rationals (or w.l.o.g. integers), it appears that there will always be some polynomial P_r,s,t (with integer coefficients) relating h_r, h_s, and h_t. E.g., 8 16 16 8 24 P = 16 h h + h h - h , 1, 2, 4 1 4 1 4 2 (Euler's "aequatio identica satis abstrusa", but our least abstruse.) 3 9 6 6 9 3 12 P = 27 h h + 9 h h + h h - h , 1, 3, 9 1 9 1 9 1 9 3 6 8 12 12 8 6 24 2 24 2 18 8 P = - 3125 h h h - 250 h h h + 256 h h + h h - h h , 1, 2, 5 1 2 5 1 2 5 2 5 1 5 1 2 12 24 48 24 24 36 36 24 24 P = 387420489 h h h + 19131876 h h h + 196830 h h h 1, 2, 3 1 2 3 1 2 3 1 2 3 72 12 24 48 12 48 24 12 72 12 60 24 - 16777216 h h - 196608 h h h - 12 h h h - h h + h h , 2 3 1 2 3 1 2 3 1 3 1 2 24 12 48 24 24 36 24 36 24 P = 387420489 h h h + 19131876 h h h + 196830 h h h 1, 2, 6 1 2 6 1 2 6 1 2 6 72 12 24 48 12 48 24 12 72 12 24 60 - 4096 h h - 12 h h h - 48 h h h - h h + h h , 2 6 1 2 6 1 2 6 1 6 1 2 12 72 12 24 48 60 24 12 48 24 P = 16777216 h h + 196608 h h h - 729 h h + 12 h h h 1, 3, 6 1 6 1 3 6 3 6 1 3 6 24 36 24 36 24 24 48 12 24 12 72 - 270 h h h - 36 h h h - h h h + h h , .... 1 3 6 1 3 6 1 3 6 1 3 These last two are interderivable with the (ever amazing) eta transformation formula that effectively takes P_a,b,c <-> P_a,ac/b,c. (The fancier modular eta transformations turn, e.g., P_1,2,3 into seriously weird (and algebraically independent) powers of q^pi^2 times e^(i pi^3) times q^(i pi ln q), a "prodigal square" (quadrigal?).) Similarly, P_1,3,9 could be derived from P_1,2,3 (= P_3,6,9) and P_1,3,6 via a resultant eliminating h_6. Thus, in principal, one need only tabulate P_1,2,p(rime). Perhaps by examining enough P_a,b,a+b, one could guess an addition formula relating eta(q), eta(p), and eta(pq). This optimism fades as the table grows: 4 12 48 12 12 40 20 12 32 P = 68719476736 h h h + 12884901888 h h h + 855638016 h h h 1, 3, 4 1 3 4 1 3 4 1 3 4 28 12 24 48 16 12 36 16 + 23068672 h h h - 6198727824 h h - 306110016 h h h 1 3 4 3 4 1 3 4 24 24 16 36 12 16 48 16 8 48 8 - 3149280 h h h + 196800 h h h - 16 h h - 387420489 h h h 1 3 4 1 3 4 1 4 1 3 4 20 36 8 32 24 8 44 12 8 56 8 52 12 - 19131876 h h h - 196830 h h h + 12 h h h - h h + h h , 1 3 4 1 3 4 1 3 4 1 4 1 3 96 108 6 36 5 288 6 P = 2 3 eta (q) eta (q ) eta (q ) 1, 5, 6 90 100 30 36 5 264 6 + 7 2 3 eta (q) eta (q ) eta (q ) + two hundred forty one similar terms 288 18 24 294 12 24 300 6 24 294 36 + 9 h h h - 6 h h h - h h h + h h . 1 5 6 1 5 6 1 5 6 1 5 I've had no luck reaching P_1,x,y, 1<x<y, y = 13, 17, or 19. (Or perhaps *not* reaching them is luck.) But notice how nice are P_1,2,4 and P_1,3,9. An alternative spanning scheme would be P_1,n,n^2, so, let's try n=6: 312 168 336 792 P = 6 h h h - 20323 * 324949 * 9624311 1, 6, 36 1 6 36 216 1008 72 * 23381814648630788600514603407190631043 h h h 1 6 36 plus 878 similar terms. Maple claims it's irreducible. Macsyma (compiled for Windows NT!) boggles. A few hundred Taylor terms are zero, but it is difficult to explore even the foothills of such a mountain. In the immortal words of Professor Doctor Bartolomeo Simpson: Eta my shorts. --rwg