I was surprised to get Jacobi's (least abstruse of all) aequatio satis abstrusa, a polynomial relating eta(q), eta(q^2), eta(q^4), abbreviated {1,2,4}, from the abstruser {1,2,3} := 0= -\[Eta][q]^60 \[Eta][q^2]^24+\[Eta][q]^72 \[Eta][q^3]^12+12 \[Eta][q]^48 \[Eta][q^2]^24 \[Eta][q^3]^12+196608 \[Eta][q]^24 \[Eta][q^2]^48 \[Eta][q^3]^12+16777216 \[Eta][q^2]^72 \[Eta][q^3]^12-196830 \[Eta][q]^36 \[Eta][q^2]^24 \[Eta][q^3]^24-19131876 \[Eta][q]^24 \[Eta][q^2]^24 \[Eta][q^3]^36-387420489 \[Eta][q]^12 \[Eta][q^2]^24 \[Eta][q^3]^48 (spurning Mma's DedekindEta, which expects period ratio tau instead of nome q.) {1,2,3} -> 6/{3,2,1} = {2,3,6} by Jacobi's imaginary transformation {1,2,3} -> {2,4,6} by q -> q^2 resultant({2,3,6},{2,4,6},eta(q^6)] = {2,3,4} (Warning: huge!) resultant({2,3,4},{1,2,3},eta(q^3)) = {1,2,4} quod erat desideratum. With the imaginary reciprocator and resultant you can apparently get any distinct triple of the form {2^a 3^b, 2^c 3^d, 2^e 3^f}, {a,b,c,d,e,f} nonnegative integers. Similarly for p and q, relatively prime, instead of 2 and 3, if you start with {1,p,q}. Strangely, with *three* relative primes, e.g., {2,3,5}, you seem to be limited to the 23 identities {{2,3,5},{4,6,9},{4,6,15},{4,6,25},{4,9,10},{4,9,15},{4,10,15},{4,10,25},{4,15,25},{6,9,10},{6,9,25}, {6,10,15},{6,10,25},{6,15,25},{9,10,15},{9,10,25},{9,15,25},{12,20,45},{12,20,75},{12,50,75}, {18,20,45},{18,45,50},{18,50,75}} The resultants are formidable. It might be easier to use undetermined coefficients on the series expansions. But not much. --rwg