RWG: For your cube root: Pick a large prime P, maybe 500 or 1000 digits, congruent to -1 (mod 15). This guarantees that cubing and ^5 are 1-1 maps, so cube/fifth roots exist and are unique. Working mod P, compute x as cbrt(stuff), and use the number recognizer to see if x is a rational function of 5^(1/3),2^(1/5),3^(1/5). That's about 150 terms; will NumRec handle that? You need terms for 5^a/3 * 2^b/5 * 3^c/5 and (same)*x. If NumRec balks, you might try using 150 different primes, and directly solving the matrix.
Rich
Rich, note that throwing in a factor of 25 reduces many days to Timing[Factor[x^3-(25*3^(3/5)-25*2^(1/5)),Extension->{2,3}^(1/5)]] {0.297, (2^(2/5) + 3^(1/5) + 2^(3/5) 3^(2/5) - 2^(1/5) 3^(3/5) - x) (5 2^(4/5) - 5 2^(2/5) 3^(1/5) - 5 3^(2/5) + (-2^(2/5) - 3^(1/5) - 2^(3/5) 3^(2/5) + 2^(1/5) 3^(3/5)) x - x^2)} < 1/3 sec. ! ? linrel is fast enough for a few dozen, but has a bug where it gets lost around 15 or so. Mma's LLL can go a few hundred, but I can guess enough structure to In[64]:= Times @@ # & /@ CP[rutset[27/2, 15], rutset[5, 3]] Out[64]:= {1, 5^(1/3), 5^(2/3), 3^(1/5)/2^(1/15), (3^(1/5) 5^(1/3))/2^(1/15), (3^(1/5) 5^(2/3))/2^(1/15), 3^(2/5)/2^(2/15), (3^(2/5) 5^(1/3))/2^(2/15), (3^(2/5) 5^(2/3))/2^(2/15),...} (CP:= CartesianProduct.) Then with LR := linrelized LLL, In[85]:= Block[{L = Prepend[%64, ((27/2)^(1/5) - 1)^(1/3)], M}, M = LR[L, LeafCount[%64]/4]; Print[L.Rest[M[[1]]]]; M][[{1, 2, 3}]] (9*3^(1/5)*5^(1/3))/2^(1/15)+9*2^(8/15)*3^(2/5)*5^(1/3)-9*2^(2/15)*3^(3/5)*5^(1/3) +9*10^(1/3)-45*(-1+3^(3/5)/2^(1/5))^(1/3) Out[85]:= {{-8, -45, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, -2, 0, 0, 0, 0}, {-787, 80, 2599, 1018, 289, 1325, -548, 119, -1386, -77, 2190, -780, -273, -967, -234, -646, 1421, -179, 761, -525, -874, -181, -1059, 151, -657, 1491, -503, 435, 624, -222, 556, -1089, 300, 1512, -2820, 928, 1539, 1864, 217, -737, -1152, -128, -1075, 756, 67, -253, 4},...} alternatively In[89]:=Block[{L = Prepend[Times @@ # & /@ CP[rutset[5, 3], rutset[18, 15]], (1 - 18^(1/5)/3)^(1/3)], M}, M = LR[L, LeafCount[L]/4]; Print[L.Rest[M[[1]]]]; M][[{1, 2, 3}]] 3*5^(1/3)+3*2^(3/5)*3^(1/5)*5^(1/3)-3*2^(1/5)*3^(2/5)*5^(1/3)+2^(2/5)*3^(4/5)*5^(1/3) -15*(1-2^(1/5)/3^(3/5))^(1/3) Out[89]:= {{-1, -15, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, -3, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {-3918, -305, 4558, -2943, 2084, 410, -231, 3243, 1724, -1800, 3734, -1203, 973, 3671, -1341, 687, 3825, -2768, -1771, -2741, 438, 1370, -1675, -2849, -177, 2675, 4046, -1408, 1185, 90, -3069, 1280, 1325, 1147, -1339, -2624, -38, -2374, 1633, -1050, 636, -1427, -513, -1876, -395, 25, 0},...} alternatively In[90]:=Block[{L = Prepend[Times @@ # & /@ CP[rutset[5, 3], rutset[2/27, 15]], (1 - 18^(1/5)/3)^(1/3)], M}, M = LR[L, LeafCount[L]/4]; Print[L.Rest[M[[1]]]]; M][[{1, 2, 3}]] -5^(1/3)-(2^(2/5) 5^(1/3))/3^(1/5)-2^(3/5) 3^(1/5) 5^(1/3)+2^(1/5) 3^(2/5) 5^(1/3) +5 (1-2^(1/5)/3^(3/5))^(1/3) Out[90]:= {{1, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 3, 0, 0, -3, 0, 0, -9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {4926, -3156, 1253, 6785, -6444, -576, -7461, 619, 3581, -972, 2114, -2847, 1155, 6019, -6505, 5731, 1870, -9425, 6052, -2433, 7267, -1805, -4650, -1785, 5949, 9199, 2865, 2969, -750, 8537, 5209, -8450, -2160, -209, 7109, -2063, 1370, -11382, -1528, 6248, 3774, 696, 1667, 927, -81, 0, 0},...} so it looks pretty good for roots of binomials, at least. But I have been running a few days of sequential search with zero new discoveries. --Bill PERIDIASTOLE DEPILATORIES