(= integer lattice reducer). I think I sent this long ago, but I just stumbled over the mail item, which I can never find when I'm looking for it.) In the late 70s, Robert Maas worked on adaptive data compression, and needed to solve these transcendental equations, In[597]:= FindRoot[{(1 - x)^(y - 2) x^(3 - 2 y) (1 - y)^(1 - y) y^y == 1, 1 == (1 - x)^(1 - x) x^x (1 - y) y^(x - 3)}, {x, 1`44/2}, {y, 1/2}, WorkingPrecision -> 44] for the coupled probabilities x and y: Out[597]= {x -> 0.64779887126104238549049829642576930694948772, y -> 0.54368901269207636157085597180174798652520330} We were curious whether they were algebraically independent, and fed their squares and crossterms into NUMREL, which Maas had implemented following an informal conversation with Rich. Here is a re-enactment with Mathematica's PSLQ. In[618]:= List @@ Expand[(1 + y + x)^2] Out[618]= {1, 2 x, x^2, 2 y, 2 x y, y^2} In[619]:= %.FindIntegerNullVector[% /. %597] Out[619]= 1 - 2 x y - y^2 Success! Is there any chance of a second relation? Try cubics: In[620]:= List @@ Expand[(1 + y + x)^3] Out[620]= {1, 3 x, 3 x^2, x^3, 3 y, 6 x y, 3 x^2 y, 3 y^2, 3 x y^2, y^3} In[621]:= %.FindIntegerNullVector[% /. %597] // Factor Out[621]= -3 y (-1 + 2 x - 2 x^2 + y) x and y are independently algebraic! Eliminating y, In[622]:= GroebnerBasis[{%/y, %%%}, {y, x}] // Factor Out[622]= {x (-1 + 2 x - 2 x^2 + 2 x^3), -1 + 2 x - 2 x^2 + y} So both satisfy cubics. In[625]:= Solve[0 == # & /@ (%622/{x, 1})][[1]] Out[625]= {x -> 1/3 (1 - (2 2^(2/3))/(13 + 3 Sqrt[33])^(1/3) + (13 + 3 Sqrt[33])^( 1/3)/2^(2/3)), y -> 1/18 (-6 + (32 2^(1/3))/(13 + 3 Sqrt[33])^(2/3) + ( 8 2^(2/3))/(13 + 3 Sqrt[33])^(1/3) - 2 (2 (13 + 3 Sqrt[33]))^(1/3) + (2 (13 + 3 Sqrt[33]))^(2/3))} In[627]:= N@% Out[627]= {x -> 0.6477988712610423, y -> 0.5436890126920761} as approximated earlier. --rwg