Re: How to do quickest

*To*: mathgroup at smc.vnet.net*Subject*: [mg123065] Re: How to do quickest*From*: Artur <grafix at csl.pl>*Date*: Tue, 22 Nov 2011 05:35:27 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <febdfdb5-c1cc-472c-afd6-e1768e8723e1@wrimail02.wolfram.com>*Reply-to*: grafix at csl.pl

Dear Daniel and Rest, Thank Your very much for improove my procedure! m have simplest form but can be + or - n is little bit more complicated but have only one sign Generally because I don't find another cases as m is root of quadratic polynopmial or 12 degree polynomial when non zero coefficient occured in even powers of variable (that mean is square root of 6 degree polynomial root) but for me aren't interesting this last cases we can uses IntegerQ[m^2] Best wishes Artur W dniu 2011-11-21 23:18, Daniel Lichtblau pisze: > ----- Original Message ----- >> From: "Artur"<grafix at csl.pl> >> To: mathgroup at smc.vnet.net >> Sent: Monday, November 21, 2011 3:29:38 AM >> Subject: How to do quickest >> >> Dear Mathematica Gurus, >> How to do quickest following procedure (which is very slowly): >> >> qq = {}; Do[y = Round[Sqrt[x^3]]; >> If[(x^3 - y^2) != 0, >> kk = m /. Solve[{4 m^2 + 6 m n + n^2 == >> x, (19 m^2 + 9 m n + n^2) Sqrt[m^2 + n^2] == y}, {m, >> n}][[1]]; >> ll = CoefficientList[MinimalPolynomial[kk][[1]], #1]; >> lll = Length[ll]; >> If[lll< 12, Print[{x/(x^3 - y^2)^2, kk, x, y, x^3 - y^2}]; >> If[Length[ll] == 3, Print[{kk, x, y}]]]], {x, 2, 1000000}]; >> qq >> >> >> (*Best wishes Artur*) > This is somewhat faster, > > rootpoly = > m /. First[ > Solve[{4 m^2 + 6 m n + n^2 - > xx, (19 m^2 + 9 m n + n^2)^2 (m^2 + n^2) - yy^2} == 0, {m, n}]] > > -Sqrt[Root[ > xx^6 - 2*xx^3*yy^2 + yy^4 + (270*xx^3 - 270*yy^2)*#1^3 - > 2916*xx*#1^5 + > 3645*#1^6& , 1]] > > I only had patience to go to 10^3. > > nn = 3; > Clear[x, y, kk, lll] > Timing[Do[ > If[! IntegerQ[Sqrt[x]], > y = Round[Sqrt[x^3]]; > kk = rootpoly /. {xx -> x, yy -> y}; > lll = Exponent[MinimalPolynomial[kk][[1]], #1]; > If[lll< 12, Print[{lll, x/(x^3 - y^2)^2, kk, x, y, x^3 - y^2}]]; > ], {x, 2, 10^nn}];] > > Faster still would be to figure out a priori conditions for which rootpoly is of degree less than 12. Then just test {x,y} pairs to see which ones qualify. Other possibilities for this might include working with > > Resultant[ > 4 m^2 + 6 m n + n^2 - x, (19 m^2 + 9 m n + n^2)^2 (m^2 + n^2) - > y^2, n] > > Out[9]= 3645*m^12 - 2916*m^10*x + 270*m^6*x^3 + x^6 - 270*m^6*y^2 - > 2*x^3*y^2 + y^4 > > Specifically one wants to know for which {x,y} pairs in the loop this will factor. > > Daniel Lichtblau > Wolfram Research > >

**Follow-Ups**:**Re: How to do quickest***From:*Artur <grafix@csl.pl>