Re: How to do quickest

*To*: mathgroup at smc.vnet.net*Subject*: [mg123080] Re: How to do quickest*From*: Artur <grafix at csl.pl>*Date*: Tue, 22 Nov 2011 07:23:40 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <febdfdb5-c1cc-472c-afd6-e1768e8723e1@wrimail02.wolfram.com> <201111221035.FAA28035@smc.vnet.net>*Reply-to*: grafix at csl.pl

P.S. Generally almost all polynomials on m are of 12 degreen. Only known exceptions are order 2 (only 23 such cases for x from 1 to 1248412) , but extremaly intersting will be to find cases of degree 3,4 and 6 (degree have to divided 12). best wishes Artur W dniu 2011-11-22 11:35, Artur pisze: > 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 >> >> >

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