Re: How to do quickest

• To: mathgroup at smc.vnet.net
• Subject: [mg123053] Re: How to do quickest
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Tue, 22 Nov 2011 05:33:08 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com

```----- 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

```

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