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Re: Re: Trinomial decics x^10+ax+b = 0; Help

  • To: mathgroup at smc.vnet.net
  • Subject: [mg93377] Re: [mg93322] Re: [mg93280] Trinomial decics x^10+ax+b = 0; Help
  • From: Artur <grafix at csl.pl>
  • Date: Wed, 5 Nov 2008 04:55:24 -0500 (EST)
  • References: <200811020657.BAA02645@smc.vnet.net> <BB6FC818-9D69-4DA2-8F6A-0A5B680ECD2C@mimuw.edu.pl> <64fc87850811020944p49edb6a2u38aa53a29b539784@mail.gmail.com> <F85E373D-0B9C-4090-BA7B-AE95575625E1@mimuw.edu.pl> <200811041115.GAA29016@smc.vnet.net>
  • Reply-to: grafix at csl.pl

Dear Tito,

My slowly working Mathematica 6 procedure:
Timing[aa = {}; Do[Print[a];
Do[pol = x^10 + a x + b; f = Length[FactorList[pol]];
   If[f > 2, , dis = Discriminant[pol, x]; data = Sqrt[dis];
    sp = data /. Sqrt[_] -> 1; sfp = data/sp /. Sqrt[x_] -> x;
    c = FactorList[pol, Extension -> Sqrt[sfp]];
    If[Length[c] > 2, AppendTo[aa, pol]; Print[c];
     Print[pol]]], {b, -100, 100}], {a, 1, 100}]; aa]

Mayby sombody know how improove these to do much quckest???

If you mean trinomials I read 10 cases available:
1) x^10+a x+b
2) x^10+a x^2+b
3) x^10+a x^3+b
4) x^10+a x^3+b
5) x^10+a x^4+b
6) x^10+a x^5+b
7) x^10+a x^6+b
8) x^10+a x^7+b
9) x^10+a x^8+b
10) x^10+a x^9+b

ad 2) we can substitute x^2=y and we have quintic y^5+a y+b but also are 
cases that are not factorizable decic is factorizable over Sqrt[] extension:
-2+8 x^2+x^10 (Galois S(5)[x]2 non-solvable) quintic -2+8 x+x^5 (Galois 
S(5) non-solvable)
-5+40 x^2+x^10 (Galois S(5)[x]2 non-solvable) quintic -5+40 x+x^5 
(Galois S(5) non-solvable)
-98+88 x^2+x^10 (Galois S(5)[x]2 non-solvable) quintic -98+88 x+x^5 
(Galois S(5) non-solvable)
no more cases in range {b,-100,100} {a,1,100}

ad 3)
no such case in range {b,-100,100} {a,1,100}

ad 4) after substitution y=x^2 we have quintic y^5+a y^2+b
no such cases of decis in range {b,-100,100} {a,1,100}

ad 5) This case is trivial because after substitution y=x^5 we have 
x^2+a x+b and almost all polynomials are factorizable over Sqrt[]

ad 6)
-32+33 x^6+x^10 (Galois S(5)[x]2 non-solvable) -32+33 x^3+x^5 (Galois 
S(5) non-solvable)
=(-4 Sqrt[2]-8 x-4 Sqrt[2] x^2-x^3+Sqrt[2] x^4-x^5)(-4 Sqrt[2]-8 x-4 
Sqrt[2] x^2-x^3+Sqrt[2] x^4-x^5)

no more cases in range {b,-100,100} {a,1,100}

ad 7)
no such case in range {b,-100,100} {a,1,100}

ad 8)
no such case in range {b,-100,100} {a,1,100}

ad 9)
no such case in range {b,-100,100} {a,1,100}

Is strange for me that all available cases have non-solvable Galois 
groups ?!?.

Best wishes
Artur







Andrzej Kozlowski pisze:
> Sorry, I forgot that one has to Catch whatever might be Thrown ;-)
>
> Unfortunately, its unlikely to make any difference :-(
>
> Anyway, the correct code is:
>
> Catch[NestWhile[({a, b} = RandomInteger[{-10^6, 10^6}, {2}];
>       If[ck[a] == ck[b] == 0 && (p[a, b] || p[b, a]), Throw[{a, b}],
>        ck[a] = ck[b] = 1]) &, 1, True &, 1, 10^6]];//Timing
>
> Andrzej Kozlowski
>
> 3 Nov 2008, at 16:08, Andrzej Kozlowski wrote:
>
>   
>> I doubt that you will have much luck with this approach. I did:
>>
>> f = -a + m^9 - 8 m^7 n + 21 m^5 n^2 - 20 m^3 n^3 + 5 m n^4 ; g = -b  
>> + m^8 n -
>>  7 m^6 n^2 + 15 m^4 n^3 - 10 m^2 n^4 + n^5;
>>
>> p[x_, y_] :=
>> MemberQ[Exponent[FactorList[h /. {a :> x, b :> y}][[All, 1]], m], 5  
>> | 10]
>>
>> (mm = Outer[List, Range[10^3], Range[10^3]]); // Timing
>> {0.012557, Null}
>>
>> (vv = Apply[p, mm, {2}];) // Timing
>> {7077.94, Null}
>>
>> Position[vv, True]
>> {}
>>
>> There is not a single example among the first 10^6. You can try this  
>> approach on larger numbers, though eventually you will find yourself  
>> short of memory to story such large arrays. Or you could try random  
>> searches, with very much larger numbers, e.g.
>>
>> ck[_] = 0;
>>
>> NestWhile[({a, b} = RandomInteger[{-10^6, 10^6}, {2}];
>>     If[ck[a] == ck[b] == 0 && (p[a, b] || p[b, a]), Throw[{a, b}],
>>      ck[a] = ck[b] = 1]) &, 1, True &, 1, 10^4]; // Timing
>>
>> {0.481902, Null}
>>
>> As you see, 10^4 random pairs between -10^6 and 10^6, and no result.  
>> The "flag" ck, by the way, is used to mark the numbers we have  
>> already tried, so when we run this again we do not apply p to them  
>> again.
>>
>> Anyway, it looks to me much harder than the proverbial looking for a  
>> needle in a haystack.
>>
>> Andrzej Kozlowski
>>
>>
>>
>> On 3 Nov 2008, at 02:44, Tito Piezas wrote:
>>
>>     
>>> Hello Andrzej,
>>>
>>> The simple trinomials x^6+3x+3 = 0 and x^8+9x+9 = 0 are solvable,  
>>> both factoring over Sqrt[-3].
>>>
>>> However, there seem to be no known decic trinomials,
>>>
>>> x^10+ax+b = 0
>>>
>>> such that it factors over a square root (or quintic) extension.
>>>
>>> The 45-deg resultant, for some {a,b}, IF it has an irreducible 5th  
>>> or 10th degree factor, will give the {a,b} of such a decic.
>>>
>>> To find one, what I did was to manually substitute one variable,  
>>> like "a" (yes, poor me), use the Table[] function for "b" (as well  
>>> as the Factor[] function), and inspect the 45-deg to see if it has  
>>> the necessary 5th or 10th deg factors. Unfortunately, there are  
>>> none for {a,|b|} < 30.
>>>
>>> So what I was thinking is to extend the range to {a,|b|} < 1000.  
>>> But that's about 10^6 cases, far more than I could do by hand.
>>>
>>> I'm sure there is a code such that we only have to give the upper  
>>> bound for {a,b}, let Mathematica run for a while, and it prints out  
>>> ONLY the {a,b} such that the 45-deg has an irreducible 5th (or  
>>> 10th) factor. I believe there might be a few {a,b} -- but doing it  
>>> by hand is like looking for a needle in a haystack.
>>>
>>> But my skills with Mathematica coding is very limited. Please help.
>>>
>>> P.S. The 45-deg resultant should be in the variable "m", so pls  
>>> eliminate the variable "n" (not m) between the two original  
>>> equations.
>>>
>>> Sincerely,
>>>
>>> Tito
>>>
>>>
>>> On Sun, Nov 2, 2008 at 6:14 AM, Andrzej Kozlowski  
>>> <akoz at mimuw.edu.pl> wrote:
>>>
>>> On 2 Nov 2008, at 15:57, tpiezas at gmail.com wrote:
>>>
>>> Hello guys,
>>>
>>> I need some help with Mathematica code.
>>>
>>> It is easy to eliminate "n" between the two eqn:
>>>
>>> -a + m^9 - 8m^7n + 21m^5n^2 - 20m^3n^3 + 5mn^4 = 0
>>> -b + m^8n - 7m^6n^2 + 15m^4n^3 - 10m^2n^4 + n^5 = 0
>>>
>>> using the Resultant[] command to find the rather simple 45-deg
>>> polynomial in "m", call it R(m).
>>>
>>> As Mathematica runs through integral values of {a,b}, if for some
>>> {a,b} the poly R(m) factors, we are interested in two cases:
>>>
>>> Case1: an irreducible decic factor
>>> Case2: an irreducible quintic factor
>>>
>>> What is the Mathematica code that tells us what {a,b} gives Case 1 or
>>> Case 2?
>>>
>>>
>>> Thanks.  :-)
>>>
>>>
>>> Tito
>>>
>>>
>>>
>>>
>>> Let f = -a + m^9 - 8 m^7 n + 21 m^5 n^2 - 20 m^3 n^3 + 5 m n^4 ; g  
>>> = -b + m^8 n -
>>> 7 m^6 n^2 + 15 m^4 n^3 - 10 m^2 n^4 + n^5;
>>>
>>> and
>>>
>>> h = Resultant[f, g, m];
>>> Exponent[h, n]
>>> 45
>>>
>>> so h is a polynomial of degree 45. Now, let
>>>
>>> p[x_, y_] := Exponent[FactorList[h /. {a :> x, b :> y}][[All, 1]], n]
>>>
>>> computing p[a,b]  gives you the exponents of the irreducible factor  
>>> for the given values of a and b. In most cases you get {0,45} - the  
>>> irreducible case. But, for example,
>>> p[a, 0]
>>> {0, 1, 9}
>>> and
>>> p[2, 1]
>>> {0, 9, 36}
>>>
>>> So now you can search for the cases you wanted.  You did not  
>>> seriously expect Mathematica would do it by itself, I hope?
>>>
>>> Andrzej Kozlowski
>>>
>>>       
>
>
>
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