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

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• Subject: [mg93321] Re: [mg93280] Trinomial decics x^10+ax+b = 0; Help with Mathematica code
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Tue, 4 Nov 2008 06:15:09 -0500 (EST)
• References: <200811020657.BAA02645@smc.vnet.net> <BB6FC818-9D69-4DA2-8F6A-0A5B680ECD2C@mimuw.edu.pl> <64fc87850811020944p49edb6a2u38aa53a29b539784@mail.gmail.com>

```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.
>
>
> 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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