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February 7th, 2014, 10:57 AM   #31
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Re: Pre-Q&A

Quote:
Originally Posted by CRGreathouse
I suspect one of the factors can be a quadratic and only one of the coefficients needs to be 10
The second part is not necessarily needed. In fact, I think I just stepped out of the barrier.

(x^2 - 3x + 3)(x^4 + 3x^3 + 6x^2 + 9x + 9) = x^6 + 27

I optimized the coefficients to 0 instead of 1
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February 7th, 2014, 11:38 AM   #32
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Re: Pre-Q&A

Quote:
Originally Posted by mathbalarka
Quote:
Originally Posted by CRGreathouse
I suspect one of the factors can be a quadratic and only one of the coefficients needs to be 10
The second part is not necessarily needed.
Of course I mean this in the context of the original problem, not the simpler variant you solved (congrats!). So I understand that it can be done with all coefficients <= 10; in fact I think it can be done with one coefficient = 10 and all other coefficients < 10.
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February 7th, 2014, 11:57 AM   #33
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Re: Pre-Q&A

Well, I figured anything can be more or less reduced to anything. For example,

(x^2 - 4x + 5)(x^5 + 4x^4 + 11x^3 + 24x^2 + 41x + 44) = x^7 + 29x + 220

But the goal of optimizing coefficients is something new to me, I am not sure how to approach that.

EDIT Just for the record, a little tweaking of the above binomial form to make f(2) prime gives x^8 + 3x^7 + 3x^6 + 3x^5 + 27, so the minimal up until now is {0, ..., 27}
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February 7th, 2014, 12:14 PM   #34
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Re: Pre-Q&A

I found an example which meets my requirements. As question-poser I won't post it (unless requested), but now I can say that it is solvable for sure!

As I understand it it is an open problem to find an instance with all coefficients below 10, or to prove that no such exists.
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February 7th, 2014, 12:20 PM   #35
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Re: Pre-Q&A

The minimal is now at {0, ..., 24}. x^8 + 9x^3 + x^2 + 24x + 3.
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February 7th, 2014, 10:48 PM   #36
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Re: Pre-Q&A

(x^2 - 3x + 3)(x^6 + 3x^5 + 6x^4 + 9x^3 + 9x^2 + 7x + 1) = x^8 + 7x^3 + 7x^2 + 18x + 3 so the record is {0, ... , 18}

EDIT Pushed to {0, ..., 16} : (x^2 - 3x + 3)(x^6 + 3x^5 + 6x^4 + 9x^3 + 9x^2 + 4x + 1) = x^8 + 4x^3 + 16x^2 + 9x + 3
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February 7th, 2014, 11:37 PM   #37
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Re: Pre-Q&A

Good work by our resident Transcendental Number Theorist!

It's not optimal but here's what i found using the method of [color=#00FF00]Mr. TNT[/color].



f(2) = 191

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February 7th, 2014, 11:44 PM   #38
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Re: Pre-Q&A

Good job! {0, ..., 27} associated trinomial is a new thing, since {0, ..., 27} was achieved by the beast x^8 + 3x^7 + 3x^6 + 3x^5 + 27. (Putting it in the notes with explicit factorization, we might need it later). I am trying hard not to code it up and search for such a polynomial, it would be quite a bit unfair to do so. As far as now, I did everything by hand, except the multiplication of polynomials.
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February 7th, 2014, 11:55 PM   #39
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Re: Pre-Q&A

[color=#C0C0C0]Using agent's factorization, by reduction of the constant term, I get

x^7 + 5x^2 + 12x + 15 = (x^5 + 3x^4 + 6x^3 + 9x^2 + 9x + 5)(x^3 - 3x + 3)

so record is {0, ..., 15} Infinitesimal improvements. Dang.[/color]

EDIT I realized that agent took the binomial I have gave before and multiplied it by x and added a 9 to get the second part going. Nice!
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February 8th, 2014, 12:03 AM   #40
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Re: Pre-Q&A

I was thinking the same thing about putting it in the notes and as a matter of fact i'm tempted to do some investigations about factorability of trinomial



I haven't looked for this form on the net but if you derive something or find something , please post.

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