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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

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.

 February 7th, 2014, 11:57 AM #33 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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}
 February 7th, 2014, 12:14 PM #34 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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.
 February 7th, 2014, 12:20 PM #35 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Pre-Q&A The minimal is now at {0, ..., 24}. x^8 + 9x^3 + x^2 + 24x + 3.
 February 7th, 2014, 10:48 PM #36 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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
 February 7th, 2014, 11:37 PM #37 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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]. $f(x) \= \ (x^2 - 3x + 3)(x^5 + 3x^4 + 6x^3 + 9x^2 + 9x + 9) \ = \ x^7 + 9x^2 + 27$ f(2) = 191
 February 7th, 2014, 11:44 PM #38 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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.
 February 7th, 2014, 11:55 PM #39 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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!
 February 8th, 2014, 12:03 AM #40 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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 $f(x) \= \ x^n + ax^m \pm r$ I haven't looked for this form on the net but if you derive something or find something , please post.

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