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February 6th, 2014, 07:31 PM   #21
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Re: Pre-Q&A

Quote:
 Originally Posted by agentredlum I haven't looked for the other reference he gives , Gross and Filaseta.
The problem is solved completely in Cole, Dunn, & Filaseta 2015 (?), but it hasn't yet been written. The authors tell me the paper should be ready in May, so if we're lucky we'll see it on the arXiv sometime in June. Gross & Filaseta solve a related problem with f(10) prime.

February 6th, 2014, 11:02 PM   #22
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Re: Pre-Q&A

Quote:
 Originally Posted by CRGreathouse Sorry to (slightly) let you down.
That's not a problem. I wasn't really looking for coefficients in {0, ..., 9}. Positive coefficients would be enough for me.

 February 7th, 2014, 06:57 AM #23 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Pre-Q&A I have made only a slight progress. $x^5 \ + \ n^4 \cdot x \ + \ n^5 \= \ (x^3 \ - \ n \cdot x^2 \ + \ n^3) \cdot (x^2 \ + \ n \cdot x \ + \ n^2)$ Is factorable with integer coefficients of the factors for every integer n (n = 0 is obviously factorable) , for our task at hand we require n > 0 This shows that for the satisfaction of our current problem , poly's of this form can be abandoned since $(x^3 \ - n \cdot x^2 \ + \ n^3) \ > \ 1$ When x = 2 and for any positive integer n. The other factor is always > 1 uder the same conditions. Note that $x^5 \ + \ n^4 \cdot x \ \pm \ n^5$ Is always factorable , with the coefficients of the factor pol's in Z[x]. This may be usefull elsewhere so put it in your notebook [color=#00FF00]Mr. TNT ![/color]
 February 7th, 2014, 07:09 AM #24 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Pre-Q&A If you ask, I'd say none of the factors of f(x) are 'small'. If you want to know, my idea is to make a first few coefficients 1 and then continue by making a sign change at h(x) so that f(x) has all of the coefficients positive. For example : (x^2 - 3x + 3)(x^4 + 4x^3 + 10x^2 + 19x + 1) = x^6 + x^5 + x^4 + x^3 - 26x^2 + 54x + 3. I'll see if I can do something through this process.
 February 7th, 2014, 07:57 AM #25 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Pre-Q&A I also used a similar approach starting with , $(2^3 \ - \ a \cdot 2^2 \ + \ \ \ ) \cdot (2^2 \ + \ a \cdot 2 \ + \ \ \ )$ One difficulty is we need non-zero 'a' so that the 2^3 is neutralized but that forces a*a all the way to the right. $(2^3 \ - \ a \cdot 2^2 \ + \ \ \ ) \cdot (2^2 \ + \ a \cdot 2 \ + \ a \cdot a)$ Which in turn forces a*a*a in the remaining slot $(2^3 \ - \ a \cdot 2^2 \ + \ a^3) \cdot (2^2 \ + \ a \cdot 2 \ + \ a^2 )$ So now you know the secret of my derivation in my previous post. 2 is too small , try a bigger prime then it may work. This particular approach using three terms for each factor is doomed IMHO because it initiates a 'reverse telescoping' process where the 'telescope' expands instead of collapsing. We may need at least one factor with many terms and large coefficients.
February 7th, 2014, 08:02 AM   #26
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Re: Pre-Q&A

Quote:
 Originally Posted by agentredlum We may need at least one factor with many terms and large coefficients.
Change the 'may' to 'will'.

 February 7th, 2014, 08:18 AM #27 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Pre-Q&A I noticed you are using $(x^2 \ - \ 3x \ + \ \ \)$ How can you be sure it will not initiate the same probems as my example , no mater how many terms the other factor has? Well , don't let me stop you , you are very close all you need to do is find the trick that eliminates -26 without turning any other coefficient negative , then check if f(2) = prime. I'm starting to suspect we may need many poly's with many terms so we can gain more freedom.
February 7th, 2014, 08:26 AM   #28
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Re: Pre-Q&A

Quote:
 Originally Posted by agentredlum How can you be sure it will not initiate the same probems as my example , no mater how many terms the other factor has?
My guess is that any polynomial with no positive root has a nonnegative coefficient polynomial multiple. I don't know how to (dis)prove that though, but I imagine it is not hard. But the actual problem here is to fins an explicit example, so it would not help.

Quote:
 Originally Posted by agentredlum I'm starting to suspect we may need many poly's with many terms so we can gain more freedom.
Yes, and we need a negative coefficient too.

 February 7th, 2014, 10:13 AM #29 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Pre-Q&A I DID IT!!! Here's a brief description of what I tried: Idea : I optimized the coefficients the first few coefficients of f(x) by setting them all to 1. This was the safest thing for me to do, as I have noted that either g or h must be very large. I then solved the linear system of equations one by one by subbing each coefficients of g to another. **I used no CAS help at this point** Precisely, the steps are : 1. I assumed g(x) = x^2 - 3x + 3 is the desired polynomial to choose. Note g(2) = 1. 2. (x^2 - 3x + 3)(x + 4) = x^3 + x^2 - 9x + 12 3. (x^2 - 3x + 3)(x^2 + 4x + 10) = x^4 + x^3 + x^2 - 18x + 30 4. (x^2 - 3x + 3)(x^3 + 4x^2 + 10x + 19) = x^5 + x^4 + x^3 + x^2 - 27x + 57 5. (x^2 - 3x + 3)(x^4 + 4x^3 + 10x^2 + 19x + 27) = x^6 + x^5 + x^4 + x^3 - 24x + 81 6. (x^2 - 3x + 3)(x^5 + 4x^4 + 10x^3 + 19x^2 + 27x + 24) = x^7 + x^6 + x^5 + x^4 + 9x + 72 [a polynomial in N[x]!] We are not done yet. We need f(2) to be prime, whereas it is now 330. This is simple to bypass, by adding a 1 to the factor (x^2 - 3x + 3)(x^5 + 4x^4 + 10x^3 + 19x^2 + 27x + 25) = x^7 + x^6 + x^5 + x^4 + x^2 + 6x + 75 So ends the first part of CRG's problem. The race is now on to push the coefficients to {0, ..., 10}.
February 7th, 2014, 10:40 AM   #30
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Re: Pre-Q&A

Quote:
 Originally Posted by mathbalarka If you ask, I'd say none of the factors of f(x) are 'small'.
I suspect one of the factors can be a quadratic and only one of the coefficients needs to be 10, but I haven't applied any silicon to the problem.

Edit: I see that the first part of my suspicion turns out to be correct!

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