February 6th, 2014, 07:31 PM  #21  
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: PreQ&A Quote:
 
February 6th, 2014, 11:02 PM  #22  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: PreQ&A Quote:
 
February 7th, 2014, 06:57 AM  #23 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: PreQ&A
I have made only a slight progress. 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 When x = 2 and for any positive integer n. The other factor is always > 1 uder the same conditions. Note that 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: PreQ&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: PreQ&A
I also used a similar approach starting with , One difficulty is we need nonzero 'a' so that the 2^3 is neutralized but that forces a*a all the way to the right. Which in turn forces a*a*a in the remaining slot 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  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: PreQ&A Quote:
 
February 7th, 2014, 08:18 AM  #27 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: PreQ&A
I noticed you are using 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  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: PreQ&A Quote:
Quote:
 
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: PreQ&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  
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: PreQ&A Quote:
Edit: I see that the first part of my suspicion turns out to be correct!  