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February 4th, 2014, 08:05 PM  #1 
Senior Member Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53  Quartic polynomial
Ok, I know I've been inactive for quite a while, so here's a little problem so that none of you forget me. Show that the polynomial is irreducible. 
February 4th, 2014, 08:51 PM  #2 
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: Quartic polynomial
I suppose you won't let Code: polisirreducible(x^4+2*x^3x1) Let's see... here's one I don't think you'll have expected. First transform x > x+1 to get x^4 + 6x^3 + 12x^2 + 9x + 1, then substitute x = 10 to get 17291. Note that this is prime, and that all coefficients are between 1 and 12. If they were between 0 and 9, Schur's theorem would tell us that the polynomial is irreducible. But recent work by Samuel Gross and Michael Filaseta shows that the same result holds with coefficients between 0 and 49598666989151226098104244512918. 
February 5th, 2014, 02:49 AM  #3 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Quartic polynomial
The question in the current form is incorrect. I can factorize that over , for example, if you forgive me the cheek. PS : Ed, I was thinking that now might be the time to continue the Math Q&A. What do you think? And others? 
February 5th, 2014, 03:41 AM  #4 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Quartic polynomial
Here's an alternative approach. By the Rational Root Theorem we know Does not have any rational roots. This means P(x) cannot be reduced to anything that has a rational linear factor. There remains only one possibility to test , can it be reduced to the product of two quadratic factors with rational coefficients? Itis clear that either a or b or both must be 0 otherwise we get a x^2 term which we don't want. If both a , b = 0 will lead to contradicting RRT. WLOG let a = 0 , then we are forced to write Comparing to P(x) we get b = 2 and 1 , impossible. Note: I'm not comfortable with this 'analysis' 
February 5th, 2014, 03:50 AM  #5  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Quartic polynomial Quote:
 
February 5th, 2014, 06:09 AM  #6 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: Quartic polynomial
Yes, that would be nice. What sort of questions could we do?

February 5th, 2014, 08:29 AM  #7  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Quartic polynomial Quote:
I nominate Eddy as leader and set up a general guideline for posting problems and proposing solutions. PM all the ones who you think are interested in participating. Let them know through this post if they are interested by posting something like "I am in". Now, someone needs to call Eddy. Ed? Edward? Can you hear me?  
February 5th, 2014, 10:04 AM  #8 
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: Quartic polynomial
Not having patience, I started a preQ&A thread here: viewtopic.php?f=18&t=46047 
February 5th, 2014, 10:56 AM  #9 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Quartic polynomial
Hmm, in general I'd have said 'Patience, DanielSan!' but this seems like a good idea. Let's use this thread as Area 51 and that as a Test Specimen for how well does a Q&A works out.

February 5th, 2014, 02:54 PM  #10 
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: Quartic polynomial
I figure we can just sort of fool around there until we come up with a real Q&A thread, however long that takes.


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