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 February 6th, 2014, 04:26 AM #11 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 227 Re: Pre-Q&A You and I are thinking alike in this matter , i also tried x^5 + x + 1 and x^4 + x^2 + 1 , then attempted to find other poly's similar , came close with x^9 + 2x + 3 but alas 519 = 3*173 ... composite Then i made a list of primes 520 < p < 588 and fooled around with $x^9 + mx^n + r$ No success. I'm doing this with no programming usage ability so by now i am quite tired of editing WIA entries and pressing = every time. BTW , I can't find anything i can see about Schur's Theorem that is relevent to CRG's claim in the other thread. Found this but i'm not sure if it applies. http://en.m.wikipedia.org/wiki/Cohn's_i ... _criterion
February 6th, 2014, 04:38 AM   #12
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Re: Pre-Q&A

Quote:
 Originally Posted by agentredlum Found this but i'm not sure if it applies.
Hmm, it does, but unfortunately, you need x = 10.

February 6th, 2014, 04:49 AM   #13
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Re: Pre-Q&A

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by agentredlum Found this but i'm not sure if it applies.
Hmm, it does, but unfortunately, you need x = 10.
Can we do anything with

$f(10)= f(2) \cdot f(5)$

Where f(10) = 5 , f(2) = 5 and f(5) = 1 ?

 February 6th, 2014, 04:50 AM #14 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Pre-Q&A I doubt.
February 6th, 2014, 04:54 AM   #15
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Re: Pre-Q&A

This is why I thought Schur's theorem could be applied here
Quote:
 Originally Posted by [color=#AA0000 CRGreathouse[/color]]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.
Though I cannot see why it could work by the theorem itself.

f(2) is prime so perhaps, all coefficients of f are in {0, 1,...,9} so the polynomial is irreducible. But perhaps since it is about x = 2 instead of x = 10 it works for coeficients in {0, 1, 2} only. You have similar observations. Maybe the answer to the bonus question would then be "to mislead "us" (me at least)"

 February 6th, 2014, 04:55 AM #16 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 227 Re: Pre-Q&A Yeah , you're right. f(10) = 5 would mean we need negative coefficients ...€¥§¿!!
February 6th, 2014, 05:05 AM   #17
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Re: Pre-Q&A

Quote:
Originally Posted by Hoempa
This is why I thought Schur's theorem could be applied here
Quote:
 Originally Posted by [color=#AA0000 CRGreathouse[/color]]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.
Though I cannot see why it could work by the theorem itself.

f(2) is prime so perhaps, all coefficients of f are in {0, 1,...,9} so the polynomial is irreducible. But perhaps since it is about x = 2 instead of x = 10 it works for coeficients in {0, 1, 2} only. You have similar observations. Maybe the answer to the bonus question would then be "to mislead "us" (me at least)"
Yes , i understood why you made the comment but i hope CRG is not making us work on a problem that has no solution , that's why i attempted to find info on the net about Schur's Theorem , so i could judge for myself . but i didn't find anything that i personally understood as relevant. I haven't looked for the other reference he gives , Gross and Filaseta.

February 6th, 2014, 05:12 AM   #18
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Re: Pre-Q&A

Quote:
 Originally Posted by agentredlum Yes , i understood why you made the comment but i hope CRG is not making us work on a problem that has no solution
He is not, I am pretty sure of that.

 February 6th, 2014, 05:17 AM #19 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 227 Re: Pre-Q&A [Deleted Joke]
February 6th, 2014, 07:24 PM   #20
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Re: Pre-Q&A

Quote:
 Originally Posted by agentredlum Found this but i'm not sure if it applies. http://en.m.wikipedia.org/wiki/Cohn's_i ... _criterion
Yes, that's it. Apparently I don't do very well without my notes.

It gets worse: it looks like I mixed up my inequalities when posting this question, and I should have allowed 0..10 rather than 0..9. Those of you who suspected the problem was unsolvable were probably right... but I have it on good authority that the problem with 0..10 is solvable.

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by agentredlum Yes , i understood why you made the comment but i hope CRG is not making us work on a problem that has no solution
He is not, I am pretty sure of that.
Sorry to (slightly) let you down. I'm pretty sure the corrected version works, though I haven't attempted to find a solution yet.

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