12 x^2 - 7x + 1 = 0 (mod p) has solution for every prime p
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From: master1729 on 29 Jun 2010 12:52 john wrote : > On Jun 29, 10:38 pm, Pubkeybreaker > <pubkeybrea...(a)aol.com> wrote: > > On Jun 29, 3:21 pm, John <to1m...(a)yahoo.com> wrote: > > > > > Prove that for every prime p the congruence 12 > x^2 - 7x + 1 = 0 (mod > > > p) has solution. > > > It seems that quadratic residue theory should > help here but first we > > > need to convert it to other form... > > > > > Any hints ? > > > > Are you sure that your polynomial is > irreducible??????.......... > > No. It equals (3x - 1)(4x - 1)... thus : (3x - 1)(4x - 1) = 0 mod p its 2 lineair equations. with solutions 1/3 mod p ,1/4 mod p ... got it ? tommy1729
From: bill on 29 Jun 2010 17:52 On Jun 29, 12:21 pm, John <to1m...(a)yahoo.com> wrote: > Prove that for every prime p the congruence 12 x^2 - 7x + 1 = 0 (mod > p) has solution. > It seems that quadratic residue theory should help here but first we > need to convert it to other form... > > Any hints ? 12 x^2 - 7x + 1 = (3x-1)*(4x-1) then 12 x^2 - 7x + 1 = 0 (mod p) if 3x-1 = 0 (mod p) or 4x-1 = 0 (mod p) for some x, where x = 1, 2, 3, ... I think that p does not have to be a prime, it can be any number.
From: Bill Dubuque on 29 Jun 2010 17:54 master1729 <tommy1729(a)gmail.com> wrote: > john wrote : >> On Jun 29, 10:38 pm, Pubkeybreaker >> <pubkeybrea...(a)aol.com> wrote: >>> On Jun 29, 3:21 pm, John <to1m...(a)yahoo.com> wrote: >>> >>>> Prove that for every prime p the congruence >>>> 12 x^2 - 7x + 1 = 0 (mod p) has solution. >>> >>>> Any hints ? >>> >>> Are you sure that your polynomial is >> irreducible??????.......... >> >> No. It equals (3x - 1)(4x - 1)... > > thus : (3x - 1)(4x - 1) = 0 mod p > its 2 lineair equations. > > with solutions 1/3 mod p ,1/4 mod p ... got it ? A set of reduced fractions has this property iff the denominators have gcd = 1 --Bill Dubuque
From: Pubkeybreaker on 29 Jun 2010 18:44 On Jun 29, 4:50 pm, John <to1m...(a)yahoo.com> wrote: > On Jun 29, 11:38 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > > > > On Jun 29, 4:14 pm, John <to1m...(a)yahoo.com> wrote: > > > > On Jun 29, 10:38 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > > > > On Jun 29, 3:21 pm, John <to1m...(a)yahoo.com> wrote: > > > > > > Prove that for every prime p the congruence 12 x^2 - 7x + 1 = 0 (mod > > > > > p) has solution. > > > > > It seems that quadratic residue theory should help here but first we > > > > > need to convert it to other form... > > > > > > Any hints ? > > > > > Are you sure that your polynomial is irreducible??????.......... > > > > No. It equals (3x - 1)(4x - 1)... > > > So now you have your answer. > > Thanks. Now I understand. One of the linear congruences has solution.- Hide quoted text - > > - Show quoted text - Now you can prove a more general result. If f(x) is irreducible over Q, then there are infinitely many p for which f(x) = 0 has no solution.
From: Gerry Myerson on 29 Jun 2010 19:47
In article <d774dc46-2aaf-4ec6-bc3f-498f1d153bce(a)c10g2000yqi.googlegroups.com>, bill <b92057(a)yahoo.com> wrote: > On Jun 29, 12:21�pm, John <to1m...(a)yahoo.com> wrote: > > Prove that for every prime p the congruence 12 x^2 - 7x + 1 = 0 (mod > > p) has solution. > > It seems that quadratic residue theory should help here but first we > > need to convert it to other form... > > > > Any hints ? > > 12 x^2 - 7x + 1 = (3x-1)*(4x-1) then > 12 x^2 - 7x + 1 = 0 (mod p) if > 3x-1 = 0 (mod p) or 4x-1 = 0 (mod p) > for some x, where x = 1, 2, 3, ... > > I think that p does not have to be a prime, it can be > any number. You have to work a little harder when p is a multiple of 6. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email) |