12 x^2 - 7x + 1 = 0 (mod p) has solution for every prime p
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From: master1729 on 30 Jun 2010 03:21 Bill Dubuque wrote : > 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 what are you talking about ? 1/3 = some integer mod p 1/4 = some integer mod p so they are not really fractions , and they dont need to be reduced. what property ? 1/3 =/= 1/4 mod p what does gcd(3,4) have to do with solving a lineair equation ? i simply solved 3x - 1 = 0 and 4x - 1 = 0. trivial. tommy1729
From: Bill Dubuque on 30 Jun 2010 10:47 master1729 <tommy1729(a)gmail.com> wrote: > Bill Dubuque wrote : >> master1729 <tommy1729(a)gmail.com> wrote: >>> john wrote: >>>>ubkeybreaker <pubkeybrea...(a)aol.com> wrote: >>>>>ohn <to1m...(a)yahoo.com>> wrote: >>>>> >>>>>> Prove that for every prime p the congruence >>>>>> 12 x^2 - 7x + 1 = 0 (mod p) has solution. >>>>> >>>>> Are you sure that your polynomial is irreducible?????? >>>> >>>> No. It equals (3x - 1)(4x - 1)... >>> >>> thus : (3x - 1)(4x - 1) = 0 mod p >>> 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 > > what are you talking about? A set of reduced fractions satisfies: for every prime p, one of them exists (mod p) <=> denominators have gcd = 1. That generalizes the original problem. > 1/3, 1/4 are some integers mod p. they aren't really fractions They are in my statement. Moreover, the subring of Q with denominators coprime to m has a natural image in Z/m. This can be verified directly, or, it is a special case of the *universal* property of fraction rings (localizations). So such fraction arithmetic still works fine in Z/m e.g. in Z/7 it remains true that 1/2 + 1/6 = 2/3, i.e. 4 - 1 = 3 Informally: fraction arithmetic is universal - it remains valid in every ring it may be interpreted, i.e. where all the fractions exist. This basic fact should be taught in every first course in abstract algebra but, alas, rarely is that the case. --Bill Dubuque
From: mjc on 1 Jul 2010 00:30 On Jun 30, 7:47 am, Bill Dubuque <w...(a)nestle.csail.mit.edu> wrote: > master1729 <tommy1...(a)gmail.com> wrote: > > Bill Dubuque wrote : > >> master1729 <tommy1...(a)gmail.com> wrote: > >>> john wrote: > >>>>ubkeybreaker <pubkeybrea...(a)aol.com> wrote: > >>>>>ohn <to1m...(a)yahoo.com>> wrote: > > >>>>>> Prove that for every prime p the congruence > >>>>>> 12 x^2 - 7x + 1 = 0 (mod p) has solution. > > >>>>> Are you sure that your polynomial is irreducible?????? > > >>>> No. It equals (3x - 1)(4x - 1)... > > >>> thus : (3x - 1)(4x - 1) = 0 mod p > >>> 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 > > > what are you talking about? > > A set of reduced fractions satisfies: for every prime p, > one of them exists (mod p) <=> denominators have gcd = 1. > > That generalizes the original problem. > > > 1/3, 1/4 are some integers mod p. they aren't really fractions > > They are in my statement. Moreover, the subring of Q with denominators > coprime to m has a natural image in Z/m. This can be verified directly, > or, it is a special case of the *universal* property of fraction rings > (localizations). So such fraction arithmetic still works fine in Z/m > e.g. in Z/7 it remains true that 1/2 + 1/6 = 2/3, i.e. 4 - 1 = 3 > Informally: fraction arithmetic is universal - it remains valid in > every ring it may be interpreted, i.e. where all the fractions exist. > This basic fact should be taught in every first course in abstract algebra > but, alas, rarely is that the case. > > --Bill Dubuque And here is an amusing result (definitely not original with me): (x^2-2)(x^2-3)(x^2-6) = 0 always has a solution mod p for any prime p. Hint: Iterated bifurcations in heterococomorphic subdomains is not the way to do it.
From: Bill Dubuque on 1 Jul 2010 10:38 mjc <mjcohen(a)acm.org> wrote: > On Jun 30, 7:47 am, Bill Dubuque <w...(a)nestle.csail.mit.edu> wrote: >> master1729 <tommy1...(a)gmail.com> wrote: >>> Bill Dubuque wrote : >>>> master1729 <tommy1...(a)gmail.com> wrote: >>>>> john wrote: >>>>>>Pubkeybreaker <pubkeybrea...(a)aol.com> wrote: >>>>>>>ohn <to1m...(a)yahoo.com>> wrote: >> >>>>>>>> Prove that for every prime p the congruence >>>>>>>> 12 x^2 - 7x + 1 = 0 (mod p) has solution. >> >>>>>>> Are you sure that your polynomial is irreducible?????? >> >>>>>> No. It equals (3x - 1)(4x - 1)... >> >>>>> thus : (3x - 1)(4x - 1) = 0 mod p >>>>> 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 >> >>> what are you talking about? >> >> A set of reduced fractions satisfies: for every prime p, >> one of them exists (mod p) <=> denominators have gcd = 1. >> >> That generalizes the original problem. >> >>> 1/3, 1/4 are some integers mod p. they aren't really fractions >> >> They are in my statement. Moreover, the subring of Q with denominators >> coprime to m has a natural image in Z/m. This can be verified directly, >> or, it is a special case of the *universal* property of fraction rings >> (localizations). So such fraction arithmetic still works fine in Z/m >> e.g. in Z/7 it remains true that 1/2 + 1/6 = 2/3, i.e. 4 - 1 = 3 >> Informally: fraction arithmetic is universal - it remains valid in >> every ring it may be interpreted, i.e. where all the fractions exist. >> This basic fact should be taught in every first course in abstract algebra >> but, alas, rarely is that the case. > > And here is an amusing result (definitely not original with me): > > (x^2-2)(x^2-3)(x^2-6) = 0 always has a solution mod p for any prime p. Generally such results follow from density theorems (Frobenius, Chebotarev). e.g. see Rolf Brandl. Integer Polynomials that are Reducible Modulo all Primes The American Mathematical Monthly, Vol. 93, No. 4 (Apr., 1986), pp. 286-288 http://www.jstor.org/stable/2323681 Stevenhagen, P.; Lenstra, H. Chebotarev and his density theorem. http://www.math.leidenuniv.nl/~hwl/papers/cheb.pdf --Bill Dubuque
From: Dan Cass on 1 Jul 2010 08:32
> On Jun 30, 7:47 am, Bill Dubuque > <w...(a)nestle.csail.mit.edu> wrote: > > master1729 <tommy1...(a)gmail.com> wrote: > > > Bill Dubuque wrote : > > >> master1729 <tommy1...(a)gmail.com> wrote: > > >>> john wrote: > > >>>>ubkeybreaker <pubkeybrea...(a)aol.com> wrote: > > >>>>>ohn <to1m...(a)yahoo.com>> wrote: > > > > >>>>>> Prove that for every prime p the congruence > > >>>>>> 12 x^2 - 7x + 1 = 0 (mod p) has solution. > > > > >>>>> Are you sure that your polynomial is > irreducible?????? > > > > >>>> No. It equals (3x - 1)(4x - 1)... > > > > >>> thus : (3x - 1)(4x - 1) = 0 mod p > > >>> 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 > > > > > what are you talking about? > > > > A set of reduced fractions satisfies: for every > prime p, > > one of them exists (mod p) <=> denominators have > gcd = 1. > > > > That generalizes the original problem. > > > > > 1/3, 1/4 are some integers mod p. they aren't > really fractions > > > > They are in my statement. Moreover, the subring of > Q with denominators > > coprime to m has a natural image in Z/m. This can > be verified directly, > > or, it is a special case of the *universal* > property of fraction rings > > (localizations). So such fraction arithmetic still > works fine in Z/m > > e.g. in Z/7 it remains true that 1/2 + 1/6 = 2/3, > i.e. 4 - 1 = 3 > > Informally: fraction arithmetic is universal - it > remains valid in > > every ring it may be interpreted, i.e. where all > the fractions exist. > > This basic fact should be taught in every first > course in abstract algebra > > but, alas, rarely is that the case. > > > > --Bill Dubuque > > And here is an amusing result (definitely not > original with me): > > (x^2-2)(x^2-3)(x^2-6) = 0 always has a solution mod p > for any prime p. > > Hint: Iterated bifurcations in heterococomorphic > subdomains is not the > way to do it. This follows from the fact that the product of two non-residues is a residue. [By residue I mean quadratic residue...] That is, if one of 2,3 is a quadratic residue, then (x^2-2)(x^2-3) = 0 has a solution mod p. On the other hand if both 2,3 are non-residues, then 6 is a residue, so that x^2-6 = 0 has a solution mod p. So in any case, (x^2-2)(x^2-3)(x^2-6) = 0 has a solution mod p. |