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[SPAM] [obm-l] Re: [obm-l] Problema com polinômios

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Subject: [SPAM] [obm-l] Re: [obm-l] Problema com polinômios
From: "Rafael Cano" <rafaelcano@xxxxxxxxxxxxx>
Date: Mon, 14 Jan 2008 16:37:09 -0300
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Ol=E1 Salhab,
N=E3o entendi muito bem...
As congru=EAncias que voc=EA usou saem do teorema das raizes racionais =
certo?
Mas por que elas valem para os outros coeficientes?
E esse m=E9todo n=E3o vai acabar num coeficiente diferente de 1 para =
x^n?

abra=E7os
  ----- Original Message -----=20
  From: Marcelo Salhab Brogliato=20
  To: obm-l@xxxxxxxxxxxxxx=20
  Sent: Monday, January 14, 2008 9:22 AM
  Subject: Re: [obm-l] Problema com polin=F4mios


  Ol=E1 Igor,

  estou tentando encontrar um contra-exemplo (pra mim, =E9 um =F3timo =
jeito de se encontrar uma demonstra=E7=E3o.. hehe!)

  p(x) =3D x^n + a_1*x^(n-1) + a_2*x^(n-2) + ... + a_(n-1)*x + a_n

  vamos supor que: p(a) =3D p(b) =3D p(c) =3D p(d) =3D 5, e p(k) =3D 8=20
  onde a, b, c, d, k sao primos entre si dois a dois.
  deste modo:
  p(a) =3D 5 =3D a_n (mod a)
  p(b) =3D 5 =3D a_n (mod b)
  p(c) =3D 5 =3D a_n (mod c)
  p(d) =3D 5 =3D a_n (mod d)
  p(k) =3D 8 =3D a_n (mod k)

  pelo teorema chines do resto, conseguimos determinar a_n (mod =
a.b.c.d.k)
  fazendo: g(x) =3D [p(x) - a_n]/x, temos que: g(a), g(b), g(c), g(d) e =
g(k) est=E3o definidos..
  ent=E3o, usando a mesma id=E9ia, determinamos a_(n-1).
  seguindo esta id=E9ia, conseguimos determinar todos os coeficientes do =
polin=F4mio!=20

  qual o erro nesta id=E9ia? n=E3o encontrei...

  abra=E7os,
  Salhab







  2008/1/12 Igor Battazza <battazza@xxxxxxxxx>:

    Ol=E1 pessoal,
    estou com d=FAvidas na seguinte quest=E3o:

    Dado o polin=F4mio p(x) =3D x^n + a_1*x^(n-1) + a_2*x^(n-2) + ... +=20
    a_(n-1)*x + a_n com coeficientes inteiros a_1, a_2, ..., a_n, e dado
    que tamb=E9m existem 4 inteiros distintos a, b, c e d tal que p(a) =
=3D
    p(b) =3D p(c) =3D p(d) =3D 5, mostre que n=E3o existe inteiro k tal =
que p(k) =3D
    8.=20

    N=E3o consigo pensar em nenhuma restri=E7=E3o que implique nisso.

    Obrigado,
    Igor.

    =
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
    Instru=E7=F5es para entrar na lista, sair da lista e usar a lista em =

    http://www.mat.puc-rio.br/~obmlistas/obm-l.html
    =
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
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<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
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<DIV><FONT face=3DArial size=3D2>Ol=E1 <FONT face=3D"Times New Roman"=20
size=3D3>Salhab,</FONT></FONT></DIV>
<DIV>N=E3o entendi muito bem...</DIV>
<DIV>As congru=EAncias que voc=EA usou saem do teorema das raizes =
racionais=20
certo?</DIV>
<DIV><FONT face=3DArial size=3D2>Mas por que elas valem para os outros=20
coeficientes?</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>E esse m=E9todo n=E3o vai acabar num =
coeficiente=20
diferente de 1 para x^n?</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>abra=E7os</FONT></DIV>
<BLOCKQUOTE=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
  <DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
  <DIV=20
  style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
  <A title=3Dmsbrogli@xxxxxxxxx =
href=3D"mailto:msbrogli@xxxxxxxxx";>Marcelo Salhab=20
  Brogliato</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Dobm-l@xxxxxxxxxxxxxx=20
  href=3D"mailto:obm-l@xxxxxxxxxxxxxx";>obm-l@xxxxxxxxxxxxxx</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Monday, January 14, 2008 =
9:22=20
  AM</DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Re: [obm-l] Problema =
com=20
  polin=F4mios</DIV>
  <DIV><FONT face=3DArial size=3D2></FONT><FONT face=3DArial =
size=3D2></FONT><FONT=20
  face=3DArial size=3D2></FONT><BR></DIV>Ol=E1 Igor,<BR><BR>estou =
tentando encontrar=20
  um contra-exemplo (pra mim, =E9 um =F3timo jeito de se encontrar uma=20
  demonstra=E7=E3o.. hehe!)<BR><BR>p(x) =3D x^n + a_1*x^(n-1) + =
a_2*x^(n-2) + ... +=20
  a_(n-1)*x + a_n<BR><BR>vamos supor que: p(a) =3D p(b) =3D p(c) =3D =
p(d) =3D 5, e p(k)=20
  =3D 8 <BR>onde a, b, c, d, k sao primos entre si dois a dois.<BR>deste =

  modo:<BR>p(a) =3D 5 =3D a_n (mod a)<BR>p(b) =3D 5 =3D a_n (mod =
b)<BR>p(c) =3D 5 =3D a_n=20
  (mod c)<BR>p(d) =3D 5 =3D a_n (mod d)<BR>p(k) =3D 8 =3D a_n (mod =
k)<BR><BR>pelo=20
  teorema chines do resto, conseguimos determinar a_n (mod=20
  a.b.c.d.k)<BR>fazendo: g(x) =3D [p(x) - a_n]/x, temos que: g(a), g(b), =
g(c),=20
  g(d) e g(k) est=E3o definidos..<BR>ent=E3o, usando a mesma id=E9ia, =
determinamos=20
  a_(n-1).<BR>seguindo esta id=E9ia, conseguimos determinar todos os =
coeficientes=20
  do polin=F4mio! <BR><BR>qual o erro nesta id=E9ia? n=E3o=20
  encontrei...<BR><BR>abra=E7os,<BR>Salhab<BR><BR><BR><BR><BR><BR><BR>
  <DIV class=3Dgmail_quote>2008/1/12 Igor Battazza &lt;<A=20
  href=3D"mailto:battazza@xxxxxxxxx";>battazza@xxxxxxxxx</A>&gt;:<BR>
  <BLOCKQUOTE class=3Dgmail_quote=20
  style=3D"PADDING-LEFT: 1ex; MARGIN: 0pt 0pt 0pt 0.8ex; BORDER-LEFT: =
rgb(204,204,204) 1px solid">Ol=E1=20
    pessoal,<BR>estou com d=FAvidas na seguinte quest=E3o:<BR><BR>Dado o =
polin=F4mio=20
    p(x) =3D x^n + a_1*x^(n-1) + a_2*x^(n-2) + ... + <BR>a_(n-1)*x + a_n =
com=20
    coeficientes inteiros a_1, a_2, ..., a_n, e dado<BR>que tamb=E9m =
existem 4=20
    inteiros distintos a, b, c e d tal que p(a) =3D<BR>p(b) =3D p(c) =3D =
p(d) =3D 5,=20
    mostre que n=E3o existe inteiro k tal que p(k) =3D<BR>8. =
<BR><BR>N=E3o consigo=20
    pensar em nenhuma restri=E7=E3o que implique=20
    =
nisso.<BR><BR>Obrigado,<BR>Igor.<BR><BR>=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D<BR>Instru=E7=F5es=20
    para entrar na lista, sair da lista e usar a lista em <BR><A=20
    href=3D"http://www.mat.puc-rio.br/%7Eobmlistas/obm-l.html"=20
    =
target=3D_blank>http://www.mat.puc-rio.br/~obmlistas/obm-l.html</A><BR>=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D<BR></B=
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