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[SPAM] [obm-l] Re: [obm-l] Teoria dos números - ajuda

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Subject: [SPAM] [obm-l] Re: [obm-l] Teoria dos números - ajuda
From: "Rafael Cano" <rafaelcano@xxxxxxxxxxxxx>
Date: Fri, 30 May 2008 00:57:45 -0300
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Ol=E1.
Acho que consegui uma solu=E7=E3o. Considerei que m e n s=E3o inteiros.
(p-1)(1 + p^n ) =3D 4m(m+1) -> m=B2 + m - (1/4)(p-1)(1 + p^n) =3D 0 e =
essa =E9 uma eq. de 2=B0 grau em m.
O discriminante =E9 1 + (p-1)(1+p^n). Se queremos m um inteiro ent=E3o a =
raiz quadrada do discriminante tamb=E9m deve ser um n=FAmero inteiro.
Seja k inteiro positivo, 1 + (p-1)(1+p^n) =3D k=B2 -> p^(n+1) - p^n + p =
=3D p(p^n - p^(n-1) + 1) =3D k=B2. Logo p divide k=B2 e isso implica que =
p divide k.
Podemos escrever k=3Dp*k_1, k_1 inteiro positivo e substituindo: p^n - =
p^(n-1) + 1 =3D p(k_1)=B2. Assim p divide p^n - p^(n-1) + 1.
(p^n - p^(n-1) + 1)/p =3D (k_1)=B2 -> p^(n-1) - p^(n-2) + 1/p =3D =
(k_1)=B2. Veja que se n>=3D2, p^(n-1) - p^(n-2) =E9 inteiro o que =E9 =
absurdo.
Ent=E3o n=3D1. Substituindo na eq. de 2=B0 grau em m o valor de n =
encontrado: m=B2 + m - (1/4)(p-1)(1 + p) =3D 0 -> m=B2 + m - (1/4)(p=B2 =
-1).
Resolvendo encontramos: m=3D [- 1 +- raiz(1 + p=B2 - 1)]/2 =3D (- 1 +- =
p)/2. Como m>0 ignoramos a raiz negativa e temos m=3D(p-1)/2 que =E9 =
sempre inteiro j=E1 que p =E9 um primo =EDmpar.
Acho que =E9 isso.

Abra=E7os.

  ----- Original Message -----=20
  From: Rhilbert Rivera=20
  To: obm-l@xxxxxxxxxxxxxx=20
  Sent: Thursday, May 29, 2008 9:12 PM
  Subject: [obm-l] Teoria dos n=FAmeros - ajuda


  Amigos, agradeceria pela ajuda na resolu=E7=E3o desse problema.
  =20

  Seja p> 2 um primo. Determine todos os valores positivos de m e n, tal =
que

  (p-1)(1 + p^n ) =3D 4m(m+1).



  Obrigado!

  =20


-------------------------------------------------------------------------=
-----
  Conhe=E7a j=E1 o Windows Live Spaces, o site de relacionamentos do =
Messenger! Crie j=E1 o seu! 
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML xmlns:o =3D "urn:schemas-microsoft-com:office:office"><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
<STYLE>.hmmessage P {
	PADDING-RIGHT: 0px; PADDING-LEFT: 0px; PADDING-BOTTOM: 0px; MARGIN: =
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<META content=3D"MSHTML 6.00.2900.3314" name=3DGENERATOR></HEAD>
<BODY class=3Dhmmessage bgColor=3D#ffffff>
<DIV><FONT face=3DArial>Ol=E1.</FONT></DIV>
<DIV><FONT face=3DArial>Acho que consegui uma solu=E7=E3o. Considerei =
que m e n s=E3o=20
inteiros.</FONT></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">(p-1)(1 + p^n ) =
=3D 4m(m+1)=20
-&gt; m=B2 + m&nbsp;- (1/4)(p-1)(1 + p^n) =3D 0&nbsp;e essa =E9 uma eq. =
de 2=B0 grau em=20
m.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">O discriminante =
=E9 1 +=20
(p-1)(1+p^n). Se queremos m um inteiro ent=E3o a raiz quadrada do =
discriminante=20
tamb=E9m deve ser um n=FAmero inteiro.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">Seja k inteiro =
positivo, 1 +=20
(p-1)(1+p^n) =3D k=B2 -&gt; p^(n+1) - p^n + p =3D p(p^n - p^(n-1) + 1) =
=3D k=B2. Logo p=20
divide k=B2 e&nbsp;isso implica que p divide k.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">Podemos escrever =
k=3Dp*k_1,=20
k_1 inteiro positivo&nbsp;e substituindo: p^n - p^(n-1) + 1 =3D =
p(k_1)=B2. Assim p=20
divide p^n - p^(n-1) + 1.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">(p^n - p^(n-1) + =
1)/p =3D=20
(k_1)=B2 -&gt; p^(n-1) - p^(n-2) + 1/p =3D (k_1)=B2. Veja que se =
n&gt;=3D2, p^(n-1) -=20
p^(n-2) =E9 inteiro o que =E9 absurdo.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">Ent=E3o n=3D1. =
Substituindo na=20
eq. de 2=B0 grau em m o valor de n encontrado: m=B2 + m&nbsp;- =
(1/4)(p-1)(1 + p) =3D 0=20
-&gt; m=B2 + m&nbsp;- (1/4)(p=B2 -1).</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">Resolvendo =
encontramos: m=3D=20
[- 1 +- raiz(1 + p=B2 - 1)]/2 =3D (- 1 +- p)/2. Como m&gt;0 ignoramos a =
raiz=20
negativa e temos m=3D(p-1)/2 que =E9 sempre inteiro j=E1 que p =E9 um =
primo=20
=EDmpar.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">Acho que =E9=20
isso.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: =
Arial"></SPAN>&nbsp;</DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: =
Arial">Abra=E7os.</SPAN></DIV>
<DIV><FONT face=3DArial></FONT>&nbsp;</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=3Drhilbert1990@xxxxxxxxxxx=20
  href=3D"mailto:rhilbert1990@xxxxxxxxxxx";>Rhilbert Rivera</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> Thursday, May 29, 2008 =
9:12=20
PM</DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> [obm-l] Teoria dos =
n=FAmeros -=20
  ajuda</DIV>
  <DIV><BR></DIV>Amigos, agradeceria pela ajuda na resolu=E7=E3o desse=20
  problema.<BR>&nbsp;<BR>
  <P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt"><SPAN=20
  style=3D"COLOR: black; FONT-FAMILY: Arial"><FONT size=3D3>Seja p&gt; 2 =
um primo.=20
  Determine todos os valores positivos de m e n, tal=20
  que<o:p></o:p></FONT></SPAN></P>
  <P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt"><SPAN=20
  style=3D"COLOR: black; FONT-FAMILY: Arial"><FONT size=3D3>(p-1)(1 + =
p^n ) =3D=20
  4m(m+1).</FONT></SPAN></P>
  <P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt"><SPAN=20
  style=3D"COLOR: black; FONT-FAMILY: Arial"><FONT =
size=3D3></FONT></SPAN>&nbsp;</P>
  <P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt"><SPAN=20
  style=3D"COLOR: black; FONT-FAMILY: Arial"><FONT=20
  size=3D3>Obrigado!</FONT></SPAN><SPAN=20
  style=3D"FONT-FAMILY: Arial"><o:p></o:p></SPAN></P>&nbsp;<BR><BR>
  <HR>
  Conhe=E7a j=E1 o Windows Live Spaces, o site de relacionamentos do =
Messenger! <A=20
  href=3D"http://www.amigosdomessenger.com.br"; target=3D_new>Crie j=E1 o =
seu!</A>=20
</BLOCKQUOTE></BODY></HTML>

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