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



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Obrigado Rafael.  Esqueci de colocar, mas os valores de m e n  s=E3o inteir=
os e positivos, como voc=EA fez. Valeu!
At=E9 a pr=F3xima


From: rafaelcano@xxxxxxxxxxxxxxx: obm-l@xxxxxxxxxxxxxxxxxxxxx: [obm-l] Re: =
[obm-l] Teoria dos n=FAmeros - ajudaDate: Fri, 30 May 2008 00:57:45 -0300



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 ra=
iz 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 div=
ide 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 encontrad=
o: 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 int=
eiro j=E1 que p =E9 um primo =EDmpar.
Acho que =E9 isso.
=20
Abra=E7os.
=20

----- 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).
=20
Obrigado!=20

Conhe=E7a j=E1 o Windows Live Spaces, o site de relacionamentos do Messenge=
r! Crie j=E1 o seu!=20
_________________________________________________________________
Cansado de espa=E7o para s=F3 50 fotos? Conhe=E7a o Spaces, o site de relac=
ionamentos com at=E9 6,000 fotos!
http://www.amigosdomessenger.com.br=

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Obrigado Rafael.&nbsp; Esqueci de colocar, mas os valores de m e n&nbsp; s=
=E3o inteiros e positivos, como voc=EA fez. Valeu!<BR>
At=E9 a pr=F3xima<BR><BR><BR>
<BLOCKQUOTE>
<HR>
From: rafaelcano@xxxxxxxxxxxxx<BR>To: obm-l@xxxxxxxxxxxxxx<BR>Subject: [obm=
-l] Re: [obm-l] Teoria dos n=FAmeros - ajuda<BR>Date: Fri, 30 May 2008 00:5=
7:45 -0300<BR><BR>
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<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 inteiros.</FONT></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">(p-1)(1 + p^n ) =3D 4=
m(m+1) -&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 m.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">O discriminante =E9 1=
 + (p-1)(1+p^n). Se queremos m um inteiro ent=E3o a raiz quadrada do discri=
minante tamb=E9m deve ser um n=FAmero inteiro.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">Seja k inteiro positi=
vo, 1 + (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 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=3D=
p*k_1, k_1 inteiro positivo&nbsp;e substituindo: p^n - p^(n-1) + 1 =3D p(k_=
1)=B2. Assim p 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 (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) - p^(n-2) =E9 inteiro o que =E9 absurdo.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">Ent=E3o n=3D1. Substi=
tuindo na eq. de 2=B0 grau em m o valor de n encontrado: m=B2 + m&nbsp;- (1=
/4)(p-1)(1 + p) =3D 0 -&gt; m=B2 + m&nbsp;- (1/4)(p=B2 -1).</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">Resolvendo encontramo=
s: m=3D [- 1 +- raiz(1 + p=B2 - 1)]/2 =3D (- 1 +- p)/2. Como m&gt;0 ignoram=
os a raiz negativa e temos m=3D(p-1)/2 que =E9 sempre inteiro j=E1 que p =
=E9 um primo =EDmpar.</SPAN></DIV>
<DIV><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial">Acho que =E9 isso.</S=
PAN></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></DI=
V>
<DIV><FONT face=3DArial></FONT>&nbsp;</DIV>
<BLOCKQUOTE style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5p=
x; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><B>=
From:</B> <A title=3Drhilbert1990@xxxxxxxxxxx href=3D"mailto:rhilbert1990@h=
otmail.com">Rhilbert Rivera</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>To:</B> <A title=3Dobm-l@xxxxxxxxxxxxxx =
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 PM=
</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> [obm-l] Teoria dos n=FAmero=
s - ajuda</DIV>
<DIV><BR></DIV>Amigos, agradeceria pela ajuda na resolu=E7=E3o desse proble=
ma.<BR>&nbsp;<BR>
<P class=3DEC_MsoNormal><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial"><F=
ONT size=3D3>Seja p&gt; 2 um primo. Determine todos os valores positivos de=
 m e n, tal que</FONT></SPAN></P>
<P class=3DEC_MsoNormal><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial"><F=
ONT size=3D3>(p-1)(1 + p^n ) =3D 4m(m+1).</FONT></SPAN></P>
<P class=3DEC_MsoNormal><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial"><F=
ONT size=3D3></FONT></SPAN>&nbsp;</P>
<P class=3DEC_MsoNormal><SPAN style=3D"COLOR: black; FONT-FAMILY: Arial"><F=
ONT size=3D3>Obrigado!</FONT></SPAN><SPAN style=3D"FONT-FAMILY: Arial"></SP=
AN></P>&nbsp;<BR><BR>
<HR>
Conhe=E7a j=E1 o Windows Live Spaces, o site de relacionamentos do Messenge=
r! <A href=3D"http://www.amigosdomessenger.com.br/"; target=3D_blank>Crie j=
=E1 o seu!</A> </BLOCKQUOTE></BLOCKQUOTE><br /><hr />Instale a Barra de Fer=
ramentas com Desktop Search e ganhe EMOTICONS para o Messenger! <a href=3D'=
http://www.msn.com.br/emoticonpack' target=3D'_new'>=C9 GR=C1TIS!</a></body=
>
</html>=

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