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[SPAM] [obm-l] Res: [obm-l] Congruência - Dúvida

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Subject: [SPAM] [obm-l] Res: [obm-l] Congruência - Dúvida
From: Danilo Nascimento <souza_danilo@xxxxxxxxxxxx>
Date: Sat, 19 May 2007 15:59:34 -0700 (PDT)
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Ola,=0A       3^11=3D=3D1 mod 23, pois (^2) -> 3^22=3D=3D1 mod 23 --> 3^23=
=3D=3D3 mod 23 o que eh verdade pela pequeno teorema de fermat. a^p=3D=3Da =
mod p, p primo. =0Avlw.=0A=0A=0A----- Mensagem original ----=0ADe: Rhilbert=
 Rivera <rhilbert1990@hotmail.com>=0APara: obm-l@mat.puc-rio.br=0AEnviadas:=
 S=E1bado, 19 de Maio de 2007 16:28:49=0AAssunto: [obm-l] Congru=EAncia - D=
=FAvida=0A=0A=0AColegas, estava olhando a solu=E7=E3o de um problema de con=
gru=EAncia e n=E3o =0Aentendi uma passagem. Est=E1 assim:=0A"sendo 23 um n=
=FAmero primo, segue que 3^11=3D=3D 1(mod 23) ou 3^11=3D=3D -1(mod 23)"=0AC=
omo n=E3o consigo ver nessa arfirma=E7=E3o o pequeno teorema de Fermat, log=
o deve =0Aser algo que ainda n=E3o estudei.=0AObrigado  pela ajuda.=0A=0AOb=
s: estou usando  =3D=3D com o significado de "=E9 congruente"=0A=0A________=
_________________________________________________________=0AO Windows Live =
Spaces =E9 seu espa=E7o na internet com fotos (500 por m=EAs), blog =0Ae ag=
ora com rede social http://spaces.live.com/=0A=0A=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=0AInstru=E7=F5es para entrar =
na lista, sair da lista e usar a lista em=0Ahttp://www.mat.puc-rio.br/~nico=
lau/olimp/obm-l.html=0A=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=0A=0A__________________________________________________=
=0AFale com seus amigos  de gra=E7a com o novo Yahoo! Messenger =0Ahttp://b=
r.messenger.yahoo.com/ 
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<html><head><style type=3D"text/css"><!-- DIV {margin:0px;} --></style></he=
ad><body><div style=3D"font-family:times new roman, new york, times, serif;=
font-size:12pt"><DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman=
, new york, times, serif">Ola,</DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-=
FAMILY: times new roman, new york, times, serif">&nbsp;&nbsp;&nbsp;&nbsp;&n=
bsp;&nbsp; 3^11=3D=3D1 mod 23, pois (^2) -&gt; 3^22=3D=3D1 mod 23 --&gt; 3^=
23=3D=3D3 mod 23 o que eh verdade pela pequeno teorema de fermat. a^p=3D=3D=
a mod p, p primo. </DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: time=
s new roman, new york, times, serif">vlw.<BR><BR></DIV>=0A<DIV style=3D"FON=
T-SIZE: 12pt; FONT-FAMILY: times new roman, new york, times, serif">----- M=
ensagem original ----<BR>De: Rhilbert Rivera &lt;rhilbert1990@hotmail.com&g=
t;<BR>Para: obm-l@mat.puc-rio.br<BR>Enviadas: S=E1bado, 19 de Maio de 2007 =
16:28:49<BR>Assunto: [obm-l] Congru=EAncia - D=FAvida<BR><BR>=0A<DIV>Colega=
s, estava olhando a solu=E7=E3o de um problema de congru=EAncia e n=E3o <BR=
>entendi uma passagem. Est=E1 assim:<BR>"sendo 23 um n=FAmero primo, segue =
que 3^11=3D=3D 1(mod 23) ou 3^11=3D=3D -1(mod 23)"<BR>Como n=E3o consigo ve=
r nessa arfirma=E7=E3o o pequeno teorema de Fermat, logo deve <BR>ser algo =
que ainda n=E3o estudei.<BR>Obrigado&nbsp;&nbsp;pela ajuda.<BR><BR>Obs: est=
ou usando&nbsp;&nbsp;=3D=3D com o significado de "=E9 congruente"<BR><BR>__=
_______________________________________________________________<BR>O Window=
s Live Spaces =E9 seu espa=E7o na internet com fotos (500 por m=EAs), blog =
<BR>e agora com rede social <A href=3D"http://spaces.live.com/" target=3D_b=
lank>http://spaces.live.com/</A><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 para entrar na lista=
, sair da lista e usar a lista em<BR><A href=3D"http://www.mat.puc-rio.br/~=
nicolau/olimp/obm-l.html"
 target=3D_blank>http://www.mat.puc-rio.br/~nicolau/olimp/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</D=
IV></DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, ne=
w york, times, serif"><BR></DIV></div><br>_________________________________=
_________________<br>Fale com seus amigos  de gra=E7a com o novo Yahoo! Mes=
senger <br>http://br.messenger.yahoo.com/ </body></html>
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