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[SPAM] [obm-l] Re: [obm-l] [obm-l] questão da OBM
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- Subject: [SPAM] [obm-l] Re: [obm-l] [obm-l] questão da OBM
- From: Carlos Yuzo Shine <cyshine@xxxxxxxxx>
- Date: Fri, 14 Dec 2007 12:25:06 -0800 (PST)
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Na verdade, n=E3o. Uma id=E9ia =E9 fazer indutivamente: suponha que (a^29 -=
1)/(a - 1) tem k fatores primos distintos. Considere, ent=E3o, a^2 no luga=
r de a: fatorando, obtemos=0A ((a^2)^29 - 1)/(a^2 - 1) =3D ((a^29 - 1)/(a =
- 1))((a^29 + 1)/(a + 1))=0A=0AO mdc de a^29 + 1 com a^29 - 1 e a - 1 =E9 n=
o m=E1ximo 2 (para observar isso, se d =E9 o mdc de a^29 + 1 e a - 1, basta=
notar que se d divide a - 1 ent=E3o divide a^29 - 1 e, como d divide a^29 =
+ 1, d divide a diferen=E7a, que =E9 2), ent=E3o deve aparecer um fator pri=
mo novo em (a^29 + 1)/(a + 1), que =E9 inteiro. Deste modo, ((a^2)^29 - 1)/=
(a^2 - 1) tem mais um fator primo novo, tendo k+1 fatores primos.=0A=0AIsso=
quser dizer que, digamos, ((2^(2^2007))^29 - 1)/(2^(2^2007) - 1) tem pelo =
menos 2007 fatores primos (neste caso, a =3D 2^(2^2007)). Na =E9poca, t=EDn=
hamos conseguido uma fatora=E7=E3o que mostra que h=E1 2007 primos distinto=
s diretamente, mas n=E3o lembro exatamente como era. Algu=E9m se habilita?=
=0A=0A[]'s=0AShine=0A=0A=0A----- Original Message ----=0AFrom: vitoriogauss=
<vitoriogauss@xxxxxxxxxx>=0ATo: obm-l <obm-l@xxxxxxxxxxxxxx>=0ASent: Frida=
y, December 14, 2007 1:34:07 PM=0ASubject: [obm-l] [obm-l] quest=E3o da OBM=
=0A=0A=0A> Colegas.... =0A> =0A> A respeito da quest=E3o (a^29 - 1)/a-1... =
para provar que h=E1 2007 fatores primos=0A =0As=F3 por congru=EAncia???=0A=
=0A> Grato =0A> =0AVit=F3rio Gauss=0A=0A=0A _________________________=
___________________________________________________________=0ANever miss a =
thing. Make Yahoo your home page. =0Ahttp://www.yahoo.com/r/hs
--0-476290008-1197663906=:36332
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Content-Transfer-Encoding: quoted-printable
<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">Na verdade, n=E3o. Uma id=E9ia =E9 fazer indutiva=
mente: suponha que (a^29 - 1)/(a - 1) tem k fatores primos distintos. Consi=
dere, ent=E3o, a^2 no lugar de a: fatorando, obtemos</DIV>=0A<DIV style=3D"=
FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new york, times, serif">&nbs=
p; ((a^2)^29 - 1)/(a^2 - 1) =3D ((a^29 - 1)/(a - 1))((a^29 + 1)/(a + 1=
))</DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new=
york, times, serif"> </DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAM=
ILY: times new roman, new york, times, serif">O mdc de a^29 + 1 com a^29 - =
1 e a - 1 =E9 no m=E1ximo 2 (para observar isso, se d =E9 o mdc de a^29 + 1=
e a - 1, basta notar que se d divide a - 1 ent=E3o divide a^29 - 1 e, como=
d divide a^29 + 1, d divide a diferen=E7a, que =E9 2), ent=E3o deve aparec=
er um fator primo novo em (a^29 + 1)/(a + 1), que =E9 inteiro. Deste modo, =
((a^2)^29 - 1)/(a^2 - 1) tem mais um fator primo novo, tendo k+1 fatores pr=
imos.</DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, =
new york, times, serif"> </DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-=
FAMILY: times new roman, new york, times, serif">Isso quser dizer que, diga=
mos, ((2^(2^2007))^29 - 1)/(2^(2^2007) - 1) tem pelo menos 2007 fatores pri=
mos (neste caso, a =3D 2^(2^2007)). Na =E9poca, t=EDnhamos conseguido uma f=
atora=E7=E3o que mostra que h=E1 2007 primos distintos diretamente, mas n=
=E3o lembro exatamente como era. Algu=E9m se habilita?</DIV>=0A<DIV style=
=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new york, times, serif">=
</DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman,=
new york, times, serif">[]'s</DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; FONT-F=
AMILY: times new roman, new york, times, serif">Shine<BR><BR></DIV>=0A<DIV =
style=3D"FONT-SIZE: 12pt; FONT-FAMILY: times new roman, new york, times, se=
rif">----- Original Message ----<BR>From: vitoriogauss <vitoriogauss@uol=
.com.br><BR>To: obm-l <obm-l@xxxxxxxxxxxxxx><BR>Sent: Friday, Dece=
mber 14, 2007 1:34:07 PM<BR>Subject: [obm-l] [obm-l] quest=E3o da OBM<BR><B=
R>=0A<DIV>=0A<DIV>> Colegas.... </DIV>=0A<DIV>> </DIV>=0A<DIV>> A =
respeito da quest=E3o (a^29 - 1)/a-1... para provar que h=E1 2007 fatores p=
rimos</DIV>=0A<DIV> </DIV>=0A<DIV>s=F3 por congru=EAncia???</DIV>=0A<D=
IV> </DIV>=0A<DIV>> Grato </DIV>=0A<DIV>> </DIV>=0A<DIV></DIV>=
=0A<DIV>Vit=F3rio Gauss</DIV></DIV></DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; =
FONT-FAMILY: times new roman, new york, times, serif"><BR></DIV></div><br>=
=0A <hr size=3D1>Be a better friend, newshound, and =0Aknow-it-all wit=
h Yahoo! Mobile. <a href=3D"http://us.rd.yahoo.com/evt=3D51733/*http://mobi=
le.yahoo.com/;_ylt=3DAhu06i62sR8HDtDypao8Wcj9tAcJ "> Try it now.</a></body>=
</html>
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