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[SPAM] Re: [obm-l] Teoria dos Números

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Muito obrigado caros colegas...
Boa semana para todos.



2008/3/9, Ralph Teixeira <ralphct@xxxxxxxxx>:
>
> p^1994+p^1995=3Dp^1994(p+1)
>
> Como p^1994 jah eh um quadrado perfeito (de p^997), a condicao pedida
> eh equivalente a p+1 ser quadrado perfeito. Mas entao:
>
> p+1=3Dk^2 (com k inteiro)
> p=3Dk^2-1=3D(k+1)(k-1)
>
> Mas se p eh primo, como eh que vai ser o produto de dois inteiros? O
> unico jeito eh se um deles for 1 e o outro for p; como nao pode ser
> k+1=3D1 (pois entao p=3D0, o que nao serve), tem de ser
>
> k-1=3D1, entao k=3D2, entao p=3D3.
>
> Assim, ha apenas um numero primo satisfazendo a dita condicao, que eh
> p=3D3. De fato:
>
> 3^1994+3^1995=3D3^1994.4=3D(3^997.2)^2.
>
> Abraco,
>    Ralph
>
> On Sun, Mar 9, 2008 at 7:01 PM, Pedro J=FAnior
> <pedromatematico06@xxxxxxxxx> wrote:
> > Determine a quantidade de n=FAmeros primos p, para que a express=E3o p^=
1994
> +
> > p^1995 seja um quadrado perfeito.
> > Desde j=E1 muito agradecido.
> > Pedro Jr
> >
>
> =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=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=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=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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<div>Muito obrigado caros colegas...</div>
<div>Boa semana para todos.<br>&nbsp;</div><br><br>
<div><span class=3D"gmail_quote">2008/3/9, Ralph Teixeira &lt;<a href=3D"ma=
ilto:ralphct@xxxxxxxxx">ralphct@xxxxxxxxx</a>&gt;:</span>
<blockquote class=3D"gmail_quote" style=3D"PADDING-LEFT: 1ex; MARGIN: 0px 0=
px 0px 0.8ex; BORDER-LEFT: #ccc 1px solid">p^1994+p^1995=3Dp^1994(p+1)<br><=
br>Como p^1994 jah eh um quadrado perfeito (de p^997), a condicao pedida<br=
>
eh equivalente a p+1 ser quadrado perfeito. Mas entao:<br><br>p+1=3Dk^2 (co=
m k inteiro)<br>p=3Dk^2-1=3D(k+1)(k-1)<br><br>Mas se p eh primo, como eh qu=
e vai ser o produto de dois inteiros? O<br>unico jeito eh se um deles for 1=
 e o outro for p; como nao pode ser<br>
k+1=3D1 (pois entao p=3D0, o que nao serve), tem de ser<br><br>k-1=3D1, ent=
ao k=3D2, entao p=3D3.<br><br>Assim, ha apenas um numero primo satisfazendo=
 a dita condicao, que eh<br>p=3D3. De fato:<br><br>3^1994+3^1995=3D3^1994.4=
=3D(3^997.2)^2.<br>
<br>Abraco,<br>&nbsp;&nbsp; Ralph<br><br>On Sun, Mar 9, 2008 at 7:01 PM, Pe=
dro J=FAnior<br>&lt;<a href=3D"mailto:pedromatematico06@xxxxxxxxx";>pedromat=
ematico06@xxxxxxxxx</a>&gt; wrote:<br>&gt; Determine a quantidade de n=FAme=
ros primos p, para que a express=E3o p^1994 +<br>
&gt; p^1995 seja um quadrado perfeito.<br>&gt; Desde j=E1 muito agradecido.=
<br>&gt; Pedro Jr<br>&gt;<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/~obmlistas/obm-l.html";>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></blockquote></div><br>

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