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[SPAM] Re: [obm-l] Inteiros!!!



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Falou nobre amigo, que Deus continue lhe dando sabedoria...
Abra=E7os


2008/4/4 Paulo Santa Rita <paulo.santarita@xxxxxxxxx>:

> Ola Pedro e demais
> colegas desta lista ... OBM-L,
>
> Obviamente que todo par (X,Y) da forma (X,-X) e solucao, pois :
> X^3 + (-X)^3 =3D 0 =3D (X+(-X))/2.  Em particular, (0,0) e solucao.
>
> Se, porem, X+Y # 0, teremos :
> X^3 + Y^3 =3D (X+Y)*(X^2 -XY + Y^2) =3D (X+Y)/2. =3D> X^2 - XY + Y^2 =3D =
1/2
> =3D> (X-Y)^2 =3D - (X^2 +Y^2) .
> A possibilidade aqui, logicamente, e :  X-Y=3D0 e X^2+Y^2 =3D 0. Mas isso
> da (X,Y)=3D(0,0)
> o que contraria a hipotese X+Y # 0
>
> Assim, todas as solucoes inteiras sao {(X,-X) / X e inteiro }
>
> Um Abracao a Todos
> Paulo Santa Rita
> 6,0A2D,040408
>
> 2008/4/4 Pedro J=FAnior <pedromatematico06@xxxxxxxxx>:
> > 02. Ache todos os pares tais de n=FAmeros inteiros (x, y) tais que:
> > x^3 + y^3 =3D (x + y)^2
>
> =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=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<http://www.mat.puc-rio.br=
/%7Eobmlistas/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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Falou nobre amigo, que Deus continue lhe dando sabedoria...<br>Abra=E7os<br=
><br><br><div class=3D"gmail_quote">2008/4/4 Paulo Santa Rita &lt;<a href=
=3D"mailto:paulo.santarita@xxxxxxxxx";>paulo.santarita@xxxxxxxxx</a>&gt;:<br=
><blockquote class=3D"gmail_quote" style=3D"border-left: 1px solid rgb(204,=
 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
Ola Pedro e demais<br>
colegas desta lista ... OBM-L,<br>
<br>
Obviamente que todo par (X,Y) da forma (X,-X) e solucao, pois :<br>
X^3 + (-X)^3 =3D 0 =3D (X+(-X))/2. &nbsp;Em particular, (0,0) e solucao.<br=
>
<br>
Se, porem, X+Y # 0, teremos :<br>
X^3 + Y^3 =3D (X+Y)*(X^2 -XY + Y^2) =3D (X+Y)/2. =3D&gt; X^2 - XY + Y^2 =3D=
 1/2<br>
=3D&gt; (X-Y)^2 =3D - (X^2 +Y^2) .<br>
A possibilidade aqui, logicamente, e : &nbsp;X-Y=3D0 e X^2+Y^2 =3D 0. Mas i=
sso<br>
da (X,Y)=3D(0,0)<br>
o que contraria a hipotese X+Y # 0<br>
<br>
Assim, todas as solucoes inteiras sao {(X,-X) / X e inteiro }<br>
<br>
Um Abracao a Todos<br>
Paulo Santa Rita<br>
6,0A2D,040408<br>
<br>
2008/4/4 Pedro J=FAnior &lt;<a href=3D"mailto:pedromatematico06@xxxxxxxxx";>=
pedromatematico06@xxxxxxxxx</a>&gt;:<br>
<div class=3D"Ih2E3d">&gt; 02. Ache todos os pares tais de n=FAmeros inteir=
os (x, y) tais que:<br>
&gt; x^3 + y^3 =3D (x + y)^2<br>
<br>
</div>=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=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/%7Eobmlistas/obm-l.html"; target=3D"_bl=
ank">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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