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[SPAM] RES: [obm-l] x^2 - xy + y^2 = Cte

To: <obm-l@xxxxxxxxxxxxxx>
Subject: [SPAM] RES: [obm-l] x^2 - xy + y^2 = Cte
From: "Bouskela" <bouskela@xxxxxxxxx>
Date: Mon, 30 Jun 2008 01:01:42 -0300
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Ol=E1, novamente,
=20
P.ex., veja que se  a=3D0 , ent=E3o a solu=E7=E3o  (-a, -b)  =E9 igual =
=E0 solu=E7=E3o  (a,
a-b)
=20
Sds.,
AB!
=20
=20


  _____ =20

De: owner-obm-l@xxxxxxxxxxxxxx [mailto:owner-obm-l@xxxxxxxxxxxxxx] Em =
nome
de Marcelo Salhab Brogliato
Enviada em: domingo, 29 de junho de 2008 23:58
Para: obm-l@xxxxxxxxxxxxxx
Assunto: Re: [obm-l] x^2 - xy + y^2 =3D Cte


Ol=E1 Bouskela,
sobre o n=FAmero de solu=E7=F5es ser menor ou igual a 4sqrt(4Cte/3) + 2, =
acredito
que esteja correto.
Mas sobre a multiplicidade, vc tem raz=E3o!
Eu cheguei que:
(a, b) ; (-a, -b) ; (b, a) ; (-b, -a) ; (a-b, a) ; (b-a, b) ; (a, a-b) ; =
(b,
b-a) ; (b-a, -a) ; (a-b, -b) ; (-a, b-a) ; (-b, a-b)
totalizando 12 solu=E7=F5es para cada encontrada.
isso vale para a !=3D b
para a=3Db, temos que eliminar: (b, a) ; (-b, -a) ; (b-a, b) ; (b, b-a) =
;
(a-b, -b) ; (-b, a-b)
sobrando apenas 6...
Mas, para a !=3D b, temos 12... e para a =3D b, temos 6... como 6*2 =3D =
12,
podemos afirmar que ser=E1 m=FAltiplo de 6 sempre..
caso eles n=E3o fossem multiplos, ai ficaria mais dif=EDcil afirmar =
alguma
coisa.

n=E3o entendi pq meu desenvolvimento precisa ser ajustado para x ou y =
igual a
zero.

abra=E7os,
Salhab






2008/6/29 Bouskela <bouskela@xxxxxxxxx>:


Ol=E1 Salhab!
=20
Voc=EA est=E1 perto da solu=E7=E3o, entretanto, ainda faltam alguns =
ajustes:
=20
P.ex., se Cte=3D100 , ent=E3o existem apenas 6 solu=E7=F5es para a eq. =
em estudo:
=20
(x, y) =3D (-10, -10) ; (10, 10) ; (10, 0) ; (-10, 0); (0, -10) e (0, =
10)
=20
O mesmo acontece para Cte =3D 1, 3, 4, 9, 12, 16, 25, 27, 36, 48, 64, =
75, 81,
100 etc.
=20
Repare que quando "x" ou "y" s=E3o iguais a "0" seu desenvolvimento =
precisa
ser ajustado...
=20
Sds.,
AB!


  _____ =20

De: owner-obm-l@xxxxxxxxxxxxxx [mailto:owner-obm-l@xxxxxxxxxxxxxx] Em =
nome
de Marcelo Salhab Brogliato
Enviada em: domingo, 29 de junho de 2008 21:16=20

Para: obm-l@xxxxxxxxxxxxxx

Assunto: Re: [obm-l] x^2 - xy + y^2 =3D Cte


Ol=E1 Bouskela,
suponha que o par (a, b) seja solu=E7=E3o de x^2 - xy + y^2 =3D Cte
ent=E3o, os pares (b, a), (-a, -b), (-b, -a) tamb=E9m s=E3o =
solu=E7=F5es.
Fatorando, temos: (x-y)^2 + xy =3D Cte.
Partindo dela, vemos que os pares (b-a, b) e (a-b, a), pois:
[(a-b)-a]^2 + (a-b)a =3D b^2 + a^2 - ab =3D Cte.
Portanto, provamos que sempre =E9 m=FAltiplo de 6.
Mas, se (b-a, b) e (a-b, a) s=E3o solu=E7=F5es, ent=E3o (b, b-a) e (a, =
a-b) tamb=E9m o
s=E3o.
E, seguindo, temos que (a-b, -b), (-b, a-b), (-a, b-a) e (b-a, -a) =
tamb=E9m o
s=E3o.
Portanto, provamos que sempre =E9 m=FAltiplo de 12.
O que estou errando?

Sobre o n=FAmero finito de solu=E7=F5es, vamos analisar esta equa=E7=E3o =
em fun=E7=E3o de
x, ent=E3o, temos:
x =3D [y +- sqrt(y^2 - 4(y^2-Cte)]/2 =3D [y +- sqrt(4Cte - 3y^2)]/2
isto =E9: 4Cte >=3D 3y^2 .... |y| <=3D sqrt(4Cte/3)
ent=E3o, podemos ir chutando todos os y's poss=EDveis e ir calculando os =
x's.
Mas, para cada y, teremos no m=E1ximo 2 x's.. portanto, temos um =
n=FAmero
limitado de solu=E7=F5es.
Ainda podemos afirmar que o n=FAmero de solu=E7=F5es =E9 <=3D =
2*[2sqrt(4Cte/3) + 1] =3D
4sqrt(4Cte/3) + 2

abra=E7os,
Salhab




2008/6/26 Bouskela <bouskela@xxxxxxxxx>:


Demonstre que a equa=E7=E3o:

x^2 - xy + y^2 =3D Cte

Onde "Cte" =E9 uma constante inteira e positiva.



Tem um n=FAmero FINITO de solu=E7=F5es inteiras; e mais: ESTE N=DAMERO =
=C9 M=DALTIPLO DE
"6".

A depender do valor da constante inteira e positiva "Cte", o n=FAmero de
solu=E7=F5es inteiras desta equa=E7=E3o =E9:

=3D 0 , p.ex.: Cte =3D 2, 5, 6, 8, 10, 11, 14, 15, 17, 18, 20, 22, 23, =
24, 26,
29, 30, 32, 33, 34, 35, 38, 40, 41, 42, 44, 45, 46, 47, 50, 51, 53, 54, =
55,
56, 58, 59, 60, 62, 65, 66, 68, 69, 70, 71, 72, 74, 77, 78, 80, 82, 83, =
85,
86, 87, 88, 89, 90, 92, 94, 95, 96, 98, 99 etc.

=3D 1 , Cte =3D 0

=3D 6 , p.ex.: Cte =3D 1, 3, 4, 9, 12, 16, 25, 27, 36, 48, 64, 75, 81, =
100 etc.

=3D 12 , p.ex.: Cte =3D 7, 13, 19, 21, 28, 31, 37, 39, 43, 52, 57, 61, =
63, 67,
73, 76, 79, 84, 93, 97 etc.

=3D 18 , p.ex.: Cte =3D 49 etc.

=3D 24 , p.ex.: Cte =3D 91 etc.

Sds.,

AB




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<BODY>
<DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana size=3D2><SPAN=20
class=3D625325203-30062008>Ol=E1, novamente,</SPAN></FONT></DIV>
<DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana size=3D2><SPAN=20
class=3D625325203-30062008></SPAN></FONT>&nbsp;</DIV>
<DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana size=3D2><SPAN=20
class=3D625325203-30062008>P.ex., veja que se&nbsp; a=3D0 , ent=E3o a =
solu=E7=E3o&nbsp;=20
(-a, -b)&nbsp; =E9 igual =E0 solu=E7=E3o&nbsp; (a, =
a-b)</SPAN></FONT></DIV>
<DIV><FONT face=3DVerdana size=3D2></FONT>&nbsp;</DIV>
<STYLE>A.psl {
	COLOR: #4e81c4; FONT-FAMILY: Verdana,Arial,fixed; TEXT-DECORATION: none
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<DIV align=3Dleft><FONT face=3DVerdana size=3D2><SPAN=20
class=3D625325203-30062008>Sds.,</SPAN></FONT></DIV>
<DIV align=3Dleft><FONT face=3DVerdana size=3D2><SPAN=20
class=3D625325203-30062008>AB!</SPAN></FONT></DIV>
<DIV align=3Dleft>&nbsp;</DIV>
<DIV>&nbsp;</DIV><BR>
<BLOCKQUOTE=20
style=3D"PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px =
solid; MARGIN-RIGHT: 0px">
  <DIV class=3DOutlookMessageHeader lang=3Dpt-br dir=3Dltr align=3Dleft>
  <HR tabIndex=3D-1>
  <FONT face=3DTahoma size=3D2><B>De:</B> owner-obm-l@xxxxxxxxxxxxxx=20
  [mailto:owner-obm-l@xxxxxxxxxxxxxx] <B>Em nome de </B>Marcelo Salhab=20
  Brogliato<BR><B>Enviada em:</B> domingo, 29 de junho de 2008=20
  23:58<BR><B>Para:</B> obm-l@xxxxxxxxxxxxxx<BR><B>Assunto:</B> Re: =
[obm-l] x^2=20
  - xy + y^2 =3D Cte<BR></FONT><BR></DIV>
  <DIV></DIV>Ol=E1 Bouskela,<BR>sobre o n=FAmero de solu=E7=F5es ser =
menor ou igual a=20
  4sqrt(4Cte/3) + 2, acredito que esteja correto.<BR>Mas sobre a =
multiplicidade,=20
  vc tem raz=E3o!<BR>Eu cheguei que:<BR>(a, b) ; (-a, -b) ; (b, a) ; =
(-b, -a) ;=20
  (a-b, a) ; (b-a, b) ; (a, a-b) ; (b, b-a) ; (b-a, -a) ; (a-b, -b) ; =
(-a, b-a)=20
  ; (-b, a-b)<BR>totalizando 12 solu=E7=F5es para cada =
encontrada.<BR>isso vale para=20
  a !=3D b<BR>para a=3Db, temos que eliminar: (b, a) ; (-b, -a) ; (b-a, =
b) ; (b,=20
  b-a) ; (a-b, -b) ; (-b, a-b)<BR>sobrando apenas 6...<BR>Mas, para a =
!=3D b,=20
  temos 12... e para a =3D b, temos 6... como 6*2 =3D 12, podemos =
afirmar que ser=E1=20
  m=FAltiplo de 6 sempre..<BR>caso eles n=E3o fossem multiplos, ai =
ficaria mais=20
  dif=EDcil afirmar alguma coisa.<BR><BR>n=E3o entendi pq meu =
desenvolvimento=20
  precisa ser ajustado para x ou y igual a=20
  zero.<BR><BR>abra=E7os,<BR>Salhab<BR><BR><BR><BR><BR><BR>
  <DIV class=3Dgmail_quote>2008/6/29 Bouskela &lt;<A=20
  href=3D"mailto:bouskela@xxxxxxxxx";>bouskela@xxxxxxxxx</A>&gt;:<BR>
  <BLOCKQUOTE class=3Dgmail_quote=20
  style=3D"PADDING-LEFT: 1ex; MARGIN: 0pt 0pt 0pt 0.8ex; BORDER-LEFT: =
rgb(204,204,204) 1px solid">
    <DIV>
    <DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana =
size=3D2><SPAN>Ol=E1=20
    Salhab!</SPAN></FONT></DIV>
    <DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana=20
    size=3D2><SPAN></SPAN></FONT>&nbsp;</DIV>
    <DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana =
size=3D2><SPAN>Voc=EA est=E1 perto da=20
    solu=E7=E3o, entretanto, ainda faltam alguns =
ajustes:</SPAN></FONT></DIV>
    <DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana=20
    size=3D2><SPAN></SPAN></FONT>&nbsp;</DIV>
    <DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana =
size=3D2><SPAN>P.ex., se Cte=3D100 ,=20
    ent=E3o existem apenas 6 solu=E7=F5es para a eq. em =
estudo:</SPAN></FONT></DIV>
    <DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana=20
    size=3D2><SPAN></SPAN></FONT>&nbsp;</DIV>
    <DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana size=3D2><SPAN>(x, =
y) =3D (-10, -10)=20
    ; (10, 10) ; (10, 0) ; (-10, 0); (0, -10) e (0, =
10)</SPAN></FONT></DIV>
    <DIV dir=3Dltr align=3Dleft><FONT face=3DVerdana=20
    size=3D2><SPAN></SPAN></FONT>&nbsp;</DIV>
    <DIV dir=3Dltr align=3Dleft><FONT =
size=3D+0><SPAN></SPAN></FONT><SPAN><FONT=20
    face=3DVerdana size=3D2>O mesmo acontece para Cte =3D 1, 3, 4, 9, =
12, 16, 25, 27,=20
    36, 48, 64, 75, 81, 100 etc.</FONT></SPAN></DIV>
    <DIV align=3Dleft><FONT face=3DVerdana size=3D2></FONT>&nbsp;</DIV>
    <DIV align=3Dleft><FONT face=3DVerdana size=3D2><SPAN>Repare que =
quando "x" ou "y"=20
    s=E3o iguais a "0" seu desenvolvimento&nbsp;precisa ser=20
    ajustado...</SPAN></FONT></DIV>
    <DIV align=3Dleft><FONT face=3DVerdana size=3D2></FONT>&nbsp;</DIV>
    <DIV><SPAN><FONT face=3DVerdana size=3D2>Sds.,</FONT></SPAN></DIV>
    <DIV><SPAN><FONT face=3DVerdana size=3D2>AB!</FONT></SPAN></DIV><BR>
    <BLOCKQUOTE=20
    style=3D"PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: =
rgb(0,0,0) 2px solid; MARGIN-RIGHT: 0px">
      <DIV lang=3Dpt-br dir=3Dltr align=3Dleft>
      <HR>
      <FONT face=3DTahoma size=3D2><B>De:</B> <A=20
      href=3D"mailto:owner-obm-l@xxxxxxxxxxxxxx"=20
      target=3D_blank>owner-obm-l@xxxxxxxxxxxxxx</A> [mailto:<A=20
      href=3D"mailto:owner-obm-l@xxxxxxxxxxxxxx"=20
      target=3D_blank>owner-obm-l@xxxxxxxxxxxxxx</A>] <B>Em nome de =
</B>Marcelo=20
      Salhab Brogliato<BR><B>Enviada em:</B> domingo, 29 de junho de =
2008 21:16
      <DIV class=3DIh2E3d><BR><B>Para:</B> <A =
href=3D"mailto:obm-l@xxxxxxxxxxxxxx"=20
      target=3D_blank>obm-l@xxxxxxxxxxxxxx</A><BR></DIV><B>Assunto:</B> =
Re:=20
      [obm-l] x^2 - xy + y^2 =3D Cte<BR></FONT><BR></DIV>
      <DIV>
      <DIV></DIV>
      <DIV class=3DWj3C7c>
      <DIV></DIV>Ol=E1 Bouskela,<BR>suponha que o par (a, b) seja =
solu=E7=E3o de x^2 -=20
      xy + y^2 =3D Cte<BR>ent=E3o, os pares (b, a), (-a, -b), (-b, -a) =
tamb=E9m s=E3o=20
      solu=E7=F5es.<BR>Fatorando, temos: (x-y)^2 + xy =3D =
Cte.<BR>Partindo dela, vemos=20
      que os pares (b-a, b) e (a-b, a), pois:<BR>[(a-b)-a]^2 + (a-b)a =
=3D b^2 +=20
      a^2 - ab =3D Cte.<BR>Portanto, provamos que sempre =E9 m=FAltiplo =
de 6.<BR>Mas,=20
      se (b-a, b) e (a-b, a) s=E3o solu=E7=F5es, ent=E3o (b, b-a) e (a, =
a-b) tamb=E9m o=20
      s=E3o.<BR>E, seguindo, temos que (a-b, -b), (-b, a-b), (-a, b-a) e =
(b-a, -a)=20
      tamb=E9m o s=E3o.<BR>Portanto, provamos que sempre =E9 m=FAltiplo =
de 12.<BR>O que=20
      estou errando?<BR><BR>Sobre o n=FAmero finito de solu=E7=F5es, =
vamos analisar=20
      esta equa=E7=E3o em fun=E7=E3o de x, ent=E3o, temos:<BR>x =3D [y =
+- sqrt(y^2 -=20
      4(y^2-Cte)]/2 =3D [y +- sqrt(4Cte - 3y^2)]/2<BR>isto =E9: 4Cte =
&gt;=3D 3y^2 ....=20
      |y| &lt;=3D sqrt(4Cte/3)<BR>ent=E3o, podemos ir chutando todos os =
y's=20
      poss=EDveis e ir calculando os x's.<BR>Mas, para cada y, teremos =
no m=E1ximo 2=20
      x's.. portanto, temos um n=FAmero limitado de =
solu=E7=F5es.<BR>Ainda podemos=20
      afirmar que o n=FAmero de solu=E7=F5es =E9 &lt;=3D =
2*[2sqrt(4Cte/3) + 1] =3D=20
      4sqrt(4Cte/3) + 2<BR><BR>abra=E7os,<BR>Salhab<BR><BR><BR><BR>
      <DIV class=3Dgmail_quote>2008/6/26 Bouskela &lt;<A=20
      href=3D"mailto:bouskela@xxxxxxxxx"=20
      target=3D_blank>bouskela@xxxxxxxxx</A>&gt;:<BR>
      <BLOCKQUOTE class=3Dgmail_quote=20
      style=3D"PADDING-LEFT: 1ex; MARGIN: 0pt 0pt 0pt 0.8ex; =
BORDER-LEFT: rgb(204,204,204) 1px solid">
        <DIV>
        <DIV>
        <P><FONT face=3DVerdana size=3D2>Demonstre que a =
equa=E7=E3o:</FONT></P>
        <P><FONT face=3DVerdana size=3D2>x^2 - xy + y^2 =3D =
Cte</FONT></P>
        <P><FONT face=3DVerdana size=3D2>Onde "Cte" =E9 uma constante =
inteira e=20
        positiva.</FONT></P>
        <P><FONT face=3DVerdana size=3D2></FONT></P>
        <P><FONT face=3DVerdana size=3D2>Tem um n=FAmero FINITO de =
solu=E7=F5es inteiras;=20
        e mais: ESTE N=DAMERO =C9 M=DALTIPLO DE "6".</FONT></P>
        <P><FONT face=3DVerdana size=3D2>A depender do valor da =
constante inteira e=20
        positiva "Cte", o n=FAmero de solu=E7=F5es inteiras desta =
equa=E7=E3o=20
=E9:</FONT></P>
        <P><FONT face=3DVerdana size=3D2>=3D 0 , p.ex.: Cte =3D 2, 5, 6, =
8, 10, 11, 14,=20
        15, 17, 18, 20, 22, 23, 24, 26, 29, 30, 32, 33, 34, 35, 38, 40, =
41, 42,=20
        44, 45, 46, 47, 50, 51, 53, 54, 55, 56, 58, 59, 60, 62, 65, 66, =
68, 69,=20
        70, 71, 72, 74, 77, 78, 80, 82, 83, 85, 86, 87, 88, 89, 90, 92, =
94, 95,=20
        96, 98, 99 etc.</FONT></P>
        <P><FONT face=3DVerdana size=3D2>=3D 1 , Cte =3D 0</FONT></P>
        <P><FONT face=3DVerdana size=3D2>=3D 6 , p.ex.: Cte =3D 1, 3, 4, =
9, 12, 16, 25,=20
        27, 36, 48, 64, 75, 81, 100 etc.</FONT></P>
        <P><FONT face=3DVerdana size=3D2>=3D 12 , p.ex.: Cte =3D 7, 13, =
19, 21, 28, 31,=20
        37, 39, 43, 52, 57, 61, 63, 67, 73, 76, 79, 84, 93, 97 =
etc.</FONT></P>
        <P><FONT face=3DVerdana size=3D2>=3D 18 , p.ex.: Cte =3D 49 =
etc.</FONT></P>
        <P><FONT face=3DVerdana size=3D2>=3D 24 , p.ex.: Cte =3D 91 =
etc.</FONT></P>
        <P><FONT face=3DVerdana size=3D2>Sds.,</FONT></P>
        <P><SPAN><FONT face=3DVerdana=20
      =
size=3D2>AB</FONT></SPAN></P></DIV></DIV></BLOCKQUOTE></DIV><BR></DIV></D=
IV></BLOCKQUOTE></DIV></BLOCKQUOTE></DIV><BR></BLOCKQUOTE></BODY></HTML>

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