Note que
x
2
−
d
y
2
=
1
⟺
(
x
−
d
y
)
(
x
+
d
y
)
=
1
{\textstyle x^{2}-dy^{2}=1\Longleftrightarrow (x-{\sqrt {d}}y)(x+{\sqrt {d}}y)=1}
Se
d
∈
N
{\displaystyle {\sqrt {d}}\in \mathbb {N} }
d
{\displaystyle d}
(
x
−
d
y
)
(
x
+
d
y
)
=
1
⟹
{
x
−
d
y
=
1
x
+
d
y
=
1
ou
{
x
−
d
y
=
−
1
x
+
d
y
=
−
1
{\displaystyle (x-{\sqrt {d}}y)(x+{\sqrt {d}}y)=1\Longrightarrow {\begin{cases}x-{\sqrt {d}}y=1\\x+{\sqrt {d}}y=1\end{cases}}\quad {\text{ou}}\quad {\begin{cases}x-{\sqrt {d}}y=-1\\x+{\sqrt {d}}y=-1\end{cases}}}
Como
d
∈
N
{\displaystyle {\sqrt {d}}\in \mathbb {N} }
n
{\displaystyle n}
n
=
d
{\displaystyle n={\sqrt {d}}}
{
x
−
n
y
=
1
x
+
n
y
=
1
⟹
(
x
−
n
y
)
+
(
x
+
n
y
)
=
1
+
1
{\displaystyle {\begin{cases}x-ny=1\\x+ny=1\end{cases}}\Longrightarrow (x-ny)+(x+ny)=1+1}
Assim,
(
x
+
x
)
+
(
n
y
−
n
y
)
=
2
⟹
x
=
1.
{\displaystyle (x+x)+(ny-ny)=2\Longrightarrow x=1.}
Substituindo
x
=
1
{\displaystyle x=1}
y
=
0
{\displaystyle y=0}
Resolvendo o caso
{
x
−
n
y
=
−
1
x
+
n
y
=
−
1
{\displaystyle {\begin{cases}x-ny=-1\\x+ny=-1\end{cases}}}
Teremos
x
=
−
1
{\displaystyle x=-1}
y
=
0
{\displaystyle y=0}
(
x
,
y
)
=
(
1
,
0
)
{\displaystyle (x,y)=(1,0)}
(
x
,
y
)
=
(
−
1
,
0
)
{\displaystyle (x,y)=(-1,0)}
soluções triviais .
Joseph Louis Lagrange provou que, contanto que
d
{\displaystyle d}
quadrado perfeito , a equação de Pell tem infinitas soluções inteiras distintas.
Os inversíveis em
Z
[
d
]
{\displaystyle Z[{\sqrt {d}}]}
editar
Como
x
2
−
d
y
2
=
(
x
+
d
y
)
(
x
−
d
y
)
{\displaystyle x^{2}-dy^{2}=(x+{\sqrt {d}}y)(x-{\sqrt {d}}y)}
a
+
d
b
{\displaystyle a+{\sqrt {d}}b}
Z
[
d
]
{\displaystyle \mathbb {Z} [{\sqrt {d}}]}
Definiremos a norma
N
:
R
→
R
{\displaystyle N:\mathbb {R} \rightarrow \mathbb {R} }
N
(
u
)
=
u
u
¯
.
{\displaystyle N(u)=u{\bar {u}}.}
Onde
u
=
x
+
d
y
{\displaystyle u=x+{\sqrt {d}}y}
u
¯
=
x
−
d
y
{\displaystyle {\bar {u}}=x-{\sqrt {d}}y}
Seja
u
=
x
+
d
y
{\displaystyle u=x+{\sqrt {d}}y}
potência
u
n
;
2
≤
n
∈
N
{\displaystyle u^{n};2\leq n\in \mathbb {N} }
Demonstração
Usaremos o processo de indução . Assim, note que para
n
=
2
{\displaystyle n=2}
N
(
u
2
)
=
N
(
(
x
+
d
y
)
(
x
+
d
y
)
)
=
N
(
x
2
+
d
y
2
+
2
d
x
y
)
=
(
x
2
+
d
y
2
+
2
d
x
y
)
(
x
2
+
y
2
−
2
d
x
y
)
=
(
x
2
+
d
y
2
)
2
−
(
2
d
x
y
)
2
=
x
4
+
2
d
x
2
y
2
+
d
2
y
4
−
4
d
x
2
y
2
=
x
4
−
2
d
x
2
y
2
+
d
2
y
4
=
(
x
2
−
d
y
2
)
(
x
2
−
d
y
2
)
=
N
(
u
)
N
(
u
)
=
1
{\displaystyle {\begin{aligned}N(u^{2})&=N{\Big (}(x+{\sqrt {d}}y)(x+{\sqrt {d}}y){\Big )}\\&=N{\Big (}x^{2}+dy^{2}+2{\sqrt {d}}xy{\Big )}\\&=\left(x^{2}+dy^{2}+2{\sqrt {d}}xy\right)\left(x^{2}+y^{2}-2{\sqrt {d}}xy\right)\\&=\left(x^{2}+dy^{2}\right)^{2}-\left(2{\sqrt {d}}xy\right)^{2}\\&=x^{4}+2dx^{2}y^{2}+d^{2}y^{4}-4dx^{2}y^{2}\\&=x^{4}-2dx^{2}y^{2}+d^{2}y^{4}\\&=(x^{2}-dy^{2})(x^{2}-dy^{2})\\&=N(u)N(u)\\&=1\end{aligned}}}
Suponha que para todo
n
≤
k
,
N
(
u
n
)
=
N
(
u
n
−
1
⋅
u
)
=
1
{\displaystyle n\leq k,N(u^{n})=N(u^{n-1}\cdot u)=1}
n
=
k
+
1
{\displaystyle n=k+1}
N
(
u
k
+
1
)
=
N
(
u
k
⋅
u
)
=
N
(
u
k
)
N
(
u
)
=
1
{\displaystyle {\begin{aligned}N(u^{k+1})&=N(u^{k}\cdot u)\\&=N(u^{k})N(u)\\&=1\end{aligned}}}
Assim, fica demonstrado que qualquer potência de uma solução é, ainda, uma solução da equação de Pell.
Seja
u
=
x
+
d
y
{\displaystyle u=x+{\sqrt {d}}y}
u
k
=
x
k
+
d
y
k
{\displaystyle u^{k}=x_{k}+{\sqrt {d}}y_{k}}
u
{\displaystyle u}
x
k
=
x
k
−
1
x
+
d
y
k
−
1
y
y
k
=
x
k
−
1
y
+
y
k
−
1
x
{\displaystyle {\begin{aligned}x_{k}&=x_{k-1}x+dy_{k-1}y\\y_{k}&=x_{k-1}y+y_{k-1}x\end{aligned}}}
Demonstração
Note que
u
k
=
u
k
−
1
u
{\displaystyle u^{k}=u^{k-1}u}
u
k
=
(
x
k
−
1
+
d
y
k
−
1
)
(
x
+
d
y
)
=
x
k
−
1
x
+
x
k
−
1
d
y
+
d
y
k
−
1
x
+
d
y
k
−
1
y
⇒
x
k
+
d
y
k
=
(
x
k
−
1
x
+
d
y
k
−
1
y
)
+
d
(
x
k
−
1
y
+
y
k
−
1
x
)
{\displaystyle {\begin{aligned}u^{k}&=(x_{k-1}+{\sqrt {d}}y_{k-1})(x+{\sqrt {d}}y)\\&=x_{k-1}x+x_{k-1}{\sqrt {d}}y+{\sqrt {d}}y_{k-1}x+dy_{k-1}y\\\Rightarrow x_{k}+{\sqrt {d}}y_{k}&=(x_{k-1}x+dy_{k-1}y)+{\sqrt {d}}(x_{k-1}y+y_{k-1}x)\end{aligned}}}
Comparando os termos, temos que
x
k
=
x
k
−
1
x
+
d
y
k
−
1
y
e
y
k
=
x
k
−
1
y
+
y
k
−
1
x
{\displaystyle x_{k}=x_{k-1}x+dy_{k-1}y{\text{ e }}y_{k}=x_{k-1}y+y_{k-1}x}
Observação : note que
u
=
x
1
+
y
1
d
{\displaystyle u=x_{1}+y_{1}{\sqrt {d}}}
O grupo dos inversíveis em
Z
[
d
]
{\displaystyle \mathbb {Z} [{\sqrt {d}}]}
editar
Seja a restrição,
N
:
Z
[
d
]
∗
→
Z
∗
{\displaystyle N:\mathbb {Z} [{\sqrt {d}}]^{*}\rightarrow \mathbb {Z} ^{*}}
Onde
Z
[
d
]
∗
=
{
u
∈
Z
[
d
]
;
N
(
u
)
=
1
}
{\displaystyle \mathbb {Z} [{\sqrt {d}}]^{*}=\{u\in \mathbb {Z} [{\sqrt {d}}];N(u)=1\}}
Z
∗
=
{
−
1
,
1
}
{\displaystyle \mathbb {Z} ^{*}=\{-1,1\}}
Z
{\displaystyle \mathbb {Z} }
Note que a norma definida dessa forma é um homomorfismo .
Agora iremos verificar que
G
=
(
Z
[
d
]
∗
,
⋅
)
{\displaystyle G=(\mathbb {Z} [{\sqrt {d}}]^{*},\cdot )}
grupo :
Associatividade: como
Z
[
d
]
⊂
R
{\displaystyle \mathbb {Z} [{\sqrt {d}}]\subset \mathbb {R} }
Elemento neutro: Note que
1
⋅
u
=
u
∀
u
∈
R
{\displaystyle 1\cdot u=u\ \forall \ u\in \mathbb {R} }
u
∈
Z
[
d
]
∗
⇒
1
⋅
u
=
u
{\displaystyle u\in \mathbb {Z} [{\sqrt {d}}]^{*}\Rightarrow 1\cdot u=u}
Inverso : Por construção, todo
u
∈
Z
[
d
]
∗
{\displaystyle u\in \mathbb {Z} [{\sqrt {d}}]^{*}}
Seja
u
∈
Z
[
d
]
∗
{\displaystyle u\in \mathbb {Z} [{\sqrt {d}}]^{*}}
u
k
{\displaystyle u^{k}}
u
{\displaystyle u}
G
{\displaystyle G}
G
{\displaystyle G}
abeliano . Assim,
G
=
{
u
0
,
u
1
,
…
}
=
{
1
,
u
,
u
2
,
.
.
.
}
{\displaystyle G=\{u^{0},u^{1},\dots \}=\{1,u,u^{2},...\}}
Da equação de Pell, temos que
x
2
=
1
+
d
y
2
{\displaystyle x^{2}=1+dy^{2}}
x
2
y
2
=
1
y
2
+
d
⟹
x
y
=
1
y
2
+
d
.
{\displaystyle {\frac {x^{2}}{y^{2}}}={\frac {1}{y^{2}}}+d\Longrightarrow {\frac {x}{y}}={\sqrt {{\frac {1}{y^{2}}}+d}}.}
Intuitivamente, podemos perceber que a razão
x
/
y
{\displaystyle x/y}
d
{\displaystyle {\sqrt {d}}}
1
/
y
2
{\displaystyle 1/y^{2}}
lim
y
→
+
∞
1
y
2
+
d
=
d
.
{\displaystyle \lim _{y\rightarrow +\infty }{\sqrt {{\frac {1}{y^{2}}}+d}}={\sqrt {d}}.}
Com isso, podemos usar frações continuadas para escrever o número irracional
d
{\displaystyle {\sqrt {d}}}
d
=
[
a
0
,
a
1
,
…
,
2
a
0
¯
]
{\displaystyle {\sqrt {d}}=[a_{0},{\overline {a_{1},\dots ,2a_{0}}}]}
Seja a equação de Pell
x
2
−
2
y
2
=
1
{\displaystyle x^{2}-2y^{2}=1}
A solução fundamental dessa equação é o par
(
3
,
2
)
{\displaystyle (3,2)}
3
2
−
2
(
2
)
2
=
9
−
8
=
1
{\displaystyle 3^{2}-2(2)^{2}=9-8=1}
u
=
3
+
2
2
{\displaystyle u=3+2{\sqrt {2}}}
u
k
=
3
x
k
−
1
+
4
y
k
−
1
+
2
(
x
k
−
1
2
+
y
k
−
1
3
)
{\displaystyle u^{k}=3x_{k-1}+4y_{k-1}+{\sqrt {2}}(x_{k-1}2+y_{k-1}3)}
u
2
=
3
x
1
+
4
y
1
+
(
2
x
1
+
3
y
1
)
2
=
17
+
12
2
u
3
=
3
x
2
+
4
y
2
+
(
2
x
2
+
3
y
2
)
2
=
99
+
70
2
u
4
=
3
x
3
+
4
y
3
+
(
2
x
3
+
3
y
3
)
2
=
577
+
408
2
{\displaystyle {\begin{aligned}u^{2}&=3x_{1}+4y_{1}+(2x_{1}+3y_{1}){\sqrt {2}}=17+12{\sqrt {2}}\\u^{3}&=3x_{2}+4y_{2}+(2x_{2}+3y_{2}){\sqrt {2}}=99+70{\sqrt {2}}\\u^{4}&=3x_{3}+4y_{3}+(2x_{3}+3y_{3}){\sqrt {2}}=577+408{\sqrt {2}}\\\end{aligned}}}
⋮
{\displaystyle \vdots }
Fazendo a razão entre os coeficientes das soluções, temos que
3
2
=
1
,
5
;
17
12
=
1
,
41
6
¯
;
99
70
=
1
,
41
428571
¯
;
577
408
=
1.4142
1568627450980392
¯
{\displaystyle {\frac {3}{2}}=1,5;\quad {\frac {17}{12}}=1,41{\overline {6}};\quad {\frac {99}{70}}=1,41{\overline {428571}};\quad {\frac {577}{408}}=1.4142{\overline {1568627450980392}}}
Essas razões estão se aproximando de
2
=
1
,
414213562373
…
{\displaystyle {\sqrt {2}}=1,414213562373\dots }
[ 3] [ 4]
![{\displaystyle Z[{\sqrt {d}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7b3fe9692a6b1b14f547dfb1256527965851520)
![{\displaystyle \mathbb {Z} [{\sqrt {d}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ded5eeae10b73c37c38418c32a12ffd7695de44)
![{\displaystyle N:\mathbb {Z} [{\sqrt {d}}]^{*}\rightarrow \mathbb {Z} ^{*}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c00c87839c26193608b8734da7051fdc263384b7)
![{\displaystyle \mathbb {Z} [{\sqrt {d}}]^{*}=\{u\in \mathbb {Z} [{\sqrt {d}}];N(u)=1\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36432f92f287a7000db603fd2094563c8a0e9fce)
![{\displaystyle G=(\mathbb {Z} [{\sqrt {d}}]^{*},\cdot )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f40e2452d9f5a98c1034a99e17956b684731837)
![{\displaystyle \mathbb {Z} [{\sqrt {d}}]\subset \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/49adfe12e314e4f32cfa2b0c15f1b9184e0a0748)
![{\displaystyle u\in \mathbb {Z} [{\sqrt {d}}]^{*}\Rightarrow 1\cdot u=u}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c214c2534547daa67458308d1b0f1c4efb3b164)
![{\displaystyle u\in \mathbb {Z} [{\sqrt {d}}]^{*}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c88a62536cff7b3d9c276812ab2ed7819a13c162)
![{\displaystyle {\sqrt {d}}=[a_{0},{\overline {a_{1},\dots ,2a_{0}}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/990f0a8a53ca7bcba7190d617b6c34ef4f91ec14)
