Zenbaki irrazional bikarratuen orrialde hau sortu da, Pellen ekuazioaren oinarrizko ebazpena determinatzeko bidean: Euler, Lagrangek eta Galoisek zenbaki irrazional bikarratu laburtuak karakterizatzen dituztelako, zatiki jarrai en bidezko garapena eginez.
Dirichletek Pellen ekuazioa bateragarria dela frogatzen du, baina ez du ebazpena determinatzen. Bere teoremaren frogapenean erabilitako metodoa jarraituz oinarrizko ebazpena determinatzen da.
Ondorengoa da prozesua: Dirichletek
x
2
−
p
y
2
=
1
{\displaystyle x^{2}-py^{2}=1}
ekuazioaren ebazpen baten existentzia frogatzen du. Ebazpen honen existentzia desberdintza batek determinatzen du:
|
x
/
y
−
p
|
<
1
/
y
2
{\displaystyle |{x}/{y}-{\sqrt {p}}|<{1}/{y^{2}}}
. Desberdintza hori betetzen duten
x
/
y
{\displaystyle x/y}
zatikiak,
p
{\displaystyle {\sqrt {p}}}
zenbakiaren
n
{\displaystyle n}
. hondar bat dira ( zatiki jarraien bidezko bere garapenaren hondarra ). Honela Pell-en ekuazioaren ebazpenak:
n
{\displaystyle n}
. hondarretara mugatzen dira.
Galoisek zenbaki irrazional bikarratu laburtuak karakterizatuko ditu, eta ondorioztatuko du: zenbaki erreal bat, irrazional bikarratu laburtua dela baldin eta soilik baldin zatiki jarraien bidezko bere garapena periodiko hutsa bada. Honela erraz ondorioztatzen da
p
{\displaystyle {\sqrt {p}}}
moduko zenbakien zatiki jarraien bidezko garapena:
T
{\displaystyle T}
-periodikoa dela:
h
e
i
n
(
1
)
{\displaystyle hein(1)}
-etik aurrera, eta ebazpenak, periodoa osatu aurreko hondarrren artean daudela:
R
T
−
1
,
R
2
T
−
1
,
.
.
.
{\displaystyle R_{T-1},R_{2T-1},...}
Hondarraren zenbakitzailea eta izendatzailea gorakorrak direnez: ebazpen minimoa edo oinarrizko ebazpena: ebazpena den periodo aurreko lehen hondarrean aurkitzen da:
R
T
−
1
{\displaystyle R_{T-1}}
, edo
R
2
T
−
1
{\displaystyle R_{2T-1}}
.
α
∈
R
{\displaystyle \alpha \in \mathbb {R} }
, zenbaki irrazional bat bikarratua izendatzen da, koefiziente osoak dituen bigarren graduko polinomio baten erroa bada.
N
2
=
{
n
2
:
n
∈
N
}
{\displaystyle \mathbb {N} ^{2}=\{n^{2}:n\in \mathbb {N} \}}
. Adieraziko da.
α
∈
Q
(
p
)
{\displaystyle \alpha \in \mathbb {Q} ({\sqrt {p}})}
, bada non
α
=
a
+
b
p
{\displaystyle \alpha =a+b{\sqrt {p}}}
, orduan
α
¯
=
a
−
b
p
{\displaystyle {\overline {\alpha }}=a-b{\sqrt {p}}}
, bere konjokatua izendatuko da.
α
∈
R
{\displaystyle \alpha \in \mathbb {R} }
ren
n
{\displaystyle n}
. hondarra:
R
n
{\displaystyle R_{n}}
, beste modu honetan ere adieraziko da:
R
n
(
α
)
{\displaystyle R_{n}(\alpha )}
, beste zenbaki irrazionalen
n
{\displaystyle n}
. hondarretik bereizteko.
α
∈
R
{\displaystyle \alpha \in \mathbb {R} }
, zenbaki irrazional bikarratua, laburtua dela esango dugu, baldin eta:
α
>
1
{\displaystyle \alpha >1}
bada eta
−
1
<
α
¯
<
0
{\displaystyle -1<{\overline {\alpha }}<0}
.
Zeinetan
α
{\displaystyle \alpha }
eta
α
¯
{\displaystyle {\overline {\alpha }}}
polinomio berdinaren erro konjokatuak diren.
Izan bedi
α
{\displaystyle \alpha }
zenbaki irrazionala, zatiki jarraien bidezko bere garapena,
α
=
[
a
0
,
a
1
,
.
.
.
,
a
n
,
.
.
.
]
{\displaystyle \alpha =[a_{0},a_{1},...,a_{n},...]}
periodikoa dela esango dugu,
(
a
n
)
n
≥
0
{\displaystyle (a_{n})_{n\geq 0}}
, segida periodikoa bada, eta segidaren periodoa:
T
{\displaystyle T}
, bada
α
{\displaystyle \alpha }
-ren zatiki jarraien bidezko garapenaren periodoa:
T
{\displaystyle T}
, dela esango dugu.
Periodoaren lehen elementua:
a
N
{\displaystyle a_{N}}
, bada
α
{\displaystyle \alpha }
-ren zatiki jarraien garapena,
T
{\displaystyle T}
-periodikoa dela esango da
h
e
i
n
(
N
)
{\displaystyle hein(N)}
-tik aurrera (heina N). Honela adieraziko da:
α
=
[
a
0
,
a
1
,
.
.
.
,
a
N
−
1
,
a
N
,
.
.
.
,
a
N
+
T
−
1
¯
]
{\displaystyle \alpha =[a_{0},a_{1},...,a_{N-1},{\overline {a_{N},...,a_{N+T-1}}}]}
.
Eulerrek frogatu zuen 1748.urtean:
α
{\displaystyle \alpha }
-ren zatiki jarraien garapena periodikoa bada, orduan
α
{\displaystyle \alpha }
zenbaki irrazionala bikarratua dela, eta Lagrangek frogatu zuen 1768.urtean:
α
{\displaystyle \alpha }
zenbaki irrazional bikarratua bada, orduan zatiki jarraien bidezko garapena periodikoa dela. Biak idatziko dira proposizio batean.
Izan bedi
α
=
[
a
0
,
a
1
,
.
.
.
,
a
n
,
.
.
.
]
{\displaystyle \alpha =[a_{0},a_{1},...,a_{n},...]}
, zatiki jarraien bidezko bere garapena. Orduan: garapena periodikoa da baldin eta soilik baldin
α
{\displaystyle \alpha }
zenbaki irrazional bikarratua bada.
Froga:
⇒
E
u
l
e
r
(
1748
)
{\displaystyle \Rightarrow Euler(1748)}
Suposa bedi
α
{\displaystyle \alpha }
-ren zatiki jarraien bidezko garapena,
T
{\displaystyle T}
-periodikoa dela,
h
e
i
n
(
N
)
{\displaystyle hein(N)}
-tik aurrera:
α
=
[
a
0
,
a
1
,
.
.
.
,
a
N
−
1
,
a
N
,
.
.
.
,
a
N
+
T
−
1
¯
]
{\displaystyle \alpha =[a_{0},a_{1},...,a_{N-1},{\overline {a_{N},...,a_{N+T-1}}}]}
.
α
{\displaystyle \alpha }
-ren
n
{\displaystyle n}
. hondarra (zatiki laburtezin gisara adierazia):
R
n
=
[
a
0
,
a
1
,
.
.
.
,
a
n
]
=
p
n
q
n
{\displaystyle R_{n}=[a_{0},a_{1},...,a_{n}]={\frac {p_{n}}{q_{n}}}}
.
α
{\displaystyle \alpha }
-ren
N
{\displaystyle N}
.zatiki osatua:
α
N
=
[
a
N
,
.
.
.
,
a
N
+
T
−
1
¯
]
{\displaystyle \alpha _{N}=[{\overline {a_{N},...,a_{N+T-1}}}]}
.
Eta
N
{\displaystyle N}
.zatiki osatuaren:
α
N
{\displaystyle \alpha _{N}}
,
n
{\displaystyle n}
. hondarra (zatiki laburtezina):
R
n
(
α
N
)
=
[
a
N
,
a
N
+
1
,
.
.
.
,
a
N
+
n
]
=
r
n
s
n
{\displaystyle R_{n}(\alpha _{N})=[a_{N},a_{N+1},...,a_{N+n}]={\frac {r_{n}}{s_{n}}}}
.
α
{\displaystyle \alpha }
zenbaki irrazional baten,
n
{\displaystyle n}
.zatiki osatuak:
α
n
{\displaystyle \alpha _{n}}
, ondorengo ezaugarriaren bidez definitzen dira:
α
=
[
a
0
,
a
1
,
.
.
.
,
a
n
−
1
,
α
n
]
{\displaystyle \alpha =[a_{0},a_{1},...,a_{n-1},\alpha _{n}]}
(ikus zatiki jarraiak).
α
N
{\displaystyle \alpha _{N}}
periodikoa dela kontuan hartuz:
α
N
=
[
a
N
,
.
.
.
,
a
N
+
T
−
1
,
a
N
,
.
.
.
,
a
N
+
T
−
1
¯
]
=
[
a
N
,
.
.
.
,
a
N
+
T
−
1
,
α
N
]
=
r
T
−
1
α
N
+
r
T
−
2
s
T
−
1
α
N
+
s
T
−
2
{\displaystyle \alpha _{N}=[a_{N},...,a_{N+T-1},{\overline {a_{N},...,a_{N+T-1}}}]=[a_{N},...,a_{N+T-1},\alpha _{N}]={\frac {r_{T-1}\alpha _{N}+r_{T-2}}{s_{T-1}\alpha _{N}+s_{T-2}}}}
. Ondorioz:
s
T
−
1
α
N
2
+
(
s
T
−
2
−
r
T
−
1
)
α
N
−
r
T
−
2
=
0
{\displaystyle s_{T-1}\alpha _{N}^{2}+(s_{T-2}-r_{T-1})\alpha _{N}-r_{T-2}=0}
.
s
T
−
1
∈
N
{\displaystyle s_{T-1}\in \mathbb {N} }
denez,
α
N
{\displaystyle \alpha _{N}}
bikarratua da.
α
N
∉
Q
{\displaystyle \alpha _{N}\not \in \mathbb {Q} }
. Absurdura bideratuz: Baldin eta:
α
N
∈
Q
⇒
α
∈
Q
{\displaystyle \alpha _{N}\in \mathbb {Q} \Rightarrow \alpha \in \mathbb {Q} }
, eta ondorioz zatiki jarraien bidezko garapena finitua da (zatiki jarraiak) eta
α
=
[
a
0
,
a
1
,
.
.
.
,
a
n
,
α
N
]
{\displaystyle \alpha =[a_{0},a_{1},...,a_{n},\alpha _{N}]}
denez,
α
{\displaystyle \alpha }
-ren garapena ere finitua da, zeina ezinezkoa den (hipotesiagatik periodikoa delako).
Ondorioz:
α
N
{\displaystyle \alpha _{N}}
, zenbaki irrazional bikarratua da:
α
N
=
a
+
b
c
{\displaystyle \alpha _{N}=a+b{\sqrt {c}}}
modukoa non:
a
,
b
∈
Q
;
b
≠
0
{\displaystyle a,b\in \mathbb {Q} ;b\neq 0}
, eta
c
∈
N
−
N
2
{\displaystyle c\in \mathbb {N} -\mathbb {N} ^{2}}
.
α
=
[
a
0
,
a
1
,
.
.
.
,
a
N
−
1
,
α
N
]
⇒
α
=
p
N
−
1
α
N
+
p
N
−
2
q
N
−
1
α
N
+
q
N
−
2
{\displaystyle \alpha =[a_{0},a_{1},...,a_{N-1},\alpha _{N}]\Rightarrow \alpha ={\frac {p_{N-1}\alpha _{N}+p_{N-2}}{q_{N-1}\alpha _{N}+q_{N-2}}}}
α
=
p
N
−
1
α
N
+
p
N
−
2
q
N
−
1
α
N
+
q
N
−
2
=
p
N
−
1
(
a
+
b
c
)
+
p
N
−
2
q
N
−
1
(
a
+
b
c
)
+
q
N
−
2
=
u
+
v
c
{\displaystyle \alpha ={\frac {p_{N-1}\alpha _{N}+p_{N-2}}{q_{N-1}\alpha _{N}+q_{N-2}}}={\frac {p_{N-1}(a+b{\sqrt {c}})+p_{N-2}}{q_{N-1}(a+b{\sqrt {c}})+q_{N-2}}}=u+v{\sqrt {c}}}
, gisakoa da, zeinetan:
u
,
v
∈
Q
{\displaystyle u,v\in \mathbb {Q} }
.
v
≠
0
{\displaystyle v\neq 0}
frogatuko da.
v
=
0
⇒
α
∈
Q
{\displaystyle v=0\Rightarrow \alpha \in \mathbb {Q} }
, eta beraz zatiki jarraien bidezko garapena finitua. Ezinezkoa.
Ondorioz:
α
{\displaystyle \alpha }
, zenbaki irrazional bikarratua da.
⇐
L
a
g
r
a
n
g
e
(
1768
)
{\displaystyle \Leftarrow Lagrange(1768)}
α
∈
R
{\displaystyle \alpha \in \mathbb {R} }
, zenbaki irraziona bikarratua bada:
∃
a
,
b
,
c
∈
Z
:
{\displaystyle \exists a,b,c\in \mathbb {Z} :}
i)
a
≠
0
{\displaystyle a\neq 0}
, ii)
a
α
2
+
b
α
+
c
=
0
{\displaystyle a\alpha ^{2}+b\alpha +c=0}
eta iii)
Δ
=
b
2
−
4
a
c
∈
N
−
N
2
{\displaystyle \Delta =b^{2}-4ac\in \mathbb {N} -\mathbb {N} ^{2}}
.
∀
n
≥
2
⇒
α
=
[
a
0
,
a
1
,
.
.
.
,
a
n
−
1
,
α
n
]
=
p
n
−
1
α
n
+
p
n
−
2
q
n
−
1
α
n
+
q
n
−
2
{\displaystyle \forall n\geq 2\Rightarrow \alpha =[a_{0},a_{1},...,a_{n-1},\alpha _{n}]={\frac {p_{n-1}\alpha _{n}+p_{n-2}}{q_{n-1}\alpha _{n}+q_{n-2}}}}
, orohar betetzen da.
Eta
α
{\displaystyle \alpha }
, bikarratua izateagatik:
a
(
p
n
−
1
α
n
+
p
n
−
2
q
n
−
1
α
n
+
q
n
−
2
)
2
+
b
p
n
−
1
α
n
+
p
n
−
2
q
n
−
1
α
n
+
q
n
−
2
+
c
=
0
⇔
{\displaystyle a({\frac {p_{n-1}\alpha _{n}+p_{n-2}}{q_{n-1}\alpha _{n}+q_{n-2}}})^{2}+b{\frac {p_{n-1}\alpha _{n}+p_{n-2}}{q_{n-1}\alpha _{n}+q_{n-2}}}+c=0\Leftrightarrow }
a
(
p
n
−
1
α
n
+
p
n
−
2
)
2
+
b
(
p
n
−
1
α
n
+
p
n
−
2
)
(
q
n
−
1
α
n
+
q
n
−
2
)
+
c
(
q
n
−
1
α
n
+
q
n
−
2
)
2
=
0
⇔
{\displaystyle a(p_{n-1}\alpha _{n}+p_{n-2})^{2}+b(p_{n-1}\alpha _{n}+p_{n-2})(q_{n-1}\alpha _{n}+q_{n-2})+c(q_{n-1}\alpha _{n}+q_{n-2})^{2}=0\Leftrightarrow }
A
n
α
n
2
+
B
n
α
n
+
C
n
=
0.
{\displaystyle A_{n}\alpha _{n}^{2}+B_{n}\alpha _{n}+C_{n}=0.}
n
≥
2
{\displaystyle n\geq 2}
rentzat:
{
A
n
=
a
p
n
−
1
2
+
b
p
n
−
1
q
n
−
1
+
c
q
n
−
1
2
B
n
=
2
a
p
n
−
1
p
n
−
2
+
b
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
+
2
c
q
n
−
1
q
n
−
2
C
n
=
a
p
n
−
2
2
+
b
p
n
−
2
q
n
−
2
+
c
q
n
−
2
2
{\displaystyle {\begin{cases}A_{n}=ap_{n-1}^{2}+bp_{n-1}q_{n-1}+cq_{n-1}^{2}\\B_{n}=2ap_{n-1}p_{n-2}+b(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})+2cq_{n-1}q_{n-2}\\C_{n}=ap_{n-2}^{2}+bp_{n-2}q_{n-2}+cq_{n-2}^{2}\end{cases}}}
Bat .
A
n
≠
0
{\displaystyle A_{n}\neq 0}
, frogatuko da.
A
n
=
a
p
n
−
1
2
+
b
p
n
−
1
q
n
−
1
+
c
q
n
−
1
2
=
0
⇔
a
(
p
n
−
1
q
n
−
1
)
2
+
b
p
n
−
1
q
n
−
1
+
c
=
0
{\displaystyle A_{n}=ap_{n-1}^{2}+bp_{n-1}q_{n-1}+cq_{n-1}^{2}=0\Leftrightarrow a({\frac {p_{n-1}}{q_{n-1}}})^{2}+b{\frac {p_{n-1}}{q_{n-1}}}+c=0}
, eta ondorioz
α
{\displaystyle \alpha }
, arrazionala, ezinezkoa.
Bi :
p
(
x
)
=
a
x
2
+
b
x
+
c
{\displaystyle p(x)=ax^{2}+bx+c}
, ren diskriminantea eta
p
n
(
x
)
=
A
n
x
2
+
B
n
x
+
C
n
{\displaystyle p_{n}(x)=A_{n}x^{2}+B_{n}x+C_{n}}
en diskrimintea berdinak direla frogatuko da.
Hots:
∀
n
≥
2
{\displaystyle \forall n\geq 2}
.
Δ
n
=
B
n
2
−
4
A
n
C
n
=
b
2
−
4
a
c
=
Δ
{\displaystyle \Delta _{n}=B_{n}^{2}-4A_{n}C_{n}=b^{2}-4ac=\Delta }
.
{
B
n
2
−
4
A
n
C
n
=
(
2
a
p
n
−
1
p
n
−
2
+
b
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
+
2
c
q
n
−
1
q
n
−
2
)
2
−
4
(
a
p
n
−
1
2
+
b
p
n
−
1
q
n
−
1
+
c
q
n
−
1
2
)
(
a
p
n
−
2
2
+
b
p
n
−
2
q
n
−
2
+
c
q
n
−
2
2
)
{\displaystyle {\begin{cases}B_{n}^{2}-4A_{n}C_{n}=(2ap_{n-1}p_{n-2}+b(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})+2cq_{n-1}q_{n-2})^{2}\\-4(ap_{n-1}^{2}+bp_{n-1}q_{n-1}+cq_{n-1}^{2})(ap_{n-2}^{2}+bp_{n-2}q_{n-2}+cq_{n-2}^{2})\end{cases}}}
Karratuak garatuz:
{
B
n
2
−
4
A
n
C
n
=
4
a
2
p
n
−
1
2
p
n
−
2
2
+
b
2
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
2
+
4
c
2
q
n
−
1
2
q
n
−
2
2
+
4
a
b
p
n
−
1
p
n
−
2
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
+
4
b
c
q
n
−
1
q
n
−
2
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
+
8
a
c
p
n
−
1
p
n
−
2
q
n
−
1
q
n
−
2
−
4
a
2
p
n
−
1
2
p
n
−
2
2
−
4
a
b
p
n
−
1
2
p
n
−
2
q
n
−
2
−
4
a
c
p
n
−
1
2
q
n
−
2
2
−
4
a
b
p
n
−
1
p
n
−
2
2
q
n
−
1
−
4
b
2
p
n
−
1
p
n
−
2
q
n
−
1
q
n
−
2
−
4
b
c
p
n
−
1
q
n
−
1
q
n
−
2
2
−
4
a
c
p
n
−
2
2
q
n
−
1
2
−
4
b
c
p
n
−
2
q
n
−
1
2
q
n
−
2
−
4
c
2
q
n
−
1
2
q
n
−
2
2
{\displaystyle {\begin{cases}B_{n}^{2}-4A_{n}C_{n}=4a^{2}p_{n-1}^{2}p_{n-2}^{2}+b^{2}(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})^{2}+4c^{2}q_{n-1}^{2}q_{n-2}^{2}\\+4abp_{n-1}p_{n-2}(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})+4bcq_{n-1}q_{n-2}(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})\\+8acp_{n-1}p_{n-2}q_{n-1}q_{n-2}\\-4a^{2}p_{n-1}^{2}p_{n-2}^{2}-4abp_{n-1}^{2}p_{n-2}q_{n-2}-4acp_{n-1}^{2}q_{n-2}^{2}\\-4abp_{n-1}p_{n-2}^{2}q_{n-1}-4b^{2}p_{n-1}p_{n-2}q_{n-1}q_{n-2}-4bcp_{n-1}q_{n-1}q_{n-2}^{2}\\-4acp_{n-2}^{2}q_{n-1}^{2}-4bcp_{n-2}q_{n-1}^{2}q_{n-2}-4c^{2}q_{n-1}^{2}q_{n-2}^{2}\end{cases}}}
Terminuak ezbatuz:
{
B
n
2
−
4
A
n
C
n
=
b
2
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
2
+
8
a
c
p
n
−
1
p
n
−
2
q
n
−
1
q
n
−
2
−
4
a
c
p
n
−
1
2
q
n
−
2
2
−
4
b
2
p
n
−
1
p
n
−
2
q
n
−
1
q
n
−
2
−
4
a
c
p
n
−
2
2
q
n
−
1
2
{\displaystyle {\begin{cases}B_{n}^{2}-4A_{n}C_{n}=b^{2}(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})^{2}\\\\+8acp_{n-1}p_{n-2}q_{n-1}q_{n-2}\\-4acp_{n-1}^{2}q_{n-2}^{2}\\-4b^{2}p_{n-1}p_{n-2}q_{n-1}q_{n-2}\\-4acp_{n-2}^{2}q_{n-1}^{2}\end{cases}}}
Terminuak batuz:
{
B
n
2
−
4
A
n
C
n
=
b
2
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
2
−
4
b
2
p
n
−
1
p
n
−
2
q
n
−
1
q
n
−
2
−
4
a
c
(
p
n
−
1
2
q
n
−
2
2
+
q
n
−
1
2
p
n
−
2
2
−
2
p
n
−
1
p
n
−
2
q
n
−
1
q
n
−
2
)
{\displaystyle {\begin{cases}B_{n}^{2}-4A_{n}C_{n}=b^{2}(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})^{2}\\-4b^{2}p_{n-1}p_{n-2}q_{n-1}q_{n-2}\\-4ac(p_{n-1}^{2}q_{n-2}^{2}+q_{n-1}^{2}p_{n-2}^{2}-2p_{n-1}p_{n-2}q_{n-1}q_{n-2})\end{cases}}}
B
n
2
−
4
A
n
C
n
=
b
2
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
2
−
4
a
c
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
2
{\displaystyle B_{n}^{2}-4A_{n}C_{n}=b^{2}(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})^{2}-4ac(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})^{2}}
B
n
2
−
4
A
n
C
n
=
(
b
2
−
4
a
c
)
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
2
{\displaystyle B_{n}^{2}-4A_{n}C_{n}=(b^{2}-4ac)(p_{n-1}q_{n-2}+q_{n-1}p_{n-2})^{2}}
Adierazpena lortu da.
Eta
(
p
n
−
1
q
n
−
2
+
q
n
−
1
p
n
−
2
)
2
=
(
−
1
)
2
n
{\displaystyle (p_{n-1}q_{n-2}+q_{n-1}p_{n-2})^{2}=(-1)^{2n}}
berdintza aplikatuz (zatiki jarraiak).
∀
n
≥
2
{\displaystyle \forall n\geq 2}
,
B
n
2
−
4
A
n
C
n
=
b
2
−
4
a
c
{\displaystyle B_{n}^{2}-4A_{n}C_{n}=b^{2}-4ac}
.
Hiru
(
A
n
)
n
≥
0
,
(
B
n
)
n
≥
0
,
(
C
n
)
n
≥
0
{\displaystyle (A_{n})_{n\geq 0},(B_{n})_{n\geq 0},(C_{n})_{n\geq 0}}
segidak barnatuak daudela frogatuko da.
α
{\displaystyle \alpha }
-ren
n
{\displaystyle n}
. hondarra:
R
n
=
p
n
q
n
=
[
a
0
,
a
1
,
.
.
.
,
a
n
−
1
,
a
n
]
{\displaystyle R_{n}={\frac {p_{n}}{q_{n}}}=[a_{0},a_{1},...,a_{n-1},a_{n}]}
.
Zatiki jarraiek zenbaki irrazionaletara duten hurbilketa egokiengatik ( zatiki jarraiak: hurbilketa egokiak ):
|
α
−
R
n
−
1
|
<
1
q
n
q
n
−
1
⇔
|
q
n
−
1
α
−
p
n
−
1
|
<
1
q
n
≤
1
q
n
−
1
{\displaystyle |\alpha -R_{n-1}|<{\frac {1}{q_{n}q_{n-1}}}\Leftrightarrow |q_{n-1}\alpha -p_{n-1}|<{\frac {1}{q_{n}}}\leq {\frac {1}{q_{n-1}}}}
betetzen da, eta ondorioz:
∀
n
≥
2
{\displaystyle \forall n\geq 2}
rentzat,
∃
δ
n
−
1
∈
(
−
1
,
1
)
:
q
n
−
1
α
−
p
n
−
1
=
−
δ
n
−
1
q
n
−
1
⇔
p
n
−
1
=
q
n
−
1
α
+
δ
n
−
1
q
n
−
1
{\displaystyle \exists \delta _{n-1}\in (-1,1):q_{n-1}\alpha -p_{n-1}=-{\frac {\delta _{n-1}}{q_{n-1}}}\Leftrightarrow p_{n-1}=q_{n-1}\alpha +{\frac {\delta _{n-1}}{q_{n-1}}}}
.
A
n
=
a
p
n
−
1
2
+
b
p
n
−
1
q
n
−
1
+
c
q
n
−
1
2
=
a
(
q
n
−
1
α
+
δ
n
−
1
q
n
−
1
)
2
+
b
(
q
n
−
1
α
+
δ
n
−
1
q
n
−
1
)
q
n
−
1
+
c
q
n
−
1
2
{\displaystyle A_{n}=ap_{n-1}^{2}+bp_{n-1}q_{n-1}+cq_{n-1}^{2}=a(q_{n-1}\alpha +{\frac {\delta _{n-1}}{q_{n-1}}})^{2}+b(q_{n-1}\alpha +{\frac {\delta _{n-1}}{q_{n-1}}})q_{n-1}+cq_{n-1}^{2}}
Eta:
a
α
2
+
b
α
+
c
=
0
{\displaystyle a\alpha ^{2}+b\alpha +c=0}
, berdintza betetzen denez:
A
n
=
q
n
−
1
2
(
a
α
2
+
b
α
+
c
)
+
2
a
δ
n
−
1
α
+
a
δ
n
−
1
2
q
n
−
1
2
+
b
δ
n
−
1
=
2
a
δ
n
−
1
α
+
a
δ
n
−
1
2
q
n
−
1
2
+
b
δ
n
−
1
{\displaystyle A_{n}=q_{n-1}^{2}(a\alpha ^{2}+b\alpha +c)+2a\delta _{n-1}\alpha +a{\frac {\delta _{n-1}^{2}}{q_{n-1}^{2}}}+b\delta _{n-1}=2a\delta _{n-1}\alpha +a{\frac {\delta _{n-1}^{2}}{q_{n-1}^{2}}}+b\delta _{n-1}}
Ondorioz:
∀
n
≥
1
,
|
A
n
|
≤
2
|
a
|
|
α
|
+
|
a
|
+
|
b
|
⇒
(
A
n
)
n
≥
0
{\displaystyle \forall n\geq 1,|A_{n}|\leq 2|a||\alpha |+|a|+|b|\Rightarrow (A_{n})_{n\geq 0}}
Barnatua.
∀
n
≥
2
,
|
C
n
|
=
|
A
n
−
1
|
≤
2
|
a
|
|
α
|
+
|
a
|
+
|
b
|
⇒
(
C
n
)
n
≥
0
{\displaystyle \forall n\geq 2,|C_{n}|=|A_{n-1}|\leq 2|a||\alpha |+|a|+|b|\Rightarrow (C_{n})_{n\geq 0}}
, Barnatua.
Eta
B
n
2
−
4
A
n
C
n
=
b
2
−
4
a
c
⇒
B
n
2
=
4
A
n
C
n
+
b
2
−
4
a
c
≤
4
|
A
n
|
|
C
n
|
+
b
2
−
4
a
c
{\displaystyle B_{n}^{2}-4A_{n}C_{n}=b^{2}-4ac\Rightarrow B_{n}^{2}=4A_{n}C_{n}+b^{2}-4ac\leq 4|A_{n}||C_{n}|+b^{2}-4ac}
.
⇒
(
B
n
)
n
≥
0
{\displaystyle \Rightarrow (B_{n})_{n\geq 0}}
, Barnatua.
Lau :
α
=
[
a
0
,
a
1
,
.
.
.
,
a
n
,
.
.
.
]
{\displaystyle \alpha =[a_{0},a_{1},...,a_{n},...]}
, periodikoa dela frogatuko da.
(
A
n
)
n
≥
0
,
(
B
n
)
n
≥
0
,
(
C
n
)
n
≥
0
{\displaystyle (A_{n})_{n\geq 0},(B_{n})_{n\geq 0},(C_{n})_{n\geq 0}}
segidak barnatuak daude hirugarren atalagatik, orduan:
∃
A
,
B
,
C
∈
N
:
∀
n
∈
N
∪
{
0
}
⇒
|
A
n
|
≤
A
,
|
B
n
|
≤
B
,
|
C
n
|
≤
C
{\displaystyle \exists A,B,C\in \mathbb {N} :\forall n\in \mathbb {N} \cup \{0\}\Rightarrow |A_{n}|\leq A,|B_{n}|\leq B,|C_{n}|\leq C}
, eta ondorioz:
{
(
A
n
,
B
n
,
C
n
)
:
n
∈
N
∪
{
0
}
}
⊂
[
1
,
A
]
×
[
1
,
B
]
×
[
1
,
C
]
{\displaystyle \{(A_{n},B_{n},C_{n}):n\in \mathbb {N} \cup \{0\}\}\subset [1,A]\times [1,B]\times [1,C]}
.
Zenbaki arrunten multzoa infinitua denez, eta
[
1
,
A
]
×
[
1
,
B
]
×
[
1
,
C
]
{\displaystyle [1,A]\times [1,B]\times [1,C]}
, multzoa berriz finitua.
⇒
∃
j
,
k
,
l
∈
N
∪
{
0
}
{\displaystyle \Rightarrow \exists j,k,l\in \mathbb {N} \cup \{0\}}
desberdinak non:
(
A
j
,
B
j
,
C
j
)
=
(
A
k
,
B
k
,
C
k
)
=
(
A
l
,
B
l
,
C
l
)
⇒
p
j
(
x
)
=
p
k
(
x
)
=
p
l
(
x
)
{\displaystyle (A_{j},B_{j},C_{j})=(A_{k},B_{k},C_{k})=(A_{l},B_{l},C_{l})\Rightarrow p_{j}(x)=p_{k}(x)=p_{l}(x)}
.
Eta
α
j
,
α
k
,
α
l
{\displaystyle \alpha _{j},\alpha _{k},\alpha _{l}}
aipaturiko polinomioen erroak badira hurrenez hurren. Polinomioak berdinak direnez, hiru erro horiek
p
j
(
x
)
{\displaystyle p_{j}(x)}
-en erroak dira. Ondorioz gutsienez bi erro berdinak dira:
α
j
=
α
k
;
k
>
j
{\displaystyle \alpha _{j}=\alpha _{k};k>j}
, suposatuko da orokortasunik galdu gabe.
T
=
k
−
j
{\displaystyle T=k-j}
, aukeratuz.
α
j
=
α
k
⇒
[
a
j
,
a
j
+
1
,
.
.
.
]
=
[
a
k
,
a
k
+
1
,
.
.
.
]
⇒
a
j
+
n
=
a
k
+
n
{\displaystyle \alpha _{j}=\alpha _{k}\Rightarrow [a_{j},a_{j+1},...]=[a_{k},a_{k+1},...]\Rightarrow a_{j+n}=a_{k+n}}
.
∀
n
≥
0
{\displaystyle \forall n\geq 0}
-rentzat.
a
j
+
n
=
a
j
+
n
+
k
−
j
=
a
j
+
n
+
T
{\displaystyle a_{j+n}=a_{j+n+k-j}=a_{j+n+T}}
⇒
{\displaystyle \Rightarrow }
-rentzat
a
n
=
a
n
+
T
{\displaystyle a_{n}=a_{n+T}}
.
Ondorioz
(
a
n
)
n
≥
0
{\displaystyle (a_{n})_{n\geq 0}}
,
T
{\displaystyle T}
-ren zatitzaile baten periodokoa da, gutsienez
h
e
i
n
(
j
)
{\displaystyle hein(j)}
-tik aurrera.
Zenbaki irrazional baten, zatiki jarraien bidezko garapena, periodiko hutsa dela esaten da,
h
e
i
n
(
0
)
{\displaystyle hein(0)}
-tik aurrera periodikoa bada.
Galoisek zenbaki irrazional bikarratu laburtuak karakterizatzen ditu.
Proposizioa2 (Galois 1828)
aldatu
Zenbaki erreal bat: irrazional bikarratu laburtua da, baldin eta soilik baldin zatiki jarraien bidezko bere garapena periodiko hutsa bada.
Froga :
⇒
{\displaystyle \Rightarrow }
Izan bedi
α
{\displaystyle \alpha }
zenbaki erreala, eta suposa bedi zenbaki irrazional bikarratu laburtua dela.
Bat :
∀
n
∈
N
∪
{
0
}
{\displaystyle \forall n\in \mathbb {N} \cup \{0\}}
,
α
n
{\displaystyle \alpha _{n}}
:
α
{\displaystyle \alpha }
-ren
n
{\displaystyle n}
. zatiki laburtua, zenbaki irrazional bikarratu laburtua dela frogatuko da.
α
{\displaystyle \alpha }
, irrazional bikarratua bada, orduan Lagrangek dio zatiki jarraien bidezko bere garapena periodikoa dela.
∀
n
∈
N
∪
{
0
}
{\displaystyle \forall n\in \mathbb {N} \cup \{0\}}
rentzat,
α
n
{\displaystyle \alpha _{n}}
-ren koefizienteak
α
{\displaystyle \alpha }
-ren koefizienteen berdinak direnez
h
e
i
n
(
n
)
{\displaystyle hein(n)}
-tik aurrera,
α
n
{\displaystyle \alpha _{n}}
-ren zatiki jarraien bidezko garapena periodikoa da, orduan Eulerrek dio
α
n
{\displaystyle \alpha _{n}}
zenbaki irrazional bikarratua dela.
Ondorioz:
∀
n
∈
N
∪
{
0
}
{\displaystyle \forall n\in \mathbb {N} \cup \{0\}}
-rentzat
α
n
{\displaystyle \alpha _{n}}
zenbaki irrazional bikarratua da.
∀
n
∈
N
∪
{
0
}
{\displaystyle \forall n\in \mathbb {N} \cup \{0\}}
-rentzat,
α
n
{\displaystyle \alpha _{n}}
laburtua dela frogatuko da indukzioz.
α
=
α
0
{\displaystyle \alpha =\alpha _{0}}
, laburtua da hipotesiagatik.
Suposa bedi
α
n
{\displaystyle \alpha _{n}}
laburtua dela, orduan:
α
n
>
1
;
α
n
¯
∈
(
−
1
,
0
)
{\displaystyle \alpha _{n}>1;{\overline {\alpha _{n}}}\in (-1,0)}
.
α
n
+
1
=
1
α
n
−
[
α
n
]
⇔
α
n
=
a
n
+
1
α
n
+
1
>
1
{\displaystyle \alpha _{n+1}={\frac {1}{\alpha _{n}-[\alpha _{n}]}}\Leftrightarrow \alpha _{n}=a_{n}+{\frac {1}{\alpha _{n+1}}}>1}
, (
[
α
n
]
=
a
n
{\displaystyle [\alpha _{n}]=a_{n}}
, parte osoa ).
α
n
−
a
n
∈
(
0
,
1
)
⇒
1
α
n
+
1
∈
(
0
,
1
)
⇒
α
n
+
1
>
1
{\displaystyle \alpha _{n}-a_{n}\in (0,1)\Rightarrow {\frac {1}{\alpha _{n+1}}}\in (0,1)\Rightarrow \alpha _{n+1}>1}
, laburtua izatearen lehen baldintza.
α
n
=
a
n
+
1
α
n
+
1
⇒
α
n
¯
=
a
n
+
1
α
n
+
1
¯
=
a
n
¯
+
(
1
α
n
+
1
)
¯
=
a
n
+
1
α
n
+
1
¯
{\displaystyle \alpha _{n}=a_{n}+{\frac {1}{\alpha _{n+1}}}\Rightarrow {\overline {\alpha _{n}}}={\overline {a_{n}+{\frac {1}{\alpha _{n+1}}}}}={\overline {a_{n}}}+{\overline {({\frac {1}{\alpha _{n+1}}})}}=a_{n}+{\frac {1}{\overline {\alpha _{n+1}}}}}
, konjokatuen ezaugarriak erabiliz.
α
n
>
1
⇒
[
α
n
]
=
a
n
≥
1
{\displaystyle \alpha _{n}>1\Rightarrow [\alpha _{n}]=a_{n}\geq 1}
α
n
+
1
¯
=
1
α
n
¯
−
a
n
⇒
1
α
n
+
1
¯
=
α
n
¯
−
a
n
≤
α
n
¯
−
1
<
−
1
⇒
α
n
+
1
¯
∈
(
−
1
,
0
)
{\displaystyle {\overline {\alpha _{n+1}}}={\frac {1}{{\overline {\alpha _{n}}}-a_{n}}}\Rightarrow {\frac {1}{\overline {\alpha _{n+1}}}}={\overline {\alpha _{n}}}-a_{n}\leq {\overline {\alpha _{n}}}-1<-1\Rightarrow {\overline {\alpha _{n+1}}}\in (-1,0)}
, laburtua izatearen bigarren baldintza.
α
n
{\displaystyle \alpha _{n}}
laburtua
⇒
α
n
+
1
{\displaystyle \Rightarrow \alpha _{n+1}}
laburtua ondorioztatu da.
Indukzio printzipioagatik:
∀
n
∈
N
∪
{
0
}
{\displaystyle \forall n\in \mathbb {N} \cup \{0\}}
-rentzat
α
n
{\displaystyle \alpha _{n}}
laburtua.
Bi:
α
{\displaystyle \alpha }
periodiko hutsa dela frogatuko da.
Absurdura bideratuz,
α
{\displaystyle \alpha }
periodiko hutsa ez bada:
α
=
[
a
0
,
a
1
,
.
.
.
,
a
N
−
1
,
a
N
,
.
.
.
,
a
N
+
T
−
1
¯
]
{\displaystyle \alpha =[a_{0},a_{1},...,a_{N-1},{\overline {a_{N},...,a_{N+T-1}}}]}
, zeinetan
T
{\displaystyle T}
periodoa den, eta
N
{\displaystyle N}
berriz lehen periodoaren lehen indizea.
Periodiko hutsa ez izateagatik:
N
≥
1
{\displaystyle N\geq 1}
.
α
=
[
a
0
,
a
1
,
.
.
.
,
a
N
−
1
,
α
N
]
{\displaystyle \alpha =[a_{0},a_{1},...,a_{N-1},\alpha _{N}]}
,
T
{\displaystyle T}
-peridikoa izateagatik ,
N
{\displaystyle N}
. zatiki laburtua:
α
N
{\displaystyle \alpha _{N}}
,
T
{\displaystyle T}
-periodikoa da.
α
N
=
[
a
0
,
a
1
,
.
.
.
,
a
N
+
T
−
1
¯
]
=
[
a
N
+
T
,
a
N
+
T
+
1
,
.
.
.
,
a
N
+
2
T
−
1
¯
]
=
α
N
+
T
{\displaystyle \alpha _{N}=[{\overline {a_{0},a_{1},...,a_{N+T-1}}}]=[{\overline {a_{N+T},a_{N+T+1},...,a_{N+2T-1}}}]=\alpha _{N+T}}
.
α
{\displaystyle \alpha }
-ren,
n
{\displaystyle n}
. hondarrak:
α
n
{\displaystyle \alpha _{n}}
laburtuak dira. Lehen baldintzagatik:
α
N
=
α
N
+
T
⇒
α
N
¯
=
α
N
+
T
¯
⇒
−
1
α
N
¯
=
−
1
α
N
+
T
¯
>
1
{\displaystyle \alpha _{N}=\alpha _{N+T}\Rightarrow {\overline {\alpha _{N}}}={\overline {\alpha _{N+T}}}\Rightarrow -{\frac {1}{\overline {\alpha _{N}}}}=-{\frac {1}{\overline {\alpha _{N+T}}}}>1}
.
{
−
1
α
N
=
a
N
−
1
−
α
N
−
1
⇒
−
1
α
N
¯
=
a
N
−
1
−
α
N
−
1
¯
>
1
−
1
α
N
+
T
=
a
N
+
T
−
1
−
α
N
+
T
−
1
⇒
−
1
α
N
+
T
¯
=
a
N
+
T
−
1
−
α
N
+
T
−
1
¯
>
1
{\displaystyle {\begin{cases}-{\frac {1}{\alpha _{N}}}=a_{N-1}-\alpha _{N-1}\Rightarrow -{\frac {1}{\overline {\alpha _{N}}}}=a_{N-1}-{\overline {\alpha _{N-1}}}>1\\-{\frac {1}{\alpha _{N+T}}}=a_{N+T-1}-\alpha _{N+T-1}\Rightarrow -{\frac {1}{\overline {\alpha _{N+T}}}}=a_{N+T-1}-{\overline {\alpha _{N+T-1}}}>1\end{cases}}}
Bigarren baldintzagatik:
{
α
N
−
1
¯
∈
(
−
1
,
0
)
;
a
N
−
1
−
α
N
−
1
¯
>
1
⇒
a
N
−
1
>
0
α
N
+
T
−
1
¯
∈
(
−
1
,
0
)
;
a
N
+
T
−
1
−
α
N
+
T
−
1
¯
>
1
⇒
a
N
+
T
−
1
>
0
{\displaystyle {\begin{cases}{\overline {\alpha _{N-1}}}\in (-1,0);a_{N-1}-{\overline {\alpha _{N-1}}}>1\Rightarrow a_{N-1}>0\\{\overline {\alpha _{N+T-1}}}\in (-1,0);a_{N+T-1}-{\overline {\alpha _{N+T-1}}}>1\Rightarrow a_{N+T-1}>0\end{cases}}}
.
[
a
N
−
1
−
α
N
−
1
¯
]
=
a
N
−
1
{\displaystyle [a_{N-1}-{\overline {\alpha _{N-1}}}]=a_{N-1}}
, eta
[
a
N
+
T
−
1
−
α
N
+
T
−
1
¯
]
=
a
N
+
T
−
1
{\displaystyle [a_{N+T-1}-{\overline {\alpha _{N+T-1}}}]=a_{N+T-1}}
. (Parte osoak kasurako)
[
−
1
α
N
¯
]
=
[
−
1
α
N
+
T
¯
]
⇒
a
N
−
1
=
[
a
N
−
1
−
α
N
−
1
¯
]
=
[
a
N
+
T
−
1
−
α
N
+
T
−
1
¯
]
=
a
N
+
T
−
1
{\displaystyle [-{\frac {1}{\overline {\alpha _{N}}}}]=[-{\frac {1}{\overline {\alpha _{N+T}}}}]\Rightarrow a_{N-1}=[a_{N-1}-{\overline {\alpha _{N-1}}}]=[a_{N+T-1}-{\overline {\alpha _{N+T-1}}}]=a_{N+T-1}}
Eta ondorioz, lehen periodoaren lehen indizea
N
−
1
{\displaystyle N-1}
, edo txikiagoa da, zeina ezinezkoa den.
⇐
{\displaystyle \Leftarrow }
Suposa dezagun
α
{\displaystyle \alpha }
-ren zatiki jarraien bidezko garapena, periodiko hutsa dela, orduan:
∃
T
∈
N
:
α
=
[
a
0
,
a
1
,
.
.
.
,
a
T
−
1
¯
]
⇒
a
0
=
a
T
∈
N
{\displaystyle \exists T\in \mathbb {N} :\alpha =[{\overline {a_{0},a_{1},...,a_{T-1}}}]\Rightarrow a_{0}=a_{T}\in \mathbb {N} }
, ondorioz:
α
>
a
0
≥
1
{\displaystyle \alpha >a_{0}\geq 1}
T
=
1
⇒
α
=
[
a
0
,
α
]
=
a
0
+
1
α
{\displaystyle T=1\Rightarrow \alpha =[a_{0},\alpha ]=a_{0}+{\frac {1}{\alpha }}}
, eta beraz
α
{\displaystyle \alpha }
,
p
(
x
)
=
x
2
−
a
0
x
−
1
{\displaystyle p(x)=x^{2}-a_{0}x-1}
, en erroa da.
p
(
x
)
{\displaystyle p(x)}
-en erroak:
α
=
a
0
+
a
0
2
+
4
2
;
α
¯
=
a
0
−
a
0
2
+
4
2
⇒
α
¯
∈
(
−
1
,
0
)
{\displaystyle \alpha ={\frac {a_{0}+{\sqrt {a_{0}^{2}+4}}}{2}};{\overline {\alpha }}={\frac {a_{0}-{\sqrt {a_{0}^{2}+4}}}{2}}\Rightarrow {\overline {\alpha }}\in (-1,0)}
,
α
{\displaystyle \alpha }
laburtua.
T
>
1
⇒
α
=
[
a
0
,
.
.
.
,
a
T
−
1
,
α
]
=
p
T
−
1
α
+
p
T
−
2
q
T
−
1
α
+
q
T
−
2
⇔
q
T
−
1
α
2
+
(
q
T
−
2
−
p
T
−
1
)
α
−
p
T
−
2
=
0
{\displaystyle T>1\Rightarrow \alpha =[a_{0},...,a_{T-1},\alpha ]={\frac {p_{T-1}\alpha +p_{T-2}}{q_{T-1}\alpha +q_{T-2}}}\Leftrightarrow q_{T-1}\alpha ^{2}+(q_{T-2}-p_{T-1})\alpha -p_{T-2}=0}
, eta ondorioz:
α
{\displaystyle \alpha }
,
p
(
x
)
=
q
T
−
1
x
2
+
(
q
T
−
2
−
p
T
−
1
)
x
−
p
T
−
2
{\displaystyle p(x)=q_{T-1}x^{2}+(q_{T-2}-p_{T-1})x-p_{T-2}}
-en erroa da, eta
α
>
a
0
≥
1
{\displaystyle \alpha >a_{0}\geq 1}
.
Bigarren erroa
(
−
1
,
0
)
{\displaystyle (-1,0)}
bitartean dagoela frogatuko da.
p
(
0
)
=
−
p
T
−
2
<
0
{\displaystyle p(0)=-p_{T-2}<0}
.
q
n
↑
,
p
n
↑
{\displaystyle q_{n}\uparrow ,p_{n}\uparrow }
, gorakorrak direnez:
p
(
−
1
)
=
q
T
−
1
−
q
T
−
2
+
p
T
−
1
−
p
T
−
2
≥
0
{\displaystyle p(-1)=q_{T-1}-q_{T-2}+p_{T-1}-p_{T-2}\geq 0}
.
p
(
−
1
)
=
0
⇒
p
T
−
1
=
q
T
−
1
;
p
T
−
2
=
q
T
−
2
⇒
R
T
−
1
=
R
T
−
2
{\displaystyle p(-1)=0\Rightarrow p_{T-1}=q_{T-1};p_{T-2}=q_{T-2}\Rightarrow R_{T-1}=R_{T-2}}
, ezinezkoa. Ondorioz:
p
(
−
1
)
>
0
{\displaystyle p(-1)>0}
.
p
(
x
)
{\displaystyle p(x)}
, jarraia denez erdiko balioaren teorema aplika daiteke.
∃
ζ
∈
(
−
1
,
0
)
:
p
(
ζ
)
=
0
{\displaystyle \exists \zeta \in (-1,0):p(\zeta )=0}
, eta
p
(
x
)
{\displaystyle p(x)}
-en erroak konjokatuak direnez:
ζ
=
α
¯
∈
(
−
1
,
0
)
{\displaystyle \zeta ={\overline {\alpha }}\in (-1,0)}
.
Ondorioz,
α
{\displaystyle \alpha }
laburtua da.
Izan bedi
p
∈
N
−
N
2
{\displaystyle p\in \mathbb {N} -\mathbb {N} ^{2}}
, zenbaki ez karratua,
α
=
[
p
]
+
p
{\displaystyle \alpha =[{\sqrt {p}}]+{\sqrt {p}}}
, eta
(
α
n
)
n
≥
0
{\displaystyle ({\alpha _{n}})_{n\geq 0}}
,
α
{\displaystyle \alpha }
-ren zatiki osatuen segida. Orduan:
1:Zatiki jarraien bidezko
p
{\displaystyle {\sqrt {p}}}
-ren garapena, periodikoa da
h
e
i
n
(
1
)
{\displaystyle hein(1)}
-etik aurrera.
2:
∃
u
n
,
v
n
∈
N
{\displaystyle \exists u_{n},v_{n}\in \mathbb {N} }
, bi segida non
∀
n
∈
N
∪
{
0
}
⇒
α
n
=
u
n
+
p
v
n
,
u
n
≤
a
0
;
p
−
u
n
2
∈
v
n
Z
{\displaystyle \forall n\in \mathbb {N} \cup \{0\}\Rightarrow \alpha _{n}={\frac {u_{n}+{\sqrt {p}}}{v_{n}}},u_{n}\leq a_{0};p-u_{n}^{2}\in v_{n}\mathbb {Z} }
.
3:
p
{\displaystyle {\sqrt {p}}}
-ren periodoa
T
=
M
i
n
{
k
∈
N
:
a
k
=
2
a
0
}
{\displaystyle T=Min\{k\in \mathbb {N} :a_{k}=2a_{0}\}}
.
4: Baldin eta,
∃
N
∈
[
1
,
T
−
1
]
:
u
N
∈
v
N
Z
⇒
T
=
2
N
{\displaystyle \exists N\in [1,T-1]:u_{N}\in v_{N}\mathbb {Z} \Rightarrow T=2N}
, eta beraz
T
{\displaystyle T}
bikoitia.
Froga
Izan bedi:
p
=
[
a
0
,
a
1
,
.
.
.
,
a
n
,
.
.
.
]
{\displaystyle {\sqrt {p}}=[a_{0},a_{1},...,a_{n},...]}
zatiki jarraien bidezko garapena, orduan
α
=
[
2
a
0
,
a
1
,
.
.
.
,
a
n
,
.
.
.
]
{\displaystyle \alpha =[2a_{0},a_{1},...,a_{n},...]}
.
1 :
α
=
a
0
+
p
{\displaystyle \alpha =a_{0}+{\sqrt {p}}}
, zenbaki irrazional bikarratu laburtua dela frogatuko da. (
a
0
=
[
p
]
{\displaystyle a_{0}=[{\sqrt {p}}]}
, parte osoa).
α
{\displaystyle \alpha }
,
p
(
x
)
=
x
2
−
2
a
0
x
+
a
0
2
−
p
{\displaystyle p(x)=x^{2}-2a_{0}x+a_{0}^{2}-p}
, polinomioaren erroa da.
α
=
[
p
]
+
p
>
1
{\displaystyle \alpha =[{\sqrt {p}}]+{\sqrt {p}}>1}
.
Beste erroa:
α
¯
=
−
(
p
−
a
0
)
∈
(
−
1
,
0
)
{\displaystyle {\overline {\alpha }}=-({\sqrt {p}}-a_{0})\in (-1,0)}
, ondorioz
α
{\displaystyle \alpha }
zenbaki irrazional bikarratu laburtua da.
Galoisek dio,
α
{\displaystyle \alpha }
-ren zatiki jarraien bidezko garapena periodiko hutsa dela:
α
=
[
2
a
0
,
a
1
,
.
.
.
,
a
T
−
1
¯
]
⇒
p
=
[
a
0
,
a
1
,
.
.
.
,
a
T
−
1
,
a
2
a
0
¯
]
{\displaystyle \alpha =[{\overline {2a_{0},a_{1},...,a_{T-1}}}]\Rightarrow {\sqrt {p}}=[a_{0},{\overline {a_{1},...,a_{T-1},a_{2a_{0}}}}]}
.
p
{\displaystyle {\sqrt {p}}}
-ren zatiki jarraien bidezko garapena periodikoa,
h
e
i
n
(
1
)
{\displaystyle hein(1)}
-etik aurrera.
2 :
u
n
,
v
n
{\displaystyle u_{n},v_{n}}
segidak errekurrentziz eraikiko dira, ondoren adierazten den bezala:
n
=
0
{\displaystyle n=0}
bada,
α
0
=
α
=
a
0
+
p
⇒
u
0
=
a
0
,
v
0
=
1
;
p
−
a
0
2
∈
1
⋅
Z
{\displaystyle \alpha _{0}=\alpha =a_{0}+{\sqrt {p}}\Rightarrow u_{0}=a_{0},v_{0}=1;p-a_{0}^{2}\in 1\cdot \mathbb {Z} }
.
Suposa bedi
u
n
;
v
n
{\displaystyle u_{n};v_{n}}
eraiki direla, ondorengo ezaugarriekin:
α
n
=
u
n
+
p
v
n
{\displaystyle \alpha _{n}={\frac {u_{n}+{\sqrt {p}}}{v_{n}}}}
,
p
−
u
n
2
∈
v
n
Z
{\displaystyle p-u_{n}^{2}\in v_{n}\mathbb {Z} }
,
u
n
,
v
n
∈
N
{\displaystyle u_{n},v_{n}\in \mathbb {N} }
, eta
u
n
≤
a
0
{\displaystyle u_{n}\leq a_{0}}
.
2.1 :
u
n
+
1
;
v
n
+
1
{\displaystyle u_{n+1};v_{n+1}}
, eraikiko dira. .
α
n
=
u
n
+
p
v
n
=
b
n
+
u
n
−
b
n
v
n
+
p
v
n
{\displaystyle \alpha _{n}={\frac {u_{n}+{\sqrt {p}}}{v_{n}}}=b_{n}+{\frac {u_{n}-b_{n}v_{n}+{\sqrt {p}}}{v_{n}}}}
, zeinetan
b
n
=
[
α
n
]
{\displaystyle b_{n}=[\alpha _{n}]}
.
u
n
+
1
=
b
n
v
n
−
u
n
{\displaystyle u_{n+1}=b_{n}v_{n}-u_{n}}
, izango da.
α
n
=
b
n
+
u
n
−
b
n
v
n
+
p
v
n
=
b
n
+
−
u
n
+
1
+
p
v
n
=
b
n
+
p
−
u
n
+
1
2
v
n
(
p
+
u
n
+
1
)
{\displaystyle \alpha _{n}=b_{n}+{\frac {u_{n}-b_{n}v_{n}+{\sqrt {p}}}{v_{n}}}=b_{n}+{\frac {-u_{n+1}+{\sqrt {p}}}{v_{n}}}=b_{n}+{\frac {p-u_{n+1}^{2}}{v_{n}({\sqrt {p}}+u_{n+1})}}}
.
α
n
=
b
n
+
1
u
n
+
1
+
p
p
−
u
n
+
1
2
v
n
⇒
v
n
+
1
=
p
−
u
n
+
1
2
v
n
{\displaystyle \alpha _{n}=b_{n}+\ {\frac {1}{\frac {u_{n+1}+{\sqrt {p}}}{\frac {p-u_{n+1}^{2}}{v_{n}}}}}\Rightarrow v_{n+1}={\frac {p-u_{n+1}^{2}}{v_{n}}}}
v
n
+
1
=
p
−
u
n
+
1
2
v
n
{\displaystyle v_{n+1}={\frac {p-u_{n+1}^{2}}{v_{n}}}}
, izango da.
2.2:
α
n
+
1
=
u
n
+
1
+
p
v
n
+
1
{\displaystyle \alpha _{n+1}={\frac {u_{n+1}+{\sqrt {p}}}{v_{n+1}}}}
,
u
n
+
1
,
v
n
+
1
∈
N
{\displaystyle u_{n+1},v_{n+1}\in \mathbb {N} }
eta
p
−
u
n
+
1
2
∈
v
n
+
1
Z
{\displaystyle p-u_{n+1}^{2}\in v_{n+1}\mathbb {Z} }
. betetzen direla frogatuko da.
α
n
+
1
=
1
α
n
−
b
n
=
1
u
n
−
p
v
n
−
b
n
=
v
n
−
u
n
+
1
+
p
=
u
n
+
1
+
p
p
−
u
n
+
1
2
v
n
=
u
n
+
1
+
p
v
n
+
1
{\displaystyle \alpha _{n+1}={\frac {1}{\alpha _{n}-b_{n}}}={\frac {1}{{\frac {u_{n}-{\sqrt {p}}}{v_{n}}}-b_{n}}}={\frac {v_{n}}{-u_{n+1}+{\sqrt {p}}}}={\frac {u_{n+1}+{\sqrt {p}}}{\frac {p-u_{n+1}^{2}}{v_{n}}}}={\frac {u_{n+1}+{\sqrt {p}}}{v_{n+1}}}}
.
u
n
+
1
=
b
n
v
n
−
u
n
∈
Z
{\displaystyle u_{n+1}=b_{n}v_{n}-u_{n}\in \mathbb {Z} }
.
v
n
+
1
=
p
−
u
n
+
1
2
v
n
=
p
−
(
b
n
v
n
−
u
n
)
2
v
n
=
p
−
u
n
2
v
n
+
2
u
n
b
n
−
b
n
2
v
n
∈
Z
{\displaystyle v_{n+1}={\frac {p-u_{n+1}^{2}}{v_{n}}}={\frac {p-(b_{n}v_{n}-u_{n})^{2}}{v_{n}}}={\frac {p-u_{n}^{2}}{v_{n}}}+2u_{n}b_{n}-b_{n}^{2}v_{n}\in \mathbb {Z} }
.
v
n
+
1
=
p
−
u
n
+
1
2
v
n
∈
Z
⇒
p
−
u
n
+
1
2
∈
v
n
+
1
Z
{\displaystyle v_{n+1}={\frac {p-u_{n+1}^{2}}{v_{n}}}\in \mathbb {Z} \Rightarrow p-u_{n+1}^{2}\in v_{n+1}\mathbb {Z} }
.
2.3 :
u
n
+
1
,
v
n
+
1
∈
N
{\displaystyle u_{n+1},v_{n+1}\in \mathbb {N} }
, frogatuko da.
α
{\displaystyle \alpha }
, zenbaki irrazional bikarratu laburtua izanagatik,
α
n
{\displaystyle \alpha _{n}}
zatiki osatu guztiak zenbaki irrazional bikarratu laburtuak dira.
{
α
n
>
1
⇒
2
[
α
n
]
>
α
n
⇒
b
n
=
[
α
n
]
>
α
n
2
α
n
=
u
n
+
p
v
n
>
u
n
+
a
0
v
n
≥
2
u
n
v
n
⇒
b
n
>
α
n
2
>
u
n
v
n
⇒
b
n
v
n
−
u
n
>
0
{\displaystyle {\begin{cases}\alpha _{n}>1\Rightarrow 2[\alpha _{n}]>\alpha _{n}\Rightarrow b_{n}=[\alpha _{n}]>{\frac {\alpha _{n}}{2}}\\\alpha _{n}={\frac {u_{n}+{\sqrt {p}}}{v_{n}}}>{\frac {u_{n}+a_{0}}{v_{n}}}\geq {\frac {2u_{n}}{v_{n}}}\end{cases}}\Rightarrow b_{n}>{\frac {\alpha _{n}}{2}}>{\frac {u_{n}}{v_{n}}}\Rightarrow b_{n}v_{n}-u_{n}>0}
.
u
n
+
1
=
b
n
v
n
−
u
n
∈
Z
{\displaystyle u_{n+1}=b_{n}v_{n}-u_{n}\in \mathbb {Z} }
eta
b
n
v
n
−
u
n
>
0
{\displaystyle b_{n}v_{n}-u_{n}>0}
, orduan:
u
n
+
1
∈
N
{\displaystyle u_{n+1}\in \mathbb {N} }
.
b
n
=
[
α
n
]
<
α
n
=
u
n
+
p
v
n
⇒
u
n
+
1
=
v
n
b
n
−
u
n
<
p
⇒
p
−
u
n
+
1
>
0
{\displaystyle b_{n}=[\alpha _{n}]<\alpha _{n}={\frac {u_{n}+{\sqrt {p}}}{v_{n}}}\Rightarrow u_{n+1}=v_{n}b_{n}-u_{n}<{\sqrt {p}}\Rightarrow {\sqrt {p}}-u_{n+1}>0}
p
−
u
n
+
1
2
>
0
;
v
n
∈
N
⇒
v
n
+
1
=
p
−
u
n
+
1
2
v
n
>
0
{\displaystyle p-u_{n+1}^{2}>0;v_{n}\in \mathbb {N} \Rightarrow v_{n+1}={\frac {p-u_{n+1}^{2}}{v_{n}}}>0}
, eta
v
n
+
1
∈
Z
{\displaystyle v_{n+1}\in \mathbb {Z} }
, orduan:
v
n
+
1
∈
N
{\displaystyle v_{n+1}\in \mathbb {N} }
.
2.4 :
u
n
+
1
≤
a
0
{\displaystyle u_{n+1}\leq a_{0}}
, frogatuko da.
u
n
≤
a
0
<
p
;
u
n
+
1
=
b
n
v
n
−
u
n
<
α
n
v
n
−
u
n
=
p
{\displaystyle u_{n}\leq a_{0}<{\sqrt {p}};u_{n+1}=b_{n}v_{n}-u_{n}<\alpha _{n}v_{n}-u_{n}={\sqrt {p}}}
,
u
n
+
1
<
p
;
u
n
+
1
∈
N
⇒
u
n
+
1
≤
[
p
]
=
a
0
{\displaystyle u_{n+1}<{\sqrt {p}};u_{n+1}\in \mathbb {N} \Rightarrow u_{n+1}\leq [{\sqrt {p}}]=a_{0}}
.
Indukzio printzipioa aplikatuz, bigarren atala ondorioztatzen da.
3 :
p
=
[
a
0
,
a
1
,
.
.
.
,
a
T
−
1
,
a
2
a
0
¯
]
{\displaystyle {\sqrt {p}}=[a_{0},{\overline {a_{1},...,a_{T-1},a_{2a_{0}}}}]}
, ondorioztatu da 1, atalean.
3.1 .
∀
j
∈
[
1
,
T
−
1
]
,
α
j
≠
α
{\displaystyle \forall j\in [1,T-1],\alpha _{j}\neq \alpha }
. Frogatuko da.
Absurdura bideratuz, baldin eta:
∃
j
∈
[
1
,
T
−
1
]
:
α
j
=
α
{\displaystyle \exists j\in [1,T-1]:\alpha _{j}=\alpha }
α
=
[
2
a
0
,
a
1
,
.
.
.
,
a
j
−
1
,
α
j
]
=
[
2
a
0
,
a
1
,
.
.
.
,
a
j
−
1
,
α
]
=
[
2
a
0
,
a
1
,
.
.
.
,
a
j
−
1
¯
]
{\displaystyle \alpha =[2a_{0},a_{1},...,a_{j-1},\alpha _{j}]=[2a_{0},a_{1},...,a_{j-1},\alpha ]=[{\overline {2a_{0},a_{1},...,a_{j-1}}}]}
Ondorioz:
p
=
[
a
0
,
a
1
,
.
.
.
,
a
j
−
1
,
a
2
a
0
¯
]
⇒
T
=
j
<
T
{\displaystyle {\sqrt {p}}=[a_{0},{\overline {a_{1},...,a_{j-1},a_{2a_{0}}}}]\Rightarrow T=j<T}
, ezinezkoa.
3.2
∀
j
∈
[
1
,
T
−
1
]
,
a
j
≠
2
a
0
{\displaystyle \forall j\in [1,T-1],a_{j}\neq 2a_{0}}
, frogatuko da.
Absurdura bideratuz:
∃
j
∈
[
1
,
T
−
1
]
:
a
j
=
2
a
0
{\displaystyle \exists j\in [1,T-1]:a_{j}=2a_{0}}
.
j
≥
1
⇒
b
j
=
a
j
=
2
a
0
{\displaystyle j\geq 1\Rightarrow b_{j}=a_{j}=2a_{0}}
, eta
u
j
+
1
=
2
a
0
v
j
−
u
j
⇒
2
a
0
v
j
=
u
j
+
u
j
+
1
≤
2
a
0
⇒
v
j
=
1
{\displaystyle u_{j+1}=2a_{0}v_{j}-u_{j}\Rightarrow 2a_{0}v_{j}=u_{j}+u_{j+1}\leq 2a_{0}\Rightarrow v_{j}=1}
.
2. atalagatik:
u
n
≤
a
0
{\displaystyle u_{n}\leq a_{0}}
, ondorioz:
2
a
0
=
u
j
+
u
j
+
1
≤
u
j
+
a
0
⇒
a
0
≤
u
j
≤
a
0
⇒
u
j
=
a
0
{\displaystyle 2a_{0}=u_{j}+u_{j+1}\leq u_{j}+a_{0}\Rightarrow a_{0}\leq u_{j}\leq a_{0}\Rightarrow u_{j}=a_{0}}
, honela:
α
j
=
u
j
+
p
v
j
=
u
j
+
p
=
a
0
+
p
=
α
{\displaystyle \alpha _{j}={\frac {u_{j}+{\sqrt {p}}}{v_{j}}}=u_{j}+{\sqrt {p}}=a_{0}+{\sqrt {p}}=\alpha }
, zeina ezinezkoa den 3.1. atalagatik.
4 : Suposa bedi
∃
N
∈
[
1
,
T
−
1
]
:
u
N
∈
v
N
Z
{\displaystyle \exists N\in [1,T-1]:u_{N}\in v_{N}\mathbb {Z} }
.
⇒
T
=
2
N
{\displaystyle \Rightarrow T=2N}
ondorioztatuko da
4.1 .
∀
j
∈
[
0
,
N
−
1
]
⇒
u
N
+
j
+
1
=
u
N
−
j
,
v
N
+
j
+
1
=
v
N
−
j
−
1
;
b
N
+
j
+
1
=
b
N
−
j
−
1
{\displaystyle \forall j\in [0,N-1]\Rightarrow u_{N+j+1}=u_{N-j},v_{N+j+1}=v_{N-j-1};b_{N+j+1}=b_{N-j-1}}
, frogatuko da.
4.1.1.
j
=
0
⇒
u
N
+
1
=
u
N
,
v
N
+
1
=
v
N
−
1
;
b
N
+
1
=
b
N
−
1
{\displaystyle j=0\Rightarrow u_{N+1}=u_{N},v_{N+1}=v_{N-1};b_{N+1}=b_{N-1}}
, frogatuko da lehenik.
{
u
N
∈
v
N
Z
⇒
2
u
N
v
N
∈
N
u
N
≤
a
0
<
p
⇒
2
u
N
v
N
<
u
N
+
p
v
N
=
α
N
{\displaystyle {\begin{cases}u_{N}\in v_{N}\mathbb {Z} \Rightarrow {\frac {2u_{N}}{v_{N}}}\in \mathbb {N} \\u_{N}\leq a_{0}<{\sqrt {p}}\end{cases}}\Rightarrow {\frac {2u_{N}}{v_{N}}}<{\frac {u_{N}+{\sqrt {p}}}{v_{N}}}=\alpha _{N}}
.
Eta
α
{\displaystyle \alpha }
zenbaki irrazional bikarratu laburtua izateagatik:
u
N
+
p
v
N
>
1
;
u
N
−
p
v
N
∈
(
−
1
,
0
)
{\displaystyle {\frac {u_{N}+{\sqrt {p}}}{v_{N}}}>1;{\frac {u_{N}-{\sqrt {p}}}{v_{N}}}\in (-1,0)}
.
α
N
=
u
N
+
p
v
N
=
2
u
N
v
N
−
u
N
−
p
v
N
<
2
u
N
v
N
+
1
{\displaystyle \alpha _{N}={\frac {u_{N}+{\sqrt {p}}}{v_{N}}}={\frac {2u_{N}}{v_{N}}}-{\frac {u_{N}-{\sqrt {p}}}{v_{N}}}<{\frac {2u_{N}}{v_{N}}}+1}
{
b
N
=
[
α
N
]
=
[
u
N
+
p
v
N
]
≤
2
u
N
v
N
∈
N
2
u
N
v
N
<
u
N
+
p
v
N
⇒
b
N
=
[
u
N
+
p
v
N
]
=
2
u
N
v
N
{\displaystyle {\begin{cases}b_{N}=[\alpha _{N}]=[{\frac {u_{N}+{\sqrt {p}}}{v_{N}}}]\leq {\frac {2u_{N}}{v_{N}}}\in \mathbb {N} \\{\frac {2u_{N}}{v_{N}}}<{\frac {u_{N}+{\sqrt {p}}}{v_{N}}}\end{cases}}\Rightarrow b_{N}=[{\frac {u_{N}+{\sqrt {p}}}{v_{N}}}]={\frac {2u_{N}}{v_{N}}}}
u
N
+
1
=
b
N
v
N
−
u
N
=
2
u
N
v
N
v
N
−
u
N
=
u
N
⇒
u
N
+
1
=
u
N
{\displaystyle u_{N+1}=b_{N}v_{N}-u_{N}={\frac {2u_{N}}{v_{N}}}v_{N}-u_{N}=u_{N}\Rightarrow u_{N+1}=u_{N}}
v
N
+
1
=
p
−
u
N
+
1
2
v
N
⇒
u
N
+
1
2
+
v
N
+
1
v
N
=
p
=
u
N
2
+
v
N
v
N
−
1
⇒
v
N
+
1
=
v
N
−
1
{\displaystyle v_{N+1}={\frac {p-u_{N+1}^{2}}{v_{N}}}\Rightarrow u_{N+1}^{2}+v_{N+1}v_{N}=p=u_{N}^{2}+v_{N}v_{N-1}\Rightarrow v_{N+1}=v_{N-1}}
.
b
N
+
1
=
[
α
N
+
1
]
=
[
u
N
+
1
+
p
v
N
+
1
]
=
[
u
N
+
p
v
N
−
1
]
{\displaystyle b_{N+1}=[\alpha _{N+1}]=[{\frac {u_{N+1}+{\sqrt {p}}}{v_{N+1}}}]=[{\frac {u_{N}+{\sqrt {p}}}{v_{N-1}}}]}
, eta
u
N
=
b
N
−
1
v
N
−
1
−
u
N
−
1
{\displaystyle u_{N}=b_{N-1}v_{N-1}-u_{N-1}}
.
b
N
+
1
=
[
α
N
+
1
]
=
[
u
N
−
p
v
N
−
1
]
=
[
b
N
−
1
−
u
N
−
1
−
p
v
N
−
1
]
=
[
b
N
−
1
−
α
N
−
1
¯
]
=
b
N
−
1
{\displaystyle b_{N+1}=[\alpha _{N+1}]=[{\frac {u_{N}-{\sqrt {p}}}{v_{N-1}}}]=[b_{N-1}-{\frac {u_{N-1}-{\sqrt {p}}}{v_{N-1}}}]=[b_{N-1}-{\overline {\alpha _{N-1}}}]=b_{N-1}}
.
4.1.2. Suposa bedi orain:
∃
j
∈
[
0
,
T
−
2
]
:
u
N
+
j
+
1
=
u
N
−
j
,
v
N
+
j
+
1
=
v
N
−
j
−
1
;
b
N
+
j
+
1
=
b
N
−
j
−
1
{\displaystyle \exists j\in [0,T-2]:u_{N+j+1}=u_{N-j},v_{N+j+1}=v_{N-j-1};b_{N+j+1}=b_{N-j-1}}
.
Orduan
(
j
+
1
)
{\displaystyle (j+1)}
entzat 4.1. erlazioa betetzen dela frogatuko da.
u
N
+
j
+
2
=
b
N
+
j
+
1
v
N
+
j
+
1
−
u
N
+
j
+
1
=
b
N
−
j
−
1
v
N
−
j
−
1
−
u
N
−
j
{\displaystyle u_{N+j+2}=b_{N+j+1}v_{N+j+1}-u_{N+j+1}=b_{N-j-1}v_{N-j-1}-u_{N-j}}
=
b
N
−
j
−
1
v
N
−
j
−
1
−
(
b
N
−
j
−
1
v
N
−
j
−
1
−
u
N
−
j
−
1
)
=
u
N
−
j
−
1
⇒
u
N
+
j
+
2
=
u
N
−
j
−
1
{\displaystyle =b_{N-j-1}v_{N-j-1}-(b_{N-j-1}v_{N-j-1}-u_{N-j-1})=u_{N-j-1}\Rightarrow u_{N+j+2}=u_{N-j-1}}
.
v
N
+
j
+
2
=
p
−
u
N
+
j
+
2
2
v
N
+
j
+
1
⇒
u
N
+
j
+
2
2
+
v
N
+
j
+
2
v
N
+
j
+
1
=
p
{\displaystyle v_{N+j+2}={\frac {p-u_{N+j+2}^{2}}{v_{N+j+1}}}\Rightarrow u_{N+j+2}^{2}+v_{N+j+2}v_{N+j+1}=p}
,
u
N
+
j
+
2
2
+
v
N
+
j
+
2
v
N
+
j
+
1
=
u
N
−
j
−
1
2
+
v
N
−
j
−
2
v
N
−
j
−
1
=
u
N
+
j
+
2
2
+
v
N
−
j
−
2
v
N
+
j
+
1
{\displaystyle u_{N+j+2}^{2}+v_{N+j+2}v_{N+j+1}=u_{N-j-1}^{2}+v_{N-j-2}v_{N-j-1}=u_{N+j+2}^{2}+v_{N-j-2}v_{N+j+1}}
,
⇒
v
N
+
j
+
2
=
v
N
−
j
−
2
{\displaystyle \Rightarrow v_{N+j+2}=v_{N-j-2}}
.
b
N
+
j
+
2
=
[
α
N
+
j
+
2
]
=
[
u
N
+
j
+
2
+
p
v
N
+
j
+
2
]
=
[
u
N
−
j
−
1
+
p
v
N
−
j
−
2
]
{\displaystyle b_{N+j+2}=[\alpha _{N+j+2}]=[{\frac {u_{N+j+2}+{\sqrt {p}}}{v_{N+j+2}}}]=[{\frac {u_{N-j-1}+{\sqrt {p}}}{v_{N-j-2}}}]}
=
[
b
N
−
j
−
2
v
N
−
j
−
2
−
u
N
−
j
−
2
+
+
p
v
N
−
j
−
2
]
=
[
b
N
−
j
−
2
−
α
N
−
j
−
2
¯
]
=
b
N
−
j
−
2
⇒
b
N
+
j
+
2
=
b
N
−
j
−
2
{\displaystyle =[{\frac {b_{N-j-2}v_{N-j-2}-u_{N-j-2}++{\sqrt {p}}}{v_{N-j-2}}}]=[b_{N-j-2}-{\overline {\alpha _{N-j-2}}}]=b_{N-j-2}\Rightarrow b_{N+j+2}=b_{N-j-2}}
4.2 .
T
=
2
N
{\displaystyle T=2N}
ondorioztatuko da.
∀
j
∈
[
0
,
N
−
1
]
⇒
u
N
+
j
+
1
=
u
N
−
j
,
v
N
+
j
+
1
=
v
N
−
j
−
1
;
b
N
+
j
+
1
=
b
N
−
j
−
1
{\displaystyle \forall j\in [0,N-1]\Rightarrow u_{N+j+1}=u_{N-j},v_{N+j+1}=v_{N-j-1};b_{N+j+1}=b_{N-j-1}}
. betetzen da 4.1 atalagatik.
j
=
N
−
1
⇒
b
2
N
=
b
0
{\displaystyle j=N-1\Rightarrow b_{2N}=b_{0}}
, betetzen da eta 3. atalagatik:
2
N
=
ζ
T
:
ζ
∈
N
{\displaystyle 2N=\zeta T:\zeta \in \mathbb {N} }
.
2
≤
2
N
≤
2
(
T
−
1
)
N
<
2
T
;
2
N
=
ζ
T
≥
T
⇒
N
<
T
≤
2
N
⇒
ζ
=
1
;
T
=
2
N
{\displaystyle 2\leq 2N\leq 2(T-1)N<2T;2N=\zeta T\geq T\Rightarrow N<T\leq 2N\Rightarrow \zeta =1;T=2N}
bikoitia.
Galoisek ondorioztatuko du:
p
{\displaystyle {\sqrt {p}}}
moduko zenbakien zatiki jarraien bidezko garapena:
T
{\displaystyle T}
-periodikoa dela:
h
e
i
n
(
1
)
{\displaystyle hein(1)}
-etik aurrera. Atal honetan frogatuko da,
p
{\displaystyle p}
parametrodun
x
2
−
p
y
2
=
1
{\displaystyle x^{2}-py^{2}=1}
edo
x
2
−
p
y
2
=
−
1
{\displaystyle x^{2}-py^{2}=-1}
ekuazioaren,
p
{\displaystyle {\sqrt {p}}}
-ren zatiki jarraien bidezko hondar bat dela eta ondar hori periodoa osatu aurrekoa dela:
R
T
−
1
,
R
2
T
−
1
,
.
.
.
{\displaystyle R_{T-1},R_{2T-1},...}
Hondarraren zenbakitzailea eta izendatzailea gorakorrak direnez:
R
T
−
1
{\displaystyle R_{T-1}}
, izango da
x
2
−
p
y
2
=
1
{\displaystyle x^{2}-py^{2}=1}
-en ebazpen minimoa,
T
{\displaystyle T}
bikoitia denean, eta
x
2
−
p
y
2
=
−
1
{\displaystyle x^{2}-py^{2}=-1}
, en ebazpena
T
{\displaystyle T}
bakoitia denean. Honela
T
{\displaystyle T}
bakoitia denean,
x
2
−
p
y
2
=
1
{\displaystyle x^{2}-py^{2}=1}
-en ebazpen minimoa
R
2
T
−
1
{\displaystyle R_{2T-1}}
izango da.
Izan bedi:
p
∈
N
−
N
2
{\displaystyle p\in \mathbb {N} -\mathbb {N} ^{2}}
, eta
m
∈
Z
−
{
0
}
:
1
≤
|
m
|
<
p
{\displaystyle m\in \mathbb {Z} -\{0\}:1\leq |m|<{\sqrt {p}}}
.
Suposa bedi
∃
(
x
,
y
)
∈
N
×
N
:
z
k
h
(
x
,
y
)
=
1
;
x
2
−
p
y
2
=
m
{\displaystyle \exists (x,y)\in \mathbb {N} \times \mathbb {N} :zkh(x,y)=1;x^{2}-py^{2}=m}
.
Orduan
x
y
{\displaystyle {\frac {x}{y}}}
,
p
{\displaystyle {\sqrt {p}}}
-ren zatiki jarraien bidezko hondar bat da.
Froga:
x
2
−
p
y
2
=
m
⇒
x
−
y
p
=
m
x
+
y
p
{\displaystyle x^{2}-py^{2}=m\Rightarrow x-y{\sqrt {p}}={\frac {m}{x+y{\sqrt {p}}}}}
.
i)
m
>
0
{\displaystyle m>0}
bada,
x
2
=
m
+
p
y
2
>
p
y
2
⇒
x
>
y
p
{\displaystyle x^{2}=m+py^{2}>py^{2}\Rightarrow x>y{\sqrt {p}}}
|
p
−
x
y
|
=
|
x
−
y
p
y
|
=
|
x
2
−
p
y
2
y
(
x
+
y
p
)
|
=
m
x
y
+
y
2
p
<
p
2
y
2
p
=
1
2
y
2
{\displaystyle |{\sqrt {p}}-{\frac {x}{y}}|=|{\frac {x-y{\sqrt {p}}}{y}}|=|{\frac {x^{2}-py^{2}}{y(x+y{\sqrt {p}})}}|={\frac {m}{xy+y^{2}{\sqrt {p}}}}<{\frac {\sqrt {p}}{2y^{2}{\sqrt {p}}}}={\frac {1}{2y^{2}}}}
, eta zatiki jarraien ataleko 3. proposizioa dela medio,
x
y
{\displaystyle {\frac {x}{y}}}
,
p
{\displaystyle {\sqrt {p}}}
-ren zatiki jarraien bidezko
n
{\displaystyle n}
. hondar bat da.
ii)
m
<
0
{\displaystyle m<0}
bada,
p
y
2
=
x
2
−
m
>
x
2
⇒
p
y
>
x
{\displaystyle py^{2}=x^{2}-m>x^{2}\Rightarrow {\sqrt {p}}y>x}
.
Lehen desberdintzan:
|
m
|
<
p
{\displaystyle |m|<{\sqrt {p}}}
, erabiliz eta bigarren desberdintzan:
p
y
>
x
{\displaystyle {\sqrt {p}}y>x}
.
|
1
p
−
y
x
|
=
|
x
−
y
p
x
p
|
=
|
x
2
−
p
y
2
x
p
(
x
+
y
p
)
|
=
|
m
|
x
2
p
+
x
y
p
<
p
p
(
x
2
+
x
y
p
)
<
1
2
x
2
{\displaystyle |{\frac {1}{\sqrt {p}}}-{\frac {y}{x}}|=|{\frac {x-y{\sqrt {p}}}{x{\sqrt {p}}}}|=|{\frac {x^{2}-py^{2}}{x{\sqrt {p}}(x+y{\sqrt {p}})}}|={\frac {|m|}{x^{2}{\sqrt {p}}+xyp}}<{\frac {\sqrt {p}}{{\sqrt {p}}(x^{2}+xy{\sqrt {p}})}}<{\frac {1}{2x^{2}}}}
Berriro ere zatiki jarraien ataleko 3. proposizioa dela medio:
y
x
{\displaystyle {\frac {y}{x}}}
,
1
p
{\displaystyle {\frac {1}{\sqrt {p}}}}
-en zatiki jarraien bidezko
n
{\displaystyle n}
. hondar bat da:
p
=
[
a
0
,
a
1
,
.
.
.
,
a
n
,
.
.
.
]
⇒
1
p
=
[
0
,
p
]
=
[
0
,
a
0
,
a
1
,
.
.
.
,
a
n
,
.
.
.
]
{\displaystyle {\sqrt {p}}=[a_{0},a_{1},...,a_{n},...]\Rightarrow {\frac {1}{\sqrt {p}}}=[0,{\sqrt {p}}]=[0,a_{0},a_{1},...,a_{n},...]}
, eta
y
x
=
[
0
,
a
0
,
a
1
,
.
.
.
,
a
n
−
1
]
{\displaystyle {\frac {y}{x}}=[0,a_{0},a_{1},...,a_{n-1}]}
, zeinetan
y
≠
0
⇒
n
≠
0
⇒
n
∈
N
{\displaystyle y\neq 0\Rightarrow n\neq 0\Rightarrow n\in \mathbb {N} }
.
y
x
=
0
+
1
[
a
0
,
a
1
,
.
.
.
,
a
n
−
1
]
⇒
x
y
=
[
a
0
,
a
1
,
.
.
.
,
a
n
−
1
]
{\displaystyle {\frac {y}{x}}=0+{\frac {1}{[a_{0},a_{1},...,a_{n-1}]}}\Rightarrow {\frac {x}{y}}=[a_{0},a_{1},...,a_{n-1}]}
,
p
{\displaystyle {\sqrt {p}}}
-ren
(
n
−
1
)
{\displaystyle (n-1)}
. hondarra.
p
∈
N
−
N
2
{\displaystyle p\in \mathbb {N} -\mathbb {N} ^{2}}
, eta
m
∈
Z
:
1
≤
|
m
|
<
p
{\displaystyle m\in \mathbb {Z} :1\leq |m|<{\sqrt {p}}}
. Emanik ondorengoak baiezta daitezke.
δ
n
=
u
n
+
p
v
n
{\displaystyle \delta _{n}={\frac {u_{n}+{\sqrt {p}}}{v_{n}}}}
,
p
{\displaystyle {\sqrt {p}}}
-ren
n
{\displaystyle n}
. zatiki osatua, zeinetan:
u
n
,
v
n
{\displaystyle u_{n},v_{n}}
proposizioa3-n bezala definituak.
p
{\displaystyle {\sqrt {p}}}
-ren
n
{\displaystyle n}
. hondarra:
p
n
q
n
{\displaystyle {\frac {p_{n}}{q_{n}}}}
,
z
k
h
(
p
n
,
q
n
)
=
1
{\displaystyle zkh(p_{n},q_{n})=1}
eta
p
{\displaystyle {\sqrt {p}}}
:
T
{\displaystyle T}
-periodikoa.
∀
l
∈
Z
{\displaystyle \forall l\in \mathbb {Z} }
-rentzat,
ξ
l
=
{
j
∈
[
1
,
T
]
:
(
−
1
)
j
v
j
=
l
}
{\displaystyle \xi _{l}=\{j\in [1,T]:(-1)^{j}v_{j}=l\}}
adieraziko da. Orduan:
1:
T
{\displaystyle T}
bikoitia bada:
x
2
−
p
y
2
=
m
{\displaystyle x^{2}-py^{2}=m}
ekuazioak ebazpen ez neutroa du
⇔
ξ
m
≠
∅
{\displaystyle \Leftrightarrow \xi _{m}\neq \emptyset }
.
Eta kasu horretan
z
k
h
(
x
,
y
)
=
1
{\displaystyle zkh(x,y)=1}
, betetzen duten ebazpen guztiak, ondorengo multzoan daude:
⋃
N
∈
ξ
m
{
(
p
N
+
k
T
−
1
,
q
N
+
k
T
−
1
)
:
k
∈
N
∪
{
0
}
}
{\displaystyle \bigcup _{N\in \xi _{m}}\{(p_{N+kT-1},q_{N+kT-1}):k\in \mathbb {N} \cup \{0\}\}}
.
2:
T
{\displaystyle T}
bakoitia bada,
x
2
−
p
y
2
=
m
{\displaystyle x^{2}-py^{2}=m}
ekuazioak ebazpena (ez neutroa) du
⇔
ξ
m
∪
ξ
−
m
≠
∅
{\displaystyle \Leftrightarrow \xi _{m}\cup \xi _{-m}\neq \emptyset }
.
Eta kasu horretan
z
k
h
(
x
,
y
)
=
1
{\displaystyle zkh(x,y)=1}
, betetzen duten ebazpen guztiak, ondorengo multzoan daude:
⋃
N
∈
ξ
m
{
(
p
N
+
2
k
T
−
1
,
q
N
+
2
k
T
−
1
)
:
k
∈
N
∪
{
0
}
}
{\displaystyle \bigcup _{N\in \xi _{m}}\{(p_{N+2kT-1},q_{N+2kT-1}):k\in \mathbb {N} \cup \{0\}\}}
, edo
⋃
M
∈
ξ
−
m
{
(
p
M
+
(
2
k
+
1
)
T
−
1
,
q
M
+
(
2
k
+
1
)
T
−
1
)
:
k
∈
N
∪
{
0
}
}
{\displaystyle \bigcup _{M\in \xi _{-m}}\{(p_{M+(2k+1)T-1},q_{M+(2k+1)T-1}):k\in \mathbb {N} \cup \{0\}\}}
3: Edozein kasutan
x
2
−
p
y
2
=
m
{\displaystyle x^{2}-py^{2}=m}
, ekuazioak infinitu soluzio ditu.
FROGA
δ
n
=
u
n
+
p
v
n
{\displaystyle \delta _{n}={\frac {u_{n}+{\sqrt {p}}}{v_{n}}}}
,
p
{\displaystyle {\sqrt {p}}}
-ren
n
{\displaystyle n}
. zatiki osatua denez:
p
=
[
a
0
,
a
1
,
.
.
.
,
a
n
−
1
,
δ
n
]
{\displaystyle {\sqrt {p}}=[a_{0},a_{1},...,a_{n-1},\delta _{n}]}
.
p
{\displaystyle {\sqrt {p}}}
-ren garapena:
T
{\displaystyle T}
-periodikoa denez,
h
e
i
n
(
1
)
{\displaystyle hein(1)}
-etik aurrera,
δ
n
{\displaystyle \delta _{n}}
ere
T
{\displaystyle T}
periodikoa da, eta modu bakarrean determinatzen dituenez:
u
n
,
v
n
{\displaystyle u_{n},v_{n}}
balioak, hauek ere gutsienez;
T
{\displaystyle T}
periodikoak dira.
Ondorioz
v
n
∈
{
v
1
,
.
.
.
,
v
T
}
{\displaystyle v_{n}\in \{v_{1},...,v_{T}\}}
.
p
{\displaystyle {\sqrt {p}}}
-ren
n
{\displaystyle n}
. hondarra:
R
n
=
p
n
q
n
=
[
a
0
,
a
1
,
.
.
.
,
a
n
]
{\displaystyle R_{n}={\frac {p_{n}}{q_{n}}}=[a_{0},a_{1},...,a_{n}]}
, adieraziko da.
∀
n
∈
N
,
p
n
−
1
2
−
p
q
n
−
1
2
=
(
−
1
)
n
v
n
{\displaystyle \forall n\in \mathbb {N} ,p_{n-1}^{2}-pq_{n-1}^{2}=(-1)^{n}v_{n}}
betetzen dela frogatuko da.
p
=
[
a
0
,
a
1
,
.
.
.
,
a
n
−
1
,
δ
n
]
=
p
n
−
1
δ
n
+
p
n
−
2
q
n
−
1
δ
n
+
q
n
−
2
=
p
n
−
1
(
u
n
+
p
)
+
p
n
−
2
v
n
q
n
−
1
(
u
n
+
p
)
+
q
n
−
2
v
n
{\displaystyle {\sqrt {p}}=[a_{0},a_{1},...,a_{n-1},\delta _{n}]={\frac {p_{n-1}\delta _{n}+p_{n-2}}{q_{n-1}\delta _{n}+q_{n-2}}}={\frac {p_{n-1}(u_{n}+{\sqrt {p}})+p_{n-2}v_{n}}{q_{n-1}(u_{n}+{\sqrt {p}})+q_{n-2}v_{n}}}}
Eta ondorioz:
∀
n
≥
2
{\displaystyle \forall n\geq 2}
-rentzat.
q
n
−
1
p
+
(
q
n
−
1
u
n
+
q
n
−
2
v
n
)
p
=
p
n
−
1
u
n
+
p
n
−
2
v
n
+
p
n
−
1
p
⇒
{\displaystyle q_{n-1}p+(q_{n-1}u_{n}+q_{n-2}v_{n}){\sqrt {p}}=p_{n-1}u_{n}+p_{n-2}v_{n}+p_{n-1}{\sqrt {p}}\Rightarrow }
q
n
−
1
p
=
p
n
−
1
u
n
+
p
n
−
2
v
n
{\displaystyle q_{n-1}p=p_{n-1}u_{n}+p_{n-2}v_{n}}
.eta
p
n
−
1
=
q
n
−
1
u
n
+
q
n
−
2
v
n
{\displaystyle p_{n-1}=q_{n-1}u_{n}+q_{n-2}v_{n}}
Honela:
p
n
−
1
2
−
p
q
n
−
1
2
=
p
n
−
1
(
q
n
−
1
u
n
+
q
n
−
2
v
n
)
−
q
n
−
1
(
p
n
−
1
u
n
+
p
n
−
2
v
n
)
{\displaystyle p_{n-1}^{2}-pq_{n-1}^{2}=p_{n-1}(q_{n-1}u_{n}+q_{n-2}v_{n})-q_{n-1}(p_{n-1}u_{n}+p_{n-2}v_{n})}
Eta:
p
n
−
1
q
n
−
2
−
q
n
−
1
p
n
−
2
=
(
−
1
)
n
{\displaystyle p_{n-1}q_{n-2}-q_{n-1}p_{n-2}=(-1)^{n}}
, betetzen denez:
p
n
−
1
2
−
p
q
n
−
1
2
=
v
n
(
p
n
−
1
q
n
−
2
−
q
n
−
1
p
n
−
2
)
=
v
n
(
−
1
)
n
{\displaystyle p_{n-1}^{2}-pq_{n-1}^{2}=v_{n}(p_{n-1}q_{n-2}-q_{n-1}p_{n-2})=v_{n}(-1)^{n}}
.
n
=
1
{\displaystyle n=1}
denean
p
0
2
−
p
q
0
2
=
a
0
2
−
p
=
(
−
1
)
v
1
{\displaystyle p_{0}^{2}-pq_{0}^{2}=a_{0}^{2}-p=(-1)v_{1}}
, izan ere:
p
0
q
0
=
a
0
,
u
1
=
a
0
v
0
−
u
0
=
a
0
;
v
1
=
p
−
u
1
2
v
0
=
p
−
a
0
2
.
{\displaystyle {\frac {p_{0}}{q_{0}}}=a_{0},u_{1}=a_{0}v_{0}-u_{0}=a_{0};v_{1}={\frac {p-u_{1}^{2}}{v_{0}}}=p-a_{0}^{2}.}
Ondorioz:
∀
n
∈
N
,
p
n
−
1
2
−
p
q
n
−
1
2
=
(
−
1
)
n
v
n
{\displaystyle \forall n\in \mathbb {N} ,p_{n-1}^{2}-pq_{n-1}^{2}=(-1)^{n}v_{n}}
1:
T
{\displaystyle T}
bikoitia bada.
Suposa bedi:
(
x
,
y
)
{\displaystyle (x,y)}
,
x
2
−
p
y
2
=
m
{\displaystyle x^{2}-py^{2}=m}
ekuazioaren ebazpen ez neutroa dela,
z
k
h
(
x
,
y
)
=
1
{\displaystyle zkh(x,y)=1}
orduan Lema3 gatik:
x
y
{\displaystyle {\frac {x}{y}}}
,
p
{\displaystyle {\sqrt {p}}}
-ren hondar bat da:
j
−
1
{\displaystyle j-1}
. hondarra:
x
y
=
R
j
−
1
{\displaystyle {\frac {x}{y}}=R_{j-1}}
,
j
∈
N
{\displaystyle j\in \mathbb {N} }
.
x
=
p
j
−
1
;
y
=
q
j
−
1
⇒
m
=
x
2
−
p
y
2
=
p
j
−
1
2
−
p
q
j
−
1
2
=
(
−
1
)
j
v
j
{\displaystyle x=p_{j-1};y=q_{j-1}\Rightarrow m=x^{2}-py^{2}=p_{j-1}^{2}-pq_{j-1}^{2}=(-1)^{j}v_{j}}
.
Zatidura euklidearra eginez, eta
h
+
1
=
N
j
{\displaystyle h+1=N_{j}}
adieraziz:
j
−
1
=
T
k
+
h
:
h
∈
[
0
,
T
−
1
]
⇔
j
=
T
k
+
h
+
1
:
h
+
1
=
N
j
∈
[
1
,
T
]
{\displaystyle j-1=Tk+h:h\in [0,T-1]\Leftrightarrow j=Tk+h+1:h+1=N_{j}\in [1,T]}
⇔
j
=
T
k
+
N
j
:
N
j
∈
[
1
,
T
]
{\displaystyle \Leftrightarrow j=Tk+N_{j}:N_{j}\in [1,T]}
(
−
1
)
j
v
j
=
(
−
1
)
T
k
+
N
j
v
T
k
+
N
j
=
(
−
1
)
N
j
v
N
j
=
m
{\displaystyle (-1)^{j}v_{j}=(-1)^{Tk+N_{j}}v_{Tk+N_{j}}=(-1)^{N_{j}}v_{N_{j}}=m}
, (
T
{\displaystyle T}
bikoitia eta
v
n
{\displaystyle v_{n}}
-
T
{\displaystyle T}
periodikoa)
⇒
N
j
∈
ξ
m
≠
∅
{\displaystyle \Rightarrow N_{j}\in \xi _{m}\neq \emptyset }
.
Eta:
(
x
,
y
)
=
(
p
N
j
+
k
T
−
1
,
q
N
j
+
k
T
−
1
)
∈
⋃
N
∈
ξ
m
{
(
p
N
+
k
T
−
1
,
q
N
+
k
T
−
1
)
:
k
∈
N
}
{\displaystyle (x,y)=(p_{N_{j}+kT-1},q_{N_{j}+kT-1})\in \bigcup _{N\in \xi _{m}}\{(p_{N+kT-1},q_{N+kT-1}):k\in \mathbb {N} \}}
.
Alderantzizkoa:
⇐
ξ
m
≠
∅
{\displaystyle \Leftarrow \xi _{m}\neq \emptyset }
bada,
∃
N
∈
[
1
,
T
]
:
(
−
1
)
N
v
N
=
m
{\displaystyle \exists N\in [1,T]:(-1)^{N}v_{N}=m}
.
Orduan:
k
∈
N
∪
{
0
}
{\displaystyle k\in \mathbb {N} \cup \{0\}}
, hartuaz:
p
N
+
k
T
−
1
2
−
p
q
N
+
k
T
−
1
2
=
(
−
1
)
N
+
k
T
v
N
+
k
T
=
(
−
1
)
N
v
N
=
m
{\displaystyle p_{N+kT-1}^{2}-pq_{N+kT-1}^{2}=(-1)^{N+kT}v_{N+kT}=(-1)^{N}v_{N}=m}
.
Ondoriz
x
=
p
N
+
k
T
−
1
,
y
=
q
N
+
k
T
−
1
{\displaystyle x=p_{N+kT-1},y=q_{N+kT-1}}
,
x
2
−
p
y
2
=
m
{\displaystyle x^{2}-py^{2}=m}
, ekuazioaren ebazpena da.
2:
T
{\displaystyle T}
bakoitia bada.
⇒
(
x
,
y
)
,
x
2
−
p
y
2
=
m
{\displaystyle \Rightarrow (x,y),x^{2}-py^{2}=m}
-ren ebazpen ez neutroa,
z
k
h
(
x
,
y
)
=
1
{\displaystyle zkh(x,y)=1}
.
Aurreko atalean bezala argudiatuz:
x
y
{\displaystyle {\frac {x}{y}}}
,
p
{\displaystyle {\sqrt {p}}}
-ren hondar bat da:
j
−
1
{\displaystyle j-1}
. hondarra:
x
y
=
R
j
−
1
{\displaystyle {\frac {x}{y}}=R_{j-1}}
,
j
∈
N
{\displaystyle j\in \mathbb {N} }
.
m
=
p
j
−
1
2
−
p
q
j
−
1
2
=
(
−
1
)
j
v
j
{\displaystyle m=p_{j-1}^{2}-pq_{j-1}^{2}=(-1)^{j}v_{j}}
.
Zatidura euklidearra eginez.
j
−
1
=
k
T
+
h
:
h
∈
[
0
,
T
−
1
]
⇔
j
=
k
T
+
N
j
:
N
j
∈
[
1
,
T
]
{\displaystyle j-1=kT+h:h\in [0,T-1]\Leftrightarrow j=kT+N_{j}:N_{j}\in [1,T]}
, zeinetan
N
j
=
h
+
1
{\displaystyle N_{j}=h+1}
.
Bi kasu eman daitezke:
2.1:
k
{\displaystyle k}
bikoitia bada
k
T
{\displaystyle kT}
bikoitia da.
k
=
2
r
:
r
∈
N
∪
{
0
}
{\displaystyle k=2r:r\in \mathbb {N} \cup \{0\}}
adieraziz:
m
=
x
2
−
p
y
2
=
p
j
−
1
2
−
p
q
j
−
1
2
=
p
N
j
+
2
r
T
−
1
2
−
p
q
N
j
+
2
r
T
−
1
2
=
(
−
1
)
j
v
j
{\displaystyle m=x^{2}-py^{2}=p_{j-1}^{2}-pq_{j-1}^{2}=p_{N_{j}+2rT-1}^{2}-pq_{N_{j}+2rT-1}^{2}=(-1)^{j}v_{j}}
.
(
−
1
)
j
=
(
−
1
)
N
j
;
v
j
=
v
N
j
{\displaystyle (-1)^{j}=(-1)^{N_{j}};v_{j}=v_{N_{j}}}
.
k
{\displaystyle k}
bikoitia eta
v
n
{\displaystyle v_{n}}
,
T
{\displaystyle T}
-periodikoa.
N
j
∈
ξ
m
≠
∅
{\displaystyle N_{j}\in \xi _{m}\neq \emptyset }
(
x
,
y
)
∈
⋃
N
∈
ξ
m
{
(
p
N
+
2
k
T
−
1
,
q
N
+
2
k
T
−
1
)
:
k
∈
N
∪
{
0
}
}
{\displaystyle (x,y)\in \bigcup _{N\in \xi _{m}}\{(p_{N+2kT-1},q_{N+2kT-1}):k\in \mathbb {N} \cup \{0\}\}}
2.2.
k
{\displaystyle k}
bakoitia bada
k
T
{\displaystyle kT}
bakoitia da.(
T
{\displaystyle T}
bakoitia denez).
k
=
2
r
+
1
:
r
∈
N
∪
{
0
}
{\displaystyle k=2r+1:r\in \mathbb {N} \cup \{0\}}
.
m
=
x
2
−
p
y
2
=
p
j
−
1
2
−
p
q
j
−
1
2
=
(
−
1
)
j
v
j
{\displaystyle m=x^{2}-py^{2}=p_{j-1}^{2}-pq_{j-1}^{2}=(-1)^{j}v_{j}}
(
−
1
)
j
v
j
=
(
−
1
)
N
j
+
(
2
r
+
1
)
T
v
N
j
+
(
2
r
+
1
)
T
=
−
(
−
1
)
N
j
v
N
j
{\displaystyle (-1)^{j}v_{j}=(-1)^{N_{j}+(2r+1)T}v_{N_{j}+(2r+1)T}=-(-1)^{N_{j}}v_{N_{j}}}
⇒
N
j
∈
ξ
−
m
≠
∅
{\displaystyle \Rightarrow N_{j}\in \xi _{-m}\neq \emptyset }
,
(
x
,
y
)
∈
⋃
N
∈
ξ
−
m
{
(
p
N
+
(
2
k
+
1
)
T
−
1
,
q
N
+
(
2
k
+
1
)
T
−
1
)
:
k
∈
N
∪
{
0
}
}
{\displaystyle (x,y)\in \bigcup _{N\in \xi _{-m}}\{(p_{N+(2k+1)T-1},q_{N+(2k+1)T-1}):k\in \mathbb {N} \cup \{0\}\}}
. alderantzizkoa
⇐
ξ
m
≠
∅
{\displaystyle \Leftarrow \xi _{m}\neq \emptyset }
bada
∃
N
∈
[
1
,
T
]
:
(
−
1
)
N
v
N
=
m
{\displaystyle \exists N\in [1,T]:(-1)^{N}v_{N}=m}
, orduan:
k
∈
N
∪
{
0
}
{\displaystyle k\in \mathbb {N} \cup \{0\}}
p
N
+
2
k
T
−
1
2
−
p
q
N
+
2
k
T
−
1
2
=
(
−
1
)
N
+
2
k
T
v
N
+
2
k
T
=
(
−
1
)
N
v
N
=
m
{\displaystyle p_{N+2kT-1}^{2}-pq_{N+2kT-1}^{2}=(-1)^{N+2kT}v_{N+2kT}=(-1)^{N}v_{N}=m}
.
ξ
−
m
≠
∅
{\displaystyle \xi _{-m}\neq \emptyset }
bada
∃
N
∈
[
1
,
T
]
:
(
−
1
)
N
v
N
=
−
m
{\displaystyle \exists N\in [1,T]:(-1)^{N}v_{N}=-m}
, orduan:
k
∈
N
∪
{
0
}
{\displaystyle k\in \mathbb {N} \cup \{0\}}
p
N
+
(
2
k
+
1
)
T
−
1
2
−
p
q
N
+
(
2
k
+
1
)
T
−
1
2
=
(
−
1
)
N
+
(
2
k
+
1
)
T
v
N
+
(
2
k
+
1
)
T
=
{\displaystyle p_{N+(2k+1)T-1}^{2}-pq_{N+(2k+1)T-1}^{2}=(-1)^{N+(2k+1)T}v_{N+(2k+1)T}=}
=
(
−
1
)
N
+
(
2
k
+
1
)
T
v
N
+
(
2
k
+
1
)
T
=
−
(
−
1
)
N
v
N
=
−
(
−
m
)
=
m
{\displaystyle =(-1)^{N+(2k+1)T}v_{N+(2k+1)T}=-(-1)^{N}v_{N}=-(-m)=m}
.
3:
p
{\displaystyle {\sqrt {p}}}
-ren garapenaren periodoa:
T
{\displaystyle T}
, bikoitia ala bakoitia denez:
x
2
−
p
y
2
=
m
{\displaystyle x^{2}-py^{2}=m}
bateragarria da eta teorema honen 1 eta 2 atalengatik infinitu ebazpen ditu.
p
∈
N
−
N
2
{\displaystyle p\in \mathbb {N} -\mathbb {N} ^{2}}
, eta
T
{\displaystyle T}
,
p
{\displaystyle {\sqrt {p}}}
-ren zatiki jarraien bidezko periodoa. Orduan:
1:
T
{\displaystyle T}
bikoitia bada,
x
2
−
p
y
2
=
1
{\displaystyle x^{2}-py^{2}=1}
-en ebazpen minimoa:
x
=
p
T
−
1
,
y
=
q
T
−
1
{\displaystyle x=p_{T-1},y=q_{T-1}}
da. Zeinetan
p
T
−
1
q
T
−
1
{\displaystyle {\frac {p_{T-1}}{q_{T-1}}}}
,
p
{\displaystyle {\sqrt {p}}}
-ren
(
T
−
1
)
{\displaystyle (T-1)}
. hondarra den.
2:
T
{\displaystyle T}
bakoitia bada,
x
2
−
p
y
2
=
1
{\displaystyle x^{2}-py^{2}=1}
-en ebazpen minimoa:
x
=
p
2
T
−
1
,
y
=
q
2
T
−
1
{\displaystyle x=p_{2T-1},y=q_{2T-1}}
da. Zeinetan
p
2
T
−
1
q
2
T
−
1
{\displaystyle {\frac {p_{2T-1}}{q_{2T-1}}}}
,
p
{\displaystyle {\sqrt {p}}}
-ren
(
2
T
−
1
)
{\displaystyle (2T-1)}
. hondarra den.
Froga : Proposizioa3-ko 3. atalagatik:
T
=
M
i
n
{
k
∈
N
:
a
k
=
2
a
0
}
{\displaystyle T=Min\{k\in \mathbb {N} :a_{k}=2a_{0}\}}
α
=
[
p
]
+
p
=
a
0
+
p
{\displaystyle \alpha =[{\sqrt {p}}]+{\sqrt {p}}=a_{0}+{\sqrt {p}}}
, periodiko hutsa da, eta
T
{\displaystyle T}
periodokoa:
α
j
=
α
j
+
T
{\displaystyle \alpha _{j}=\alpha _{j+T}}
.
α
T
=
α
0
⇒
u
T
+
p
v
T
=
a
0
+
p
{\displaystyle \alpha _{T}=\alpha _{0}\Rightarrow {\frac {u_{T}+{\sqrt {p}}}{v_{T}}}=a_{0}+{\sqrt {p}}}
, honela
v
T
=
1
{\displaystyle v_{T}=1}
.
Honela aurreko teorema aplikatuz.
T
{\displaystyle T}
bikoitia bada,
(
−
1
)
T
v
T
=
1
⇒
T
∈
ξ
1
≠
∅
{\displaystyle (-1)^{T}v_{T}=1\Rightarrow T\in \xi _{1}\neq \emptyset }
.
Eta ebazpen txikiena:
x
=
p
T
−
1
,
y
=
q
T
−
1
⇒
x
y
=
R
T
−
1
{\displaystyle x=p_{T-1},y=q_{T-1}\Rightarrow {\frac {x}{y}}=R_{T-1}}
.
T
{\displaystyle T}
bakoitia bada,
(
−
1
)
T
v
T
=
−
1
⇒
T
∈
ξ
−
1
≠
∅
{\displaystyle (-1)^{T}v_{T}=-1\Rightarrow T\in \xi _{-1}\neq \emptyset }
.
Eta beraz ebazpen txikiena:
x
=
p
2
T
−
1
,
y
=
q
2
T
−
1
⇒
x
y
=
R
2
T
−
1
{\displaystyle x=p_{2T-1},y=q_{2T-1}\Rightarrow {\frac {x}{y}}=R_{2T-1}}
.