Sarrera
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Zenbaki irrazional bikarratuen orrialde hau sortu da, Pellen ekuazioaren oinarrizko ebazpena determinatzeko bidean: Euler, Lagrangek eta Galoisek zenbaki irrazional bikarratu laburtuak karakterizatzen dituztelako, zatiki jarraien 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 {\displaystyle x^{2}-py^{2}=1} ekuazioaren ebazpen baten existentzia frogatzen du. Ebazpen honen existentzia desberdintza batek determinatzen du: {\displaystyle |{x}/{y}-{\sqrt {p}}|<{1}/{y^{2}}}. Desberdintza hori betetzen duten {\displaystyle x/y} zatikiak, {\displaystyle {\sqrt {p}}} zenbakiaren {\displaystyle n}. hondar bat dira ( zatiki jarraien bidezko bere garapenaren hondarra ). Honela Pell-en ekuazioaren ebazpenak: {\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 {\displaystyle {\sqrt {p}}} moduko zenbakien zatiki jarraien bidezko garapena: {\displaystyle T}-periodikoa dela: {\displaystyle hein(1)}-etik aurrera, eta ebazpenak, periodoa osatu aurreko hondarrren artean daudela: {\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: {\displaystyle R\_{T-1}}, edo {\displaystyle R\_{2T-1}}.

### **Definizioa**1:

{\displaystyle \alpha \in \mathbb {R} }, zenbaki irrazional bat bikarratua izendatzen da, koefiziente osoak dituen bigarren graduko polinomio baten erroa bada.

### **Notazioa**:

{\displaystyle \mathbb {N} ^{2}=\{n^{2}:n\in \mathbb {N} \}}. Adieraziko da.

{\displaystyle \alpha \in \mathbb {Q} ({\sqrt {p}})}, bada non {\displaystyle \alpha =a+b{\sqrt {p}}}, orduan {\displaystyle {\overline {\alpha }}=a-b{\sqrt {p}}} , bere konjokatua izendatuko da.

{\displaystyle \alpha \in \mathbb {R} } ren {\displaystyle n}. hondarra: {\displaystyle R\_{n}}, beste modu honetan ere adieraziko da: {\displaystyle R\_{n}(\alpha )}, beste zenbaki irrazionalen {\displaystyle n}. hondarretik bereizteko.

### **Definizioa**2:

{\displaystyle \alpha \in \mathbb {R} }, zenbaki irrazional bikarratua, laburtua dela esango dugu, baldin eta: {\displaystyle \alpha >1} bada eta {\displaystyle -1<{\overline {\alpha }}<0}.

Zeinetan {\displaystyle \alpha } eta {\displaystyle {\overline {\alpha }}} polinomio berdinaren erro konjokatuak diren.

### **Definizioa3**

Izan bedi {\displaystyle \alpha } zenbaki irrazionala, zatiki jarraien bidezko bere garapena, {\displaystyle \alpha =[a\_{0},a\_{1},...a\_{n},...]} periodikoa dela esango dugu, {\displaystyle (a\_{n})\_{n\geq 0}}, segida periodikoa bada, eta segidaren periodoa: {\displaystyle T}, bada {\displaystyle \alpha }-ren zatiki jarraien bidezko garapenaren periodoa: {\displaystyle T}, dela esango dugu.

Periodoaren lehen elementua: {\displaystyle a\_{N}}, bada {\displaystyle \alpha }-ren zatiki jarraien garapena, {\displaystyle T}-periodikoa dela esango da {\displaystyle hein(N)}-tik aurrera (heina N). Honela adieraziko da:

{\displaystyle \alpha =[a\_{0},a\_{1},...a\_{N-1},{\overline {a\_{N},...a\_{N+T-1}}}]}.

* Eulerrek frogatu zuen 1748.urtean: {\displaystyle \alpha }-ren zatiki jarraien garapena periodikoa bada, orduan {\displaystyle \alpha } zenbaki irrazionala bikarratua dela, eta Lagrangek frogatu zuen 1768.urtean: {\displaystyle \alpha } zenbaki irrazional bikarratua bada, orduan zatiki jarraien bidezko garapena periodikoa dela. Biak idatziko dira proposizio batean.

### **Proposizioa1:**

Izan bedi {\displaystyle \alpha =[a\_{0},a\_{1},...a\_{n},...]}, zatiki jarraien bidezko bere garapena. Orduan: garapena periodikoa da baldin eta soilik baldin {\displaystyle \alpha } zenbaki irrazional bikarratua bada.

**Froga:** {\displaystyle \Rightarrow Euler(1748)}

Suposa bedi {\displaystyle \alpha }-ren zatiki jarraien bidezko garapena, {\displaystyle T}-periodikoa dela, {\displaystyle hein(N)}-tik aurrera: {\displaystyle \alpha =[a\_{0},a\_{1},...a\_{N-1},{\overline {a\_{N},...a\_{N+T-1}}}]} .

{\displaystyle \alpha }-ren {\displaystyle n}. hondarra (zatiki laburtezin gisara adierazia): {\displaystyle R\_{n}=[a\_{0},a\_{1},...a\_{n}]={\frac {p\_{n}}{q\_{n}}}}.

{\displaystyle \alpha }-ren {\displaystyle N}.zatiki osatua: {\displaystyle \alpha \_{N}=[{\overline {a\_{N},...a\_{N+T-1}}}]}.

Eta {\displaystyle N}.zatiki osatuaren: {\displaystyle \alpha \_{N}}, {\displaystyle n}. hondarra (zatiki laburtezina): {\displaystyle R\_{n}(\alpha \_{N})=[a\_{N},a\_{N+1},...a\_{N+n}]={\frac {r\_{n}}{s\_{n}}}}.

{\displaystyle \alpha } zenbaki irrazional baten, {\displaystyle n}.zatiki osatuak: {\displaystyle \alpha \_{n}}, ondorengo ezaugarriaren bidez definitzen dira:

{\displaystyle \alpha =[a\_{0},a\_{1},...a\_{n-1},\alpha \_{n}]} (ikus zatiki jarraiak). {\displaystyle \alpha \_{N}} periodikoa dela kontuan hartuz:

{\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:

{\displaystyle s\_{T-1}\alpha \_{N}^{2}+(s\_{T-2}-r\_{T-1})\alpha \_{N}-r\_{T-2}=0}.

{\displaystyle s\_{T-1}\in \mathbb {N} } denez, {\displaystyle \alpha \_{N}} bikarratua da.

{\displaystyle \alpha \_{N}\not \in \mathbb {Q} }. Absurdura bideratuz: Baldin eta: {\displaystyle \alpha \_{N}\in \mathbb {Q} \Rightarrow \alpha \in \mathbb {Q} } , eta ondorioz zatiki jarraien bidezko garapena finitua da (zatiki jarraiak) eta {\displaystyle \alpha =[a\_{0},a\_{1},...a\_{n},\alpha \_{N}]} denez, {\displaystyle \alpha }-ren garapena ere finitua da, zeina ezinezkoa den (hipotesiagatik periodikoa delako).

Ondorioz: {\displaystyle \alpha \_{N}}, zenbaki irrazional bikarratua da: {\displaystyle \alpha \_{N}=a+b{\sqrt {c}}} modukoa non: {\displaystyle a,b\in \mathbb {Q} ;b\neq 0}, eta {\displaystyle c\in \mathbb {N} -\mathbb {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}}}}

{\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: {\displaystyle u,v\in \mathbb {Q} }.

{\displaystyle v\neq 0} frogatuko da.

{\displaystyle v=0\Rightarrow \alpha \in \mathbb {Q} }, eta beraz zatiki jarraien bidezko garapena finitua. Ezinezkoa.

Ondorioz: {\displaystyle \alpha }, zenbaki irrazional bikarratua da.

{\displaystyle \Leftarrow Lagrange(1768)}

{\displaystyle \alpha \in \mathbb {R} }, zenbaki irraziona bikarratua bada: {\displaystyle \exists a,b,c\in \mathbb {Z} :}i) {\displaystyle a\neq 0}, ii) {\displaystyle a\alpha ^{2}+b\alpha +c=0} eta iii) {\displaystyle \Delta =b^{2}-4ac\in \mathbb {N} -\mathbb {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 {\displaystyle \alpha }, bikarratua izateagatik: {\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 }

{\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 } {\displaystyle A\_{n}\alpha \_{n}^{2}+B\_{n}\alpha \_{n}+C\_{n}=0.}

{\displaystyle n\geq 2} rentzat:

{\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**. {\displaystyle A\_{n}\neq 0}, frogatuko da.

{\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 {\displaystyle \alpha }, arrazionala, ezinezkoa.

**Bi**: {\displaystyle p(x)=ax^{2}+bx+c}, ren diskriminantea eta {\displaystyle p\_{n}(x)=A\_{n}x^{2}+B\_{n}x+C\_{n}}en diskrimintea berdinak direla frogatuko da.

Hots: {\displaystyle \forall n\geq 2}. {\displaystyle \Delta \_{n}=B\_{n}^{2}-4A\_{n}C\_{n}=b^{2}-4ac=\Delta }.

{\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:

{\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:

{\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:

{\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}}}

{\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}}

{\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 {\displaystyle (p\_{n-1}q\_{n-2}+q\_{n-1}p\_{n-2})^{2}=(-1)^{2n}}berdintza aplikatuz (zatiki jarraiak).

{\displaystyle \forall n\geq 2}, {\displaystyle B\_{n}^{2}-4A\_{n}C\_{n}=b^{2}-4ac}.

**Hiru** {\displaystyle (A\_{n})\_{n\geq 0},(B\_{n})\_{n\geq 0},(C\_{n})\_{n\geq 0}}segidak barnatuak daudela frogatuko da.

{\displaystyle \alpha }-ren {\displaystyle n}. hondarra: {\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 ):

{\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:

{\displaystyle \forall n\geq 2} rentzat, {\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}}}}.

{\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: {\displaystyle a\alpha ^{2}+b\alpha +c=0}, berdintza betetzen denez:

{\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: {\displaystyle \forall n\geq 1,|A\_{n}|\leq 2|a||\alpha |+|a|+|b|\Rightarrow (A\_{n})\_{n\geq 0}} Barnatua.

{\displaystyle \forall n\geq 2,|C\_{n}|=|A\_{n-1}|\leq 2|a||\alpha |+|a|+|b|\Rightarrow (C\_{n})\_{n\geq 0}}, Barnatua.

Eta {\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}.

{\displaystyle \Rightarrow (B\_{n})\_{n\geq 0}}, Barnatua.

**Lau**: {\displaystyle \alpha =[a\_{0},a\_{1},...a\_{n},...]}, periodikoa dela frogatuko da.

{\displaystyle (A\_{n})\_{n\geq 0},(B\_{n})\_{n\geq 0},(C\_{n})\_{n\geq 0}} segidak barnatuak daude hirugarren atalagatik, orduan:

{\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:

{\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 {\displaystyle [1,A]\times [1,B]\times [1,C]}, multzoa berriz finitua.

{\displaystyle \Rightarrow \exists j,k,l\in \mathbb {N} \cup \{0\}}desberdinak non: {\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 {\displaystyle \alpha \_{j},\alpha \_{k},\alpha \_{l}} aipaturiko polinomioen erroak badira hurrenez hurren. Polinomioak berdinak direnez, hiru erro horiek {\displaystyle p\_{j}(x)}-en erroak dira. Ondorioz gutsienez bi erro berdinak dira: {\displaystyle \alpha \_{j}=\alpha \_{k};k>j}, suposatuko da orokortasunik galdu gabe.

{\displaystyle T=k-j}, aukeratuz.{\displaystyle \alpha \_{j}=\alpha \_{k}\Rightarrow [a\_{j},a\_{j+1},...]=[a\_{k},a\_{k+1},...]\Rightarrow a\_{j+n}=a\_{k+n}}.

{\displaystyle \forall n\geq 0}-rentzat. {\displaystyle a\_{j+n}=a\_{j+n+k-j}=a\_{j+n+T}} {\displaystyle \Rightarrow } -rentzat {\displaystyle a\_{n}=a\_{n+T}}.

Ondorioz {\displaystyle (a\_{n})\_{n\geq 0}}, {\displaystyle T}-ren zatitzaile baten periodokoa da, gutsienez {\displaystyle hein(j)}-tik aurrera.

### **Definizioa4**

Zenbaki irrazional baten, zatiki jarraien bidezko garapena, periodiko hutsa dela esaten da, {\displaystyle hein(0)}-tik aurrera periodikoa bada.

Galoisek zenbaki irrazional bikarratu laburtuak karakterizatzen ditu.

### **Proposizioa2** (Galois 1828)

Zenbaki erreal bat: irrazional bikarratu laburtua da, baldin eta soilik baldin zatiki jarraien bidezko bere garapena periodiko hutsa bada.

**Froga**: {\displaystyle \Rightarrow }

Izan bedi {\displaystyle \alpha } zenbaki erreala, eta suposa bedi zenbaki irrazional bikarratu laburtua dela.

**Bat**: {\displaystyle \forall n\in \mathbb {N} \cup \{0\}}, {\displaystyle \alpha \_{n}}: {\displaystyle \alpha }-ren {\displaystyle n}. zatiki laburtua, zenbaki irrazional bikarratu laburtua dela frogatuko da.

{\displaystyle \alpha }, irrazional bikarratua bada, orduan Lagrangek dio zatiki jarraien bidezko bere garapena periodikoa dela.

{\displaystyle \forall n\in \mathbb {N} \cup \{0\}}rentzat, {\displaystyle \alpha \_{n}}-ren koefizienteak {\displaystyle \alpha }-ren koefizienteen berdinak direnez {\displaystyle hein(n)}-tik aurrera, {\displaystyle \alpha \_{n}}-ren zatiki jarraien bidezko garapena periodikoa da, orduan Eulerrek dio {\displaystyle \alpha \_{n}} zenbaki irrazional bikarratua dela.

Ondorioz: {\displaystyle \forall n\in \mathbb {N} \cup \{0\}}-rentzat {\displaystyle \alpha \_{n}}zenbaki irrazional bikarratua da.

{\displaystyle \forall n\in \mathbb {N} \cup \{0\}}-rentzat, {\displaystyle \alpha \_{n}} laburtua dela frogatuko da indukzioz.

{\displaystyle \alpha =\alpha \_{0}}, laburtua da hipotesiagatik.

Suposa bedi {\displaystyle \alpha \_{n}}laburtua dela, orduan: {\displaystyle \alpha \_{n}>1;{\overline {\alpha \_{n}}}\in (-1,0)}.

{\displaystyle \alpha \_{n+1}={\frac {1}{\alpha \_{n}-[\alpha \_{n}]}}\Leftrightarrow \alpha \_{n}=a\_{n}+{\frac {1}{\alpha \_{n+1}}}>1}, ( {\displaystyle [\alpha \_{n}]=a\_{n}}, parte osoa ).

{\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. {\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.

{\displaystyle \alpha \_{n}>1\Rightarrow [\alpha \_{n}]=a\_{n}\geq 1}

{\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.

{\displaystyle \alpha \_{n}}laburtua {\displaystyle \Rightarrow \alpha \_{n+1}}laburtua ondorioztatu da.

Indukzio printzipioagatik: {\displaystyle \forall n\in \mathbb {N} \cup \{0\}}-rentzat {\displaystyle \alpha \_{n}}laburtua.

**Bi:** {\displaystyle \alpha } periodiko hutsa dela frogatuko da.

Absurdura bideratuz, {\displaystyle \alpha } periodiko hutsa ez bada:

{\displaystyle \alpha =[a\_{0},a\_{1},...a\_{N-1},{\overline {a\_{N},...a\_{N+T-1}}}]}, zeinetan {\displaystyle T} periodoa den, eta {\displaystyle N} berriz lehen periodoaren lehen indizea.

Periodiko hutsa ez izateagatik: {\displaystyle N\geq 1}.

{\displaystyle \alpha =[a\_{0},a\_{1},...a\_{N-1},\alpha \_{N}]}, {\displaystyle T}-peridikoa izateagatik , {\displaystyle N}. zatiki laburtua: {\displaystyle \alpha \_{N}}, {\displaystyle T}-periodikoa da.

{\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}}.

{\displaystyle \alpha }-ren, {\displaystyle n}. hondarrak: {\displaystyle \alpha \_{n}} laburtuak dira. Lehen baldintzagatik:

{\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}.

{\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:

{\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}}}.

{\displaystyle [a\_{N-1}-{\overline {\alpha \_{N-1}}}]=a\_{N-1}}, eta {\displaystyle [a\_{N+T-1}-{\overline {\alpha \_{N+T-1}}}]=a\_{N+T-1}}. (Parte osoak kasurako)

{\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 {\displaystyle N-1} , edo txikiagoa da, zeina ezinezkoa den.

{\displaystyle \Leftarrow }

Suposa dezagun {\displaystyle \alpha }-ren zatiki jarraien bidezko garapena, periodiko hutsa dela, orduan:

{\displaystyle \exists T\in \mathbb {N} :\alpha =[{\overline {a\_{0},a\_{1},...a\_{T-1}}}]\Rightarrow a\_{0}=a\_{T}\in \mathbb {N} }, ondorioz: {\displaystyle \alpha >a\_{0}\geq 1}

{\displaystyle T=1\Rightarrow \alpha =[a\_{0},\alpha ]=a\_{0}+{\frac {1}{\alpha }}}, eta beraz {\displaystyle \alpha }, {\displaystyle p(x)=x^{2}-a\_{0}x-1}, en erroa da.

{\displaystyle p(x)}-en erroak: {\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)}, {\displaystyle \alpha } laburtua.

{\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:

{\displaystyle \alpha }, {\displaystyle p(x)=q\_{T-1}x^{2}+(q\_{T-2}-p\_{T-1})x-p\_{T-2}}-en erroa da, eta {\displaystyle \alpha >a\_{0}\geq 1}.

Bigarren erroa {\displaystyle (-1,0)} bitartean dagoela frogatuko da.

{\displaystyle p(0)=-p\_{T-2}<0}.

{\displaystyle q\_{n}\uparrow ,p\_{n}\uparrow }, gorakorrak direnez: {\displaystyle p(-1)=q\_{T-1}-q\_{T-2}+p\_{T-1}-p\_{T-2}\geq 0}.

{\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:

{\displaystyle p(-1)>0}.

{\displaystyle p(x)}, jarraia denez erdiko balioaren teorema aplika daiteke.

{\displaystyle \exists \zeta \in (-1,0):p(\zeta )=0}, eta {\displaystyle p(x)}-en erroak konjokatuak direnez: {\displaystyle \zeta ={\overline {\alpha }}\in (-1,0)}.

Ondorioz, {\displaystyle \alpha } laburtua da.

### **Proposizioa3**:

Izan bedi {\displaystyle p\in \mathbb {N} -\mathbb {N} ^{2}}, zenbaki ez karratua, {\displaystyle \alpha =[{\sqrt {p}}]+{\sqrt {p}}}, eta {\displaystyle ({\alpha \_{n}})\_{n\geq 0}}, {\displaystyle \alpha }-ren zatiki osatuen segida. Orduan:

1:Zatiki jarraien bidezko {\displaystyle {\sqrt {p}}}-ren garapena, periodikoa da {\displaystyle hein(1)}-etik aurrera.

2: {\displaystyle \exists u\_{n},v\_{n}\in \mathbb {N} } , bi segida non {\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:{\displaystyle {\sqrt {p}}}-ren periodoa {\displaystyle T=Min\{k\in \mathbb {N} :a\_{k}=2a\_{0}\}}.

4: Baldin eta, {\displaystyle \exists N\in [1,T-1]:u\_{N}\in v\_{N}\mathbb {Z} \Rightarrow T=2N}, eta beraz {\displaystyle T} bikoitia.

**Froga**

Izan bedi:{\displaystyle {\sqrt {p}}=[a\_{0},a\_{1},...a\_{n},...]} zatiki jarraien bidezko garapena, orduan {\displaystyle \alpha =[2a\_{0},a\_{1},...a\_{n},...]}.

**1**: {\displaystyle \alpha =a\_{0}+{\sqrt {p}}} , zenbaki irrazional bikarratu laburtua dela frogatuko da. ({\displaystyle a\_{0}=[{\sqrt {p}}]}, parte osoa).

{\displaystyle \alpha }, {\displaystyle p(x)=x^{2}-2a\_{0}x+a\_{0}^{2}-p}, polinomioaren erroa da. {\displaystyle \alpha =[{\sqrt {p}}]+{\sqrt {p}}>1} .

Beste erroa: {\displaystyle {\overline {\alpha }}=-({\sqrt {p}}-a\_{0})\in (-1,0)}, ondorioz {\displaystyle \alpha } zenbaki irrazional bikarratu laburtua da.

Galoisek dio, {\displaystyle \alpha }-ren zatiki jarraien bidezko garapena periodiko hutsa dela:

{\displaystyle \alpha =[{\overline {2a\_{0},a\_{1},...a\_{T-1}}}]\Rightarrow {\sqrt {p}}=[a\_{0},{\overline {a\_{1},...a\_{T-1},a\_{2a\_{0}}}}]}.

{\displaystyle {\sqrt {p}}}-ren zatiki jarraien bidezko garapena periodikoa, {\displaystyle hein(1)}-etik aurrera.

**2**: {\displaystyle u\_{n},v\_{n}} segidak errekurrentziz eraikiko dira, ondoren adierazten den bezala:

{\displaystyle n=0} bada, {\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 {\displaystyle u\_{n};v\_{n}}eraiki direla, ondorengo ezaugarriekin:

{\displaystyle \alpha \_{n}={\frac {u\_{n}+{\sqrt {p}}}{v\_{n}}}}, {\displaystyle p-u\_{n}^{2}\in v\_{n}\mathbb {Z} }, {\displaystyle u\_{n},v\_{n}\in \mathbb {N} }, eta {\displaystyle u\_{n}\leq a\_{0}}.

**2.1**: {\displaystyle u\_{n+1};v\_{n+1}}, eraikiko dira. .

{\displaystyle \alpha \_{n}={\frac {u\_{n}+{\sqrt {p}}}{v\_{n}}}=b\_{n}+{\frac {u\_{n}-b\_{n}v\_{n}+{\sqrt {p}}}{v\_{n}}}}, zeinetan {\displaystyle b\_{n}=[\alpha \_{n}]}.

{\displaystyle u\_{n+1}=b\_{n}v\_{n}-u\_{n}}, izango da.

{\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})}}}. {\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}}}}

{\displaystyle v\_{n+1}={\frac {p-u\_{n+1}^{2}}{v\_{n}}}}, izango da.

**2.2:** {\displaystyle \alpha \_{n+1}={\frac {u\_{n+1}+{\sqrt {p}}}{v\_{n+1}}}}, {\displaystyle u\_{n+1},v\_{n+1}\in \mathbb {N} } eta {\displaystyle p-u\_{n+1}^{2}\in v\_{n+1}\mathbb {Z} }. betetzen direla frogatuko da.

{\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}}}}.

{\displaystyle u\_{n+1}=b\_{n}v\_{n}-u\_{n}\in \mathbb {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} }.

{\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**: {\displaystyle u\_{n+1},v\_{n+1}\in \mathbb {N} }, frogatuko da.

{\displaystyle \alpha }, zenbaki irrazional bikarratu laburtua izanagatik, {\displaystyle \alpha \_{n}} zatiki osatu guztiak zenbaki irrazional bikarratu laburtuak dira.

{\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}.

{\displaystyle u\_{n+1}=b\_{n}v\_{n}-u\_{n}\in \mathbb {Z} } eta {\displaystyle b\_{n}v\_{n}-u\_{n}>0}, orduan:

{\displaystyle u\_{n+1}\in \mathbb {N} }.

{\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}

{\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 {\displaystyle v\_{n+1}\in \mathbb {Z} }, orduan:

{\displaystyle v\_{n+1}\in \mathbb {N} }.

**2.4**: {\displaystyle u\_{n+1}\leq a\_{0}}, frogatuko da.

{\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}}},

{\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**:{\displaystyle {\sqrt {p}}=[a\_{0},{\overline {a\_{1},...a\_{T-1},a\_{2a\_{0}}}}]}, ondorioztatu da 1, atalean.

**3.1**. {\displaystyle \forall j\in [1,T-1],\alpha \_{j}\neq \alpha }. Frogatuko da.

Absurdura bideratuz, baldin eta:{\displaystyle \exists j\in [1,T-1]:\alpha \_{j}=\alpha }

{\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:{\displaystyle {\sqrt {p}}=[a\_{0},{\overline {a\_{1},...a\_{j-1},a\_{2a\_{0}}}}]\Rightarrow T=j<T}, ezinezkoa.

**3.2** {\displaystyle \forall j\in [1,T-1],a\_{j}\neq 2a\_{0}}, frogatuko da.

Absurdura bideratuz: {\displaystyle \exists j\in [1,T-1]:a\_{j}=2a\_{0}}.

{\displaystyle j\geq 1\Rightarrow b\_{j}=a\_{j}=2a\_{0}}, eta {\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: {\displaystyle u\_{n}\leq a\_{0}}, ondorioz:

{\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:

{\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 {\displaystyle \exists N\in [1,T-1]:u\_{N}\in v\_{N}\mathbb {Z} }. {\displaystyle \Rightarrow T=2N} ondorioztatuko da

**4.1**. {\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.** {\displaystyle j=0\Rightarrow u\_{N+1}=u\_{N},v\_{N+1}=v\_{N-1};b\_{N+1}=b\_{N-1}}, frogatuko da lehenik.

{\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 {\displaystyle \alpha } zenbaki irrazional bikarratu laburtua izateagatik:

{\displaystyle {\frac {u\_{N}+{\sqrt {p}}}{v\_{N}}}>1;{\frac {u\_{N}-{\sqrt {p}}}{v\_{N}}}\in (-1,0)}.

{\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}

{\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}}}}

{\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}}

{\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}}.

{\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 {\displaystyle u\_{N}=b\_{N-1}v\_{N-1}-u\_{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: {\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 {\displaystyle (j+1)}entzat 4.1. erlazioa betetzen dela frogatuko da.

{\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}}

{\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}}.

{\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},

{\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}}, {\displaystyle \Rightarrow v\_{N+j+2}=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}}}]}

{\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**. {\displaystyle T=2N} ondorioztatuko da.

{\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.

{\displaystyle j=N-1\Rightarrow b\_{2N}=b\_{0}}, betetzen da eta 3. atalagatik: {\displaystyle 2N=\zeta T:\zeta \in \mathbb {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.

Pellen ekuazioaren oinarrizko ebazpena
--------------------------------------

Galoisek ondorioztatuko du: {\displaystyle {\sqrt {p}}} moduko zenbakien zatiki jarraien bidezko garapena: {\displaystyle T}-periodikoa dela: {\displaystyle hein(1)}-etik aurrera. Atal honetan frogatuko da, {\displaystyle p} parametrodun {\displaystyle x^{2}-py^{2}=1} edo {\displaystyle x^{2}-py^{2}=-1} ekuazioaren, {\displaystyle {\sqrt {p}}} -ren zatiki jarraien bidezko hondar bat dela eta ondar hori periodoa osatu aurrekoa dela: {\displaystyle R\_{T-1},R\_{2T-1},...} Hondarraren zenbakitzailea eta izendatzailea gorakorrak direnez: {\displaystyle R\_{T-1}}, izango da {\displaystyle x^{2}-py^{2}=1}-en ebazpen minimoa, {\displaystyle T} bikoitia denean, eta {\displaystyle x^{2}-py^{2}=-1}, en ebazpena {\displaystyle T} bakoitia denean. Honela {\displaystyle T} bakoitia denean, {\displaystyle x^{2}-py^{2}=1}-en ebazpen minimoa {\displaystyle R\_{2T-1}}izango da. 

### **Lema1**

Izan bedi: {\displaystyle p\in \mathbb {N} -\mathbb {N} ^{2}}, eta {\displaystyle m\in \mathbb {Z} -\{0\}:1\leq |m|<{\sqrt {p}}}. 

Suposa bedi {\displaystyle \exists (x,y)\in \mathbb {N} \times \mathbb {N} :zkh(x,y)=1;x^{2}-py^{2}=m}.

Orduan {\displaystyle {\frac {x}{y}}}, {\displaystyle {\sqrt {p}}}-ren zatiki jarraien bidezko hondar bat da.

**Froga:**

{\displaystyle x^{2}-py^{2}=m\Rightarrow x-y{\sqrt {p}}={\frac {m}{x+y{\sqrt {p}}}}}.

**i)** {\displaystyle m>0} bada, {\displaystyle x^{2}=m+py^{2}>py^{2}\Rightarrow x>y{\sqrt {p}}}

{\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, {\displaystyle {\frac {x}{y}}}, {\displaystyle {\sqrt {p}}} -ren zatiki jarraien bidezko {\displaystyle n}. hondar bat da.

**ii)** {\displaystyle m<0} bada, {\displaystyle py^{2}=x^{2}-m>x^{2}\Rightarrow {\sqrt {p}}y>x}.

Lehen desberdintzan: {\displaystyle |m|<{\sqrt {p}}}, erabiliz eta bigarren desberdintzan: {\displaystyle {\sqrt {p}}y>x}.{\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: {\displaystyle {\frac {y}{x}}}, {\displaystyle {\frac {1}{\sqrt {p}}}}-en zatiki jarraien bidezko {\displaystyle n}. hondar bat da:

{\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 {\displaystyle {\frac {y}{x}}=[0,a\_{0},a\_{1},...a\_{n-1}]}, zeinetan {\displaystyle y\neq 0\Rightarrow n\neq 0\Rightarrow n\in \mathbb {N} }.

{\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}]}, {\displaystyle {\sqrt {p}}}-ren {\displaystyle (n-1)}. hondarra.

### Teorema.

{\displaystyle p\in \mathbb {N} -\mathbb {N} ^{2}}, eta {\displaystyle m\in \mathbb {Z} :1\leq |m|<{\sqrt {p}}}. Emanik ondorengoak baiezta daitezke.

{\displaystyle \delta \_{n}={\frac {u\_{n}+{\sqrt {p}}}{v\_{n}}}}, {\displaystyle {\sqrt {p}}}-ren {\displaystyle n}. zatiki osatua, zeinetan: {\displaystyle u\_{n},v\_{n}}proposizioa3-n bezala definituak.

{\displaystyle {\sqrt {p}}}-ren {\displaystyle n}. hondarra: {\displaystyle {\frac {p\_{n}}{q\_{n}}}}, {\displaystyle zkh(p\_{n},q\_{n})=1} eta {\displaystyle {\sqrt {p}}}: {\displaystyle T}-periodikoa.

{\displaystyle \forall l\in \mathbb {Z} }-rentzat, {\displaystyle \xi \_{l}=\{j\in [1,T]:(-1)^{j}v\_{j}=l\}} adieraziko da. Orduan:

1: {\displaystyle T} bikoitia bada: {\displaystyle x^{2}-py^{2}=m} ekuazioak ebazpen ez neutroa du {\displaystyle \Leftrightarrow \xi \_{m}\neq \emptyset }. 

Eta kasu horretan {\displaystyle zkh(x,y)=1}, betetzen duten ebazpen guztiak, ondorengo multzoan daude:{\displaystyle \bigcup \_{N\in \xi \_{m}}\{(p\_{N+kT-1},q\_{N+kT-1}):k\in \mathbb {N} \cup \{0\}\}}.

2: {\displaystyle T} bakoitia bada,{\displaystyle x^{2}-py^{2}=m} ekuazioak ebazpena (ez neutroa) du {\displaystyle \Leftrightarrow \xi \_{m}\cup \xi \_{-m}\neq \emptyset }. 

Eta kasu horretan {\displaystyle zkh(x,y)=1} , betetzen duten ebazpen guztiak, ondorengo multzoan daude: {\displaystyle \bigcup \_{N\in \xi \_{m}}\{(p\_{N+2kT-1},q\_{N+2kT-1}):k\in \mathbb {N} \cup \{0\}\}}, edo

{\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 {\displaystyle x^{2}-py^{2}=m}, ekuazioak infinitu soluzio ditu.

**FROGA**

{\displaystyle \delta \_{n}={\frac {u\_{n}+{\sqrt {p}}}{v\_{n}}}}, {\displaystyle {\sqrt {p}}}-ren {\displaystyle n}. zatiki osatua denez: {\displaystyle {\sqrt {p}}=[a\_{0},a\_{1},...a\_{n-1},\delta \_{n}]}.

{\displaystyle {\sqrt {p}}}-ren garapena: {\displaystyle T}-periodikoa denez, {\displaystyle hein(1)}-etik aurrera, {\displaystyle \delta \_{n}}ere {\displaystyle T} periodikoa da, eta modu bakarrean determinatzen dituenez: {\displaystyle u\_{n},v\_{n}}balioak, hauek ere gutsienez; {\displaystyle T} periodikoak dira.

Ondorioz {\displaystyle v\_{n}\in \{v\_{1},...v\_{T}\}}.

{\displaystyle {\sqrt {p}}}-ren {\displaystyle n}. hondarra: {\displaystyle R\_{n}={\frac {p\_{n}}{q\_{n}}}=[a\_{0},a\_{1},...a\_{n}]}, adieraziko da.

{\displaystyle \forall n\in \mathbb {N} ,p\_{n-1}^{2}-pq\_{n-1}^{2}=(-1)^{n}v\_{n}} betetzen dela frogatuko da.

{\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: {\displaystyle \forall n\geq 2}-rentzat.

{\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 }{\displaystyle q\_{n-1}p=p\_{n-1}u\_{n}+p\_{n-2}v\_{n}}.eta {\displaystyle p\_{n-1}=q\_{n-1}u\_{n}+q\_{n-2}v\_{n}}

Honela:

{\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: {\displaystyle p\_{n-1}q\_{n-2}-q\_{n-1}p\_{n-2}=(-1)^{n}}, betetzen denez:{\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}}.

{\displaystyle n=1} denean {\displaystyle p\_{0}^{2}-pq\_{0}^{2}=a\_{0}^{2}-p=(-1)v\_{1}}, izan ere: {\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:{\displaystyle \forall n\in \mathbb {N} ,p\_{n-1}^{2}-pq\_{n-1}^{2}=(-1)^{n}v\_{n}}

**1:** {\displaystyle T} bikoitia bada.

Suposa bedi: {\displaystyle (x,y)} , {\displaystyle x^{2}-py^{2}=m} ekuazioaren ebazpen ez neutroa dela, {\displaystyle zkh(x,y)=1} orduan Lema3 gatik:

{\displaystyle {\frac {x}{y}}}, {\displaystyle {\sqrt {p}}}-ren hondar bat da: {\displaystyle j-1}. hondarra: {\displaystyle {\frac {x}{y}}=R\_{j-1}}, {\displaystyle j\in \mathbb {N} }.

{\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 {\displaystyle h+1=N\_{j}} adieraziz:

{\displaystyle j-1=Tk+h:h\in [0,T-1]\Leftrightarrow j=Tk+h+1:h+1=N\_{j}\in [1,T]}{\displaystyle \Leftrightarrow j=Tk+N\_{j}:N\_{j}\in [1,T]}

{\displaystyle (-1)^{j}v\_{j}=(-1)^{Tk+N\_{j}}v\_{Tk+N\_{j}}=(-1)^{N\_{j}}v\_{N\_{j}}=m}, ({\displaystyle T} bikoitia eta {\displaystyle v\_{n}}- {\displaystyle T} periodikoa){\displaystyle \Rightarrow N\_{j}\in \xi \_{m}\neq \emptyset }.

Eta: {\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: {\displaystyle \Leftarrow \xi \_{m}\neq \emptyset } bada, {\displaystyle \exists N\in [1,T]:(-1)^{N}v\_{N}=m}.

Orduan: {\displaystyle k\in \mathbb {N} \cup \{0\}}, hartuaz:

{\displaystyle p\_{N+kT-1}^{2}-pq\_{N+kT-1}^{2}=(-1)^{N+kT}v\_{N+kT}=(-1)^{N}v\_{N}=m}.

Ondoriz {\displaystyle x=p\_{N+kT-1},y=q\_{N+kT-1}}, {\displaystyle x^{2}-py^{2}=m}, ekuazioaren ebazpena da. 

2: {\displaystyle T} bakoitia bada.

{\displaystyle \Rightarrow (x,y),x^{2}-py^{2}=m}-ren ebazpen ez neutroa, {\displaystyle zkh(x,y)=1}. 

Aurreko atalean bezala argudiatuz:

{\displaystyle {\frac {x}{y}}}, {\displaystyle {\sqrt {p}}}-ren hondar bat da: {\displaystyle j-1}. hondarra: {\displaystyle {\frac {x}{y}}=R\_{j-1}}, {\displaystyle j\in \mathbb {N} }.

{\displaystyle m=p\_{j-1}^{2}-pq\_{j-1}^{2}=(-1)^{j}v\_{j}}.

Zatidura euklidearra eginez.

{\displaystyle j-1=kT+h:h\in [0,T-1]\Leftrightarrow j=kT+N\_{j}:N\_{j}\in [1,T]}, zeinetan {\displaystyle N\_{j}=h+1}.

Bi kasu eman daitezke:

2.1: {\displaystyle k} bikoitia bada {\displaystyle kT} bikoitia da. {\displaystyle k=2r:r\in \mathbb {N} \cup \{0\}}adieraziz:

{\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}}.

{\displaystyle (-1)^{j}=(-1)^{N\_{j}};v\_{j}=v\_{N\_{j}}}. {\displaystyle k} bikoitia eta {\displaystyle v\_{n}}, {\displaystyle T}-periodikoa. {\displaystyle N\_{j}\in \xi \_{m}\neq \emptyset }

{\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. {\displaystyle k} bakoitia bada {\displaystyle kT} bakoitia da.( {\displaystyle T} bakoitia denez). {\displaystyle k=2r+1:r\in \mathbb {N} \cup \{0\}}.

{\displaystyle m=x^{2}-py^{2}=p\_{j-1}^{2}-pq\_{j-1}^{2}=(-1)^{j}v\_{j}}

{\displaystyle (-1)^{j}v\_{j}=(-1)^{N\_{j}+(2r+1)T}v\_{N\_{j}+(2r+1)T}=-(-1)^{N\_{j}}v\_{N\_{j}}}

{\displaystyle \Rightarrow N\_{j}\in \xi \_{-m}\neq \emptyset }, {\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

{\displaystyle \Leftarrow \xi \_{m}\neq \emptyset } bada {\displaystyle \exists N\in [1,T]:(-1)^{N}v\_{N}=m}, orduan: {\displaystyle k\in \mathbb {N} \cup \{0\}}

{\displaystyle p\_{N+2kT-1}^{2}-pq\_{N+2kT-1}^{2}=(-1)^{N+2kT}v\_{N+2kT}=(-1)^{N}v\_{N}=m}.

{\displaystyle \xi \_{-m}\neq \emptyset } bada {\displaystyle \exists N\in [1,T]:(-1)^{N}v\_{N}=-m}, orduan: {\displaystyle k\in \mathbb {N} \cup \{0\}}

{\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}=}{\displaystyle =(-1)^{N+(2k+1)T}v\_{N+(2k+1)T}=-(-1)^{N}v\_{N}=-(-m)=m}.

3: {\displaystyle {\sqrt {p}}}-ren garapenaren periodoa: {\displaystyle T} , bikoitia ala bakoitia denez: {\displaystyle x^{2}-py^{2}=m} bateragarria da eta teorema honen 1 eta 2 atalengatik infinitu ebazpen ditu.

### **Korolarioa.**

{\displaystyle p\in \mathbb {N} -\mathbb {N} ^{2}}, eta {\displaystyle T}, {\displaystyle {\sqrt {p}}}-ren zatiki jarraien bidezko periodoa. Orduan:

1: {\displaystyle T} bikoitia bada, {\displaystyle x^{2}-py^{2}=1} -en ebazpen minimoa: {\displaystyle x=p\_{T-1},y=q\_{T-1}}da. Zeinetan {\displaystyle {\frac {p\_{T-1}}{q\_{T-1}}}}, {\displaystyle {\sqrt {p}}}-ren {\displaystyle (T-1)}. hondarra den.

2: {\displaystyle T} bakoitia bada, {\displaystyle x^{2}-py^{2}=1} -en ebazpen minimoa: {\displaystyle x=p\_{2T-1},y=q\_{2T-1}}da. Zeinetan {\displaystyle {\frac {p\_{2T-1}}{q\_{2T-1}}}}, {\displaystyle {\sqrt {p}}}-ren {\displaystyle (2T-1)}. hondarra den.

**Froga**: Proposizioa3-ko 3. atalagatik: {\displaystyle T=Min\{k\in \mathbb {N} :a\_{k}=2a\_{0}\}}

{\displaystyle \alpha =[{\sqrt {p}}]+{\sqrt {p}}=a\_{0}+{\sqrt {p}}}, periodiko hutsa da, eta {\displaystyle T} periodokoa: {\displaystyle \alpha \_{j}=\alpha \_{j+T}}.

{\displaystyle \alpha \_{T}=\alpha \_{0}\Rightarrow {\frac {u\_{T}+{\sqrt {p}}}{v\_{T}}}=a\_{0}+{\sqrt {p}}}, honela {\displaystyle v\_{T}=1}.

Honela aurreko teorema aplikatuz.

{\displaystyle T} bikoitia bada, {\displaystyle (-1)^{T}v\_{T}=1\Rightarrow T\in \xi \_{1}\neq \emptyset }.

Eta ebazpen txikiena: {\displaystyle x=p\_{T-1},y=q\_{T-1}\Rightarrow {\frac {x}{y}}=R\_{T-1}}.

{\displaystyle T} bakoitia bada, {\displaystyle (-1)^{T}v\_{T}=-1\Rightarrow T\in \xi \_{-1}\neq \emptyset }.

Eta beraz ebazpen txikiena: {\displaystyle x=p\_{2T-1},y=q\_{2T-1}\Rightarrow {\frac {x}{y}}=R\_{2T-1}}.