Pell-en ekuazioa bateragarria da.
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**Pellen ekuazioa**: {\displaystyle x^{2}-py^{2}=1} , bateragarria da, {\displaystyle p} edozein zenbaki arrunt eta ez karratu izanik. Pell-en ekuazioak beti onartzen du ebazpen neutroa: {\displaystyle x=1,y=0}, horregatik bateragarria dela diogunean, neutroa ez den ebazpen baten existentziaz mintzo gara.

Peter Gustav Lejeune Dirichlet (1805-1859)
Pell-en ekuazioa bateragarria dela frogatuko da, {\displaystyle p} edozein zenbaki arrunt eta ez karratu izanik. Honetarako Dirichlet-en bidea jarraituz: Lema bat, korolario bat eta proposizio bat frogatuz.

**Oharra:** Notazioetan, {\displaystyle [a,b]} bitartea erabiliko da, {\displaystyle [a,b]\cap \mathbb {Z} }multzoa adierazteko, eta {\displaystyle [x]}, berriz, {\displaystyle x}-ren zati osoa adierazteko.

{\displaystyle \aleph }, aleph zero sinboloak infinitu kontagarria adierazten du, eta {\displaystyle \left\vert A\right\vert }-k, {\displaystyle A} multzoaren elementu kopurua. 

Zenbaki arrazionalak: {\displaystyle \mathbb {Q} =\{{\frac {p}{q}}:p\in \mathbb {Z} ,q\in \mathbb {N} \}}

### Lema

{\displaystyle \alpha \in \mathbb {R} } emanik, {\displaystyle \forall n\in \mathbb {N} }-rentzat {\displaystyle \exists {\frac {p}{q}}\in \mathbb {Q} } non {\displaystyle \left\vert \alpha -{\frac {p}{q}}\right\vert <{\frac {1}{q\cdot n}}} eta {\displaystyle q\in [1,n]}.

#### Froga.

{\displaystyle {\frac {p}{q}}} zenbaki arrazionala, {\displaystyle {p}\in \mathbb {Z} } eta {\displaystyle {q}\in \mathbb {N} } suposatzen da orokortasuna galdu gabe.

{\displaystyle {n}\in \mathbb {N} } emanik, ondorengo segida eraikiko da: {\displaystyle x\_{j}=j\alpha -[j\alpha ]} non {\displaystyle j\in [0,n]}.

{\displaystyle x\_{j}=j\alpha -[j\alpha ]\in [0,1{\bigr )}=\bigcup \_{k=0}^{n-1}[{\frac {k}{n}},{\frac {k+1}{n}}{\bigr )}}, {\displaystyle \forall j\in [0,n]}-rentzat.

Honela {\displaystyle n+1} zenbaki , {\displaystyle n} bitarte disjuntutan banatu dira, eta usategi printzipioa erabiliz, existitzen da bitarte bat, gutsienez bi zenbaki bere baitan dituena. 

{\displaystyle \exists {r},{s}\in [0,n]}, eta {\displaystyle {r}<{s}} non {\displaystyle \left\vert x\_{r}-x\_{s}\right\vert <{\frac {1}{n}}}

{\displaystyle \left\vert x\_{r}-x\_{s}\right\vert =\left\vert r\alpha -s\alpha -([r\alpha ]-[s\alpha ])\right\vert =}{\displaystyle (r-s)\left\vert \alpha -{\frac {[r\alpha ]-[s\alpha ]}{r-s}}\right\vert <{\frac {1}{n}}}

{\displaystyle {r},{s}\in [0,n]}, eta {\displaystyle {r}<{s}} {\displaystyle \Rightarrow 1\leq r-s\leq n}. Honela {\displaystyle q=r-s} eta {\displaystyle p=[r\alpha ]-[s\alpha ]} aukeratuz.

{\displaystyle \exists {\frac {p}{q}}\in Q} non {\displaystyle \left\vert \alpha -{\frac {p}{q}}\right\vert <{\frac {1}{q\cdot n}}} eta {\displaystyle q\in [1,n]}.

### Korolarioa (Dirichleten teorema)

{\displaystyle \alpha \in \mathbb {R} -\mathbb {Q} } eta {\displaystyle \Re =\left\{{\frac {p}{q}}\in \mathbb {Q} :\left\vert \alpha -{\frac {p}{q}}\right\vert <{\frac {1}{q^{2}}}\right\}}{\displaystyle \Rightarrow \left\vert \Re \right\vert =\aleph }

#### Froga

{\displaystyle p=\left[\alpha \right],q=1}, hartuaz {\displaystyle \left\vert \alpha -{\frac {[\alpha ]}{1}}\right\vert <{\frac {1}{1^{2}}}\Rightarrow [\alpha ]\in \Re }. Ondorioz {\displaystyle \Re \neq \emptyset }.

Absurdura bideratuz suposa bedi, {\displaystyle \left\vert \Re \right\vert <\aleph } dela (finitua).

{\displaystyle \left\vert \Re \right\vert <\aleph \Rightarrow \exists \epsilon =Min\left\{\left\vert \alpha -r\right\vert :r\in \Re \right\}}

Zenbaki arrazionalen arkimedesen ezaugarriagatik:{\displaystyle \exists n\in \mathbb {N} :{\frac {1}{n}}<\epsilon } .

{\displaystyle \alpha } eta {\displaystyle n} zenbakiei, aurreko Lema aplikatuz: {\displaystyle \exists {\frac {p}{q}}\in \mathbb {Q} :\left\vert \alpha -{\frac {p}{q}}\right\vert <{\frac {1}{q\cdot n}};q\in [1,n]}

Ondorioz, {\displaystyle \left\vert \alpha -{\frac {p}{q}}\right\vert <{\frac {1}{q\cdot n}}\leq {\frac {1}{n}}<\epsilon \Rightarrow {\frac {p}{q}}\not \in \Re }

Eta bestalde {\displaystyle \left\vert \alpha -{\frac {p}{q}}\right\vert <{\frac {1}{q\cdot n}}\leq {\frac {1}{q^{2}}}\Rightarrow {\frac {p}{q}}\in \Re }

Absurdua denez ezinezkoa da {\displaystyle \Re } finitua izatea{\displaystyle \Rightarrow \left\vert \Re \right\vert =\aleph }

#### Proposizioa

{\displaystyle p} zenbaki arrunt eta ez karratua bada, Pell-en ekuazioak: {\displaystyle x^{2}-py^{2}=1} , badu ebazpen ez neutro bat.

#### Froga

{\displaystyle p} zenbaki arrunt eta ez karratua bada, {\displaystyle {\sqrt {p}}} ez da zenbaki arrazionala: {\displaystyle {\sqrt {p}}\in \mathbb {R} -\mathbb {Q} }.

{\displaystyle \alpha ={\sqrt {p}}\in \mathbb {R} -\mathbb {Q} } zenbakiari aurreko korolarioa aplikatuz, {\displaystyle \left\vert \Re \right\vert =\aleph } , zeinetan: 

{\displaystyle \Re =\left\{{\frac {x}{y}}\in \mathbb {Q} :\left\vert {\sqrt {p}}-{\frac {x}{y}}\right\vert <{\frac {1}{y^{2}}}\right\}}. 

* Ondorengo emaitza frogatuko da hiru pausotan:

{\displaystyle \exists m\in Z;\left\vert m\right\vert <1+2{\sqrt {p}}} non {\displaystyle \left\vert {\mathcal {A\_{m}}}\right\vert =\aleph }.

Zeinetan {\displaystyle {\mathcal {A}}\_{m}=\{(x,y)\in N\times N:\left\vert x^{2}-py^{2}\right\vert =m\}} multzoa den.

**Bat**: {\displaystyle \left\vert \Re \_{1}\right\vert =\aleph }, forgatuko da, zeinetan {\displaystyle \Re \_{1}=\left\{{\frac {x}{y}}\in \mathbb {Q} :\left\vert x^{2}-y^{2}p\right\vert <1+2{\sqrt {p}}\right\}}. Ondorengo desberdintzak betetzen dituzte {\displaystyle \Re } multzoko zatikiek: {\displaystyle {\frac {x}{y}}\in \Re } emanik: {\displaystyle \left\vert {\sqrt {p}}-{\frac {x}{y}}\right\vert <{\frac {1}{y^{2}}}\Leftrightarrow \left\vert y{\sqrt {p}}-x\right\vert <{\frac {1}{y}}}, eta desberdintza triangeluarra erabiliz:

{\displaystyle \left\vert y{\sqrt {p}}+x\right\vert =\left\vert 2y{\sqrt {p}}+x-y{\sqrt {p}}\right\vert \leq } {\displaystyle 2y{\sqrt {p}}+\left\vert x-y{\sqrt {p}}\right\vert <{\frac {1}{y}}+2y{\sqrt {p}}} .

Bi zenbakien biderketa eginez:

{\displaystyle \left\vert x-y{\sqrt {p}}\right\vert \cdot \left\vert x+y{\sqrt {p}}\right\vert =\left\vert x^{2}-py^{2}\right\vert } {\displaystyle <{\frac {1}{y}}\cdot ({\frac {1}{y}}+2y{\sqrt {p}})={\frac {1}{y^{2}}}+2{\sqrt {p}}\leq 1+2{\sqrt {p}}}. Honela: {\displaystyle {\frac {x}{y}}\in \Re \Rightarrow {\frac {x}{y}}\in \Re \_{1}}. Eta emaitza frogatzen da: {\displaystyle \Re \subset \Re \_{1};|\Re |=\aleph \Rightarrow |\Re \_{1}|=\aleph }.

**Bi** {\displaystyle \left\vert {\mathcal {A}}\right\vert =\aleph } zeinetan {\displaystyle {\mathcal {A}}=\{(x,y)\in \mathbb {N} \times \mathbb {N} :\left\vert x^{2}-py^{2}\right\vert <1+2{\sqrt {p}}\}}. 

Ondorengo aplikazioa sortuko da: {\displaystyle f:{\mathcal {A}}\rightarrow \Re \_{1}}, zeinetan {\displaystyle f(x,y)={\frac {x}{y}}}.

Erraz frogatzen da ondo definitutako aplikazioa dela, eta supraiektiboa dela.

{\displaystyle f} supraiektiboa {\displaystyle \Rightarrow \left\vert \Re \_{1}\right\vert \leq \left\vert {\mathcal {A}}\right\vert }.

{\displaystyle \Re \_{1}}infinitua denez, {\displaystyle {\mathcal {A}}} ere infinitua da: {\displaystyle \left\vert \Re \_{1}\right\vert =\aleph \Rightarrow |{\mathcal {A}}|=\aleph }.

**Hiru**: {\displaystyle \exists m\in Z;\left\vert m\right\vert <1+2{\sqrt {p}}} non {\displaystyle \left\vert {\mathcal {A}}\_{m}\right\vert =\aleph }.

Zeinetan {\displaystyle {\mathcal {A}}\_{m}=\{(x,y)\in \mathbb {N} \times \mathbb {N} :\left\vert x^{2}-py^{2}\right\vert =m\}}multzoa den.

Multzoen arteko ondorengo berdintza betetzen da:

{\displaystyle {\mathcal {A}}=\bigcup \_{|m|<1+2{\sqrt {p}}}{\mathcal {A}}\_{m}}.

Absurdura bideratuz {\displaystyle \forall m\in Z;\left\vert m\right\vert <1+2{\sqrt {p}}\Rightarrow \left\vert {\mathcal {A}}\_{m}\right\vert <\aleph } suposatzen bada.

{\displaystyle |{\mathcal {A}}|=|\bigcup \_{|m|<1+2{\sqrt {p}}}{\mathcal {A}}\_{m}|\leq \sum \_{|m|<1+2{\sqrt {p}}}|{\mathcal {A}}\_{m}|<\aleph }. Multzo finituen batura finitua finitua izateagatik.

Honela, {\displaystyle \left\vert {\mathcal {A}}\right\vert <\aleph } ondorioztatu da, zeinak {\displaystyle \left\vert {\mathcal {A}}\right\vert =\aleph }, ukatzen duen: absurdua.

Existitzen da beraz {\displaystyle {\mathcal {A}}\_{m}}multzoren bat infinitu elementu dituena.

* Behin {\displaystyle m\in Z} aukeratu dugularik ( {\displaystyle \left\vert m\right\vert <1+2{\sqrt {p}}}) eta {\displaystyle \left\vert {\mathcal {A}}\_{m}\right\vert =\aleph }, {\displaystyle {\mathcal {A}}\_{m}} multzoan Pellen ekuazioa betetzen duen ebazpen ez neutro bat existitzen dela frogatuko da. Ondorengo atalak frogatuz:

**Bat**: Ondorengo emaitza frogatuko da:

{\displaystyle \exists (x\_{1},y\_{1}),(x\_{2},y\_{2})\in {\mathcal {A}}\_{m}}: **{\displaystyle (x\_{1},y\_{1})\neq (x\_{2},y\_{2})}, {\displaystyle x\_{1}\equiv x\_{2}{\pmod {|m|}};y\_{1}\equiv y\_{2}{\pmod {|m|}}} .**

{\displaystyle f:{\mathcal {A}}\_{m}\rightarrow Z\_{|m|}\times Z\_{|m|}}, aplikazioa eraikiko da zeinetan {\displaystyle f(x,y)=(x+\left\vert m\right\vert Z,y+\left\vert m\right\vert Z)}, {\displaystyle x+\left\vert m\right\vert Z\in Z\_{|m|}=\{{{\bar {0}},{\bar {1}},...{\overline {m-1}}}\}} eraztuneko elementuak izanik.

{\displaystyle {\mathcal {A}}\_{m}} Multzoa infinitua izateagatik eta {\displaystyle Z\_{|m|}\times Z\_{|m|}}, multzoa berriz finitua, irudi berdineko bi elementu desberdin existitzen direla ondoriozta daiteke ( zentzu zorrotzean infinitu ere exititu arren). {\displaystyle \left\vert {\mathcal {A}}\_{m}\right\vert =\aleph }, eta {\displaystyle \left\vert Z\_{|m|}\times Z\_{|m|}\right\vert =m^{2}\Rightarrow } {\displaystyle \exists (x\_{1},y\_{1}),(x\_{2},y\_{2})\in {\mathcal {A}}\_{m}}.

{\displaystyle f(x\_{1},y\_{1})=f(x\_{2},y\_{2})} eta {\displaystyle (x\_{1},y\_{1})\neq (x\_{2},y\_{2})}. {\displaystyle \exists (x\_{1},y\_{1}),(x\_{2},y\_{2})\in {\mathcal {A}}\_{m}} non

{\displaystyle (x\_{1}+\left\vert m\right\vert Z,y\_{1}+\left\vert m\right\vert Z)=(x\_{2}+\left\vert m\right\vert Z,y\_{2}+\left\vert m\right\vert Z)} , honela {\displaystyle \exists (x\_{1},y\_{1}),(x\_{2},y\_{2})\in {\mathcal {A}}\_{m}}:

{\displaystyle (x\_{1},y\_{1})\neq (x\_{2},y\_{2})} eta {\displaystyle x\_{1}\equiv x\_{2}{\pmod {|m|}};y\_{1}\equiv y\_{2}{\pmod {|m|}}}.

**Bi**: Ondorengo erlazioa frogatuko da: **{\displaystyle {zkh}(x\_{1},y\_{1})={zkh}(x\_{2},y\_{2})=d}**.

{\displaystyle d/{zkh}(x\_{1},y\_{1})\Rightarrow d/{zkh}(x\_{2},y\_{2})}frogatuko da, lehenik. {\displaystyle d/{zkh}(x\_{1},y\_{1})\Rightarrow d/x\_{1};d/y\_{1}\Rightarrow {\begin{cases}d^{2}/x\_{1}^{2};d^{2}/y\_{1}^{2}\\x\_{1}^{2}-py\_{1}^{2}=m\end{cases}}\Rightarrow d^{2}/m}{\displaystyle {\begin{cases}{\begin{cases}d/x1;d/m\\x\_{1}\equiv x\_{2}{\pmod {|m|}}\end{cases}}\Rightarrow d/x\_{2}\\{\begin{cases}d/y\_{1};d/m\\y\_{1}\equiv y\_{2}{\pmod {|m|}}\end{cases}}\Rightarrow d/y\_{2}\end{cases}}\Rightarrow d/{zkt}(x\_{2},y\_{2})}. 

Eta modu berean argudiatzen da: {\displaystyle d/{zkh}(x\_{2},y\_{2})\Rightarrow d/{zkh}(x\_{1},y\_{1})}.

Ondorioz:**{\displaystyle {zkh}(x\_{1},y\_{1})={zkh}(x\_{2},y\_{2})=d}**.

**Hiru**: {\displaystyle (x\_{1}x\_{2}-py\_{1}y\_{2},x\_{1}y\_{2}-x\_{2}y\_{1})\in {\mathcal {A}}\_{m}}frogatuko da.

{\displaystyle (x\_{1},y\_{1}),(x\_{2},y\_{2})\in {\mathcal {A}}\_{m}\Rightarrow (x\_{1}x\_{2}-py\_{1}y\_{2},x\_{1}y\_{2}-x\_{2}y\_{1})\in {\mathcal {A}}\_{m}}.

{\displaystyle (x\_{1}^{2}-py\_{1}^{2})(x\_{2}^{2}-py\_{2}^{2})=(x\_{1}x\_{2}-py\_{1}y\_{2})^{2}-p(x\_{1}y\_{2}-x\_{2}y\_{1})^{2}=m^{2}}.

**Lau**: {\displaystyle x\_{1}y\_{2}-x\_{2}y\_{1}\equiv 0{\pmod {|m|}}} eta {\displaystyle x\_{1}x\_{2}-py\_{1}y\_{2}\equiv 0{\pmod {|m|}}}frogatuko da.

{\displaystyle {\begin{cases}x\_{2}-x\_{1}\equiv 0{\pmod {\left\vert m\right\vert }}\Rightarrow y\_{2}(x\_{2}-x\_{1})\equiv 0{\pmod {\left\vert m\right\vert }}\\y\_{2}-y\_{1}\equiv 0{\pmod {\left\vert m\right\vert }}\Rightarrow x\_{2}(y\_{2}-y\_{1})\equiv 0{\pmod {\left\vert m\right\vert }}\end{cases}}}

Kenketa eginez: {\displaystyle x\_{1}y\_{2}-x\_{2}y\_{1}\equiv 0{\pmod {|m|}}}

{\displaystyle x\_{1}y\_{2}-x\_{2}y\_{1}\equiv 0{\pmod {|m|}}\Rightarrow (x\_{1}y\_{2}-x\_{2}y\_{1})^{2}\equiv 0{\pmod {m^{2}}}}{\displaystyle {\begin{cases}(x\_{1}x\_{2}-py\_{1}y\_{2})^{2}-p(x\_{1}y\_{2}-x\_{2}y\_{1})^{2}=m^{2}\\(x\_{1}y\_{2}-x\_{2}y\_{1})^{2}\equiv 0{\pmod {m^{2}}}\end{cases}}\Rightarrow (x\_{1}x\_{2}-py\_{1}y\_{2})^{2}\equiv 0{\pmod {m^{2}}}}  
{\displaystyle (x\_{1}x\_{2}-py\_{1}y\_{2})^{2}\equiv 0{\pmod {m^{2}}}\Rightarrow x\_{1}x\_{2}-py\_{1}y\_{2}\equiv 0{\pmod {|m|}}}.

{\displaystyle x\_{1}x\_{2}-py\_{1}y\_{2}\equiv 0{\pmod {|m|}}}.

**Bost**: Pellen ekuazioaren ebazpen bat existitzen dela frogatuko da. 

**{\displaystyle \exists (u,v)\in \mathbb {Z} \times \mathbb {Z} :u^{2}-pv^{2}=1}**

{\displaystyle {\begin{cases}x\_{1}y\_{2}-x\_{2}y\_{1}\equiv 0{\pmod {|m|}}\\x\_{1}x\_{2}-py\_{1}y\_{2}\equiv 0{\pmod {|m|}}\end{cases}}\Rightarrow \exists u,v\in \mathbb {Z} } non {\displaystyle {\begin{cases}x\_{1}y\_{2}-x\_{2}y\_{1}=vm\\x\_{1}x\_{2}-py\_{1}y\_{2}=um\end{cases}}}

{\displaystyle (x\_{1}^{2}-py\_{1}^{2})(x\_{2}^{2}-py\_{2}^{2})=(x\_{1}x\_{2}-py\_{1}y\_{2})^{2}-p(x\_{1}y\_{2}-x\_{2}y\_{1})^{2}=m^{2}}.

{\displaystyle (x\_{1}x\_{2}-py\_{1}y\_{2})^{2}-p(x\_{1}y\_{2}-x\_{2}y\_{1})^{2}=m^{2}\Leftrightarrow (um)^{2}-p(vm)^{2}=m^{2}\Leftrightarrow u^{2}-pv^{2}=1}

Ondorioz Pell ekuazioaren ebazpen bat existitzen da: {\displaystyle x=u,y=v}.

**Sei:** Ebazpena ez dela neutroa frogatuko da: **{\displaystyle v\neq 0}**.

Absurdura bideratuz, {\displaystyle v=0} baldin bada: {\displaystyle x\_{1}y\_{2}=x\_{2}y\_{1}}.

{\displaystyle {zkt}(x\_{1},y\_{1})={zkt}(x\_{2},y\_{2})=d} denez ondorengo zenbakiak sortuko dira:

{\displaystyle x\_{1}=dx'\_{1},x\_{2}=dx'\_{2},y\_{1}=dy'\_{1},y\_{2}=dy'\_{2}}.

Zenbaki hauen zatitzaile komunetako handiena: 

{\displaystyle {zkt}(x\_{1},y\_{1})={zkt}(x\_{2},y\_{2})=d\Rightarrow {zkt}(x'\_{1},y'\_{1})={zkt}(x'\_{2},y'\_{2})=1}.

Eta: {\displaystyle x\_{1}y\_{2}=x\_{2}y\_{1}\Rightarrow x'\_{1}y'\_{2}=x'\_{2}y'\_{1}}

{\displaystyle {\begin{cases}{\begin{cases}x'\_{1}y'\_{2}=x'\_{2}y'\_{1}\\{zkt}(x'\_{1},y'\_{1})=1\end{cases}}\Rightarrow {\frac {y'\_{2}}{y'\_{1}}}={\frac {x'\_{2}}{x'\_{1}}}\in \mathbb {Z} \\{\begin{cases}x'\_{1}y'\_{2}=x'\_{2}y'\_{1}\\{zkt}(x'\_{2},y'\_{2})=1\end{cases}}\Rightarrow {\frac {x'\_{1}}{x'\_{2}}}={\frac {y'\_{1}}{y'\_{2}}}\in \mathbb {Z} \end{cases}}}. 

Zenbaki oso bat bere alderantzizkoaren berdina bada, zenbaki oso hori: 1 edo -1 da.

{\displaystyle {\frac {x'\_{2}}{x'\_{1}}}=-1\Rightarrow x\_{2}=-x\_{1}} bada, bietako bat negatiboa da, eta aukeraketa {\displaystyle x+\left\vert m\right\vert Z\in Z\_{|m|}=\{{{\bar {0}},{\bar {1}},...{\overline {m-1}}}\}} multzotik egin da ezinezkoa.

{\displaystyle {\frac {x'\_{2}}{x'\_{1}}}=1}, bada {\displaystyle (x'\_{1},y'\_{1})=(x'\_{2},y'\_{2})\Rightarrow (x\_{1},y\_{1})=(x\_{2},y\_{2})}. Zeinak osagaien aukeraketa ukatzen duen, ezinezkoa.

Ondorioz: **{\displaystyle v\neq 0}**.