Pellen ekuazioaren bateragarritasuna

Pell-en ekuazioa bateragarria da.

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.

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: ${\displaystyle v\neq 0}$.