Pellen ekuazioaren bateragarritasuna

Pell-en ekuazioa bateragarria da.

Pellen ekuazioa: $x^{2}-py^{2}=1$ , bateragarria da, $p$ edozein zenbaki arrunt eta ez karratu izanik. Pell-en ekuazioak beti onartzen du ebazpen neutroa: $x=1,y=0$, 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, $p$ edozein zenbaki arrunt eta ez karratu izanik. Honetarako Dirichlet-en bidea jarraituz: Lema bat, korolario bat eta proposizio bat frogatuz.

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

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

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

Lema

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

Froga.

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

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

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

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

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

$\left\vert x_{r}-x_{s}\right\vert =\left\vert r\alpha -s\alpha -([r\alpha ]-[s\alpha ])\right\vert =$$(r-s)\left\vert \alpha -{\frac {[r\alpha ]-[s\alpha ]}{r-s}}\right\vert <{\frac {1}{n}}$

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

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

Korolarioa (Dirichleten teorema)

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

Froga

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

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

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

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

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

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

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

Absurdua denez ezinezkoa da $\Re$ finitua izatea$\Rightarrow \left\vert \Re \right\vert =\aleph$

Proposizioa

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

Froga

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

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

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

• Ondorengo emaitza frogatuko da hiru pausotan:

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

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

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

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

Bi zenbakien biderketa eginez:

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

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

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

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

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

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

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

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

Multzoen arteko ondorengo berdintza betetzen da:

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

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

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

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

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

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

Bat: Ondorengo emaitza frogatuko da:

$\exists (x_{1},y_{1}),(x_{2},y_{2})\in {\mathcal {A}}_{m}$: $(x_{1},y_{1})\neq (x_{2},y_{2})$, $x_{1}\equiv x_{2}{\pmod {|m|}};y_{1}\equiv y_{2}{\pmod {|m|}}$ .

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

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

$f(x_{1},y_{1})=f(x_{2},y_{2})$ eta $(x_{1},y_{1})\neq (x_{2},y_{2})$. $\exists (x_{1},y_{1}),(x_{2},y_{2})\in {\mathcal {A}}_{m}$ non

$(x_{1}+\left\vert m\right\vert Z,y_{1}+\left\vert m\right\vert Z)=(x_{2}+\left\vert m\right\vert Z,y_{2}+\left\vert m\right\vert Z)$ , honela $\exists (x_{1},y_{1}),(x_{2},y_{2})\in {\mathcal {A}}_{m}$:

$(x_{1},y_{1})\neq (x_{2},y_{2})$ eta $x_{1}\equiv x_{2}{\pmod {|m|}};y_{1}\equiv y_{2}{\pmod {|m|}}$.

Bi: Ondorengo erlazioa frogatuko da: ${zkh}(x_{1},y_{1})={zkh}(x_{2},y_{2})=d$.

$d/{zkh}(x_{1},y_{1})\Rightarrow d/{zkh}(x_{2},y_{2})$frogatuko da, lehenik. $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$${\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: $d/{zkh}(x_{2},y_{2})\Rightarrow d/{zkh}(x_{1},y_{1})$.

Ondorioz:${zkh}(x_{1},y_{1})={zkh}(x_{2},y_{2})=d$.

Hiru: $(x_{1}x_{2}-py_{1}y_{2},x_{1}y_{2}-x_{2}y_{1})\in {\mathcal {A}}_{m}$frogatuko da.

$(x_{1},y_{1}),(x_{2},y_{2})\in {\mathcal {A}}_{m}\Rightarrow (x_{1}x_{2}-py_{1}y_{2},x_{1}y_{2}-x_{2}y_{1})\in {\mathcal {A}}_{m}$.

$(x_{1}^{2}-py_{1}^{2})(x_{2}^{2}-py_{2}^{2})=(x_{1}x_{2}-py_{1}y_{2})^{2}-p(x_{1}y_{2}-x_{2}y_{1})^{2}=m^{2}$.

Lau: $x_{1}y_{2}-x_{2}y_{1}\equiv 0{\pmod {|m|}}$ eta $x_{1}x_{2}-py_{1}y_{2}\equiv 0{\pmod {|m|}}$frogatuko da.

${\begin{cases}x_{2}-x_{1}\equiv 0{\pmod {\left\vert m\right\vert }}\Rightarrow y_{2}(x_{2}-x_{1})\equiv 0{\pmod {\left\vert m\right\vert }}\\y_{2}-y_{1}\equiv 0{\pmod {\left\vert m\right\vert }}\Rightarrow x_{2}(y_{2}-y_{1})\equiv 0{\pmod {\left\vert m\right\vert }}\end{cases}}$

Kenketa eginez: $x_{1}y_{2}-x_{2}y_{1}\equiv 0{\pmod {|m|}}$

$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}}}$${\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}}}$
$(x_{1}x_{2}-py_{1}y_{2})^{2}\equiv 0{\pmod {m^{2}}}\Rightarrow x_{1}x_{2}-py_{1}y_{2}\equiv 0{\pmod {|m|}}$.

$x_{1}x_{2}-py_{1}y_{2}\equiv 0{\pmod {|m|}}$.

Bost: Pellen ekuazioaren ebazpen bat existitzen dela frogatuko da.

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

${\begin{cases}x_{1}y_{2}-x_{2}y_{1}\equiv 0{\pmod {|m|}}\\x_{1}x_{2}-py_{1}y_{2}\equiv 0{\pmod {|m|}}\end{cases}}\Rightarrow \exists u,v\in \mathbb {Z}$ non ${\begin{cases}x_{1}y_{2}-x_{2}y_{1}=vm\\x_{1}x_{2}-py_{1}y_{2}=um\end{cases}}$

$(x_{1}^{2}-py_{1}^{2})(x_{2}^{2}-py_{2}^{2})=(x_{1}x_{2}-py_{1}y_{2})^{2}-p(x_{1}y_{2}-x_{2}y_{1})^{2}=m^{2}$.

$(x_{1}x_{2}-py_{1}y_{2})^{2}-p(x_{1}y_{2}-x_{2}y_{1})^{2}=m^{2}\Leftrightarrow (um)^{2}-p(vm)^{2}=m^{2}\Leftrightarrow u^{2}-pv^{2}=1$

Ondorioz Pell ekuazioaren ebazpen bat existitzen da: $x=u,y=v$.

Sei: Ebazpena ez dela neutroa frogatuko da: $v\neq 0$.

Absurdura bideratuz, $v=0$ baldin bada: $x_{1}y_{2}=x_{2}y_{1}$.

${zkt}(x_{1},y_{1})={zkt}(x_{2},y_{2})=d$ denez ondorengo zenbakiak sortuko dira:

$x_{1}=dx'_{1},x_{2}=dx'_{2},y_{1}=dy'_{1},y_{2}=dy'_{2}$.

Zenbaki hauen zatitzaile komunetako handiena:

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

Eta: $x_{1}y_{2}=x_{2}y_{1}\Rightarrow x'_{1}y'_{2}=x'_{2}y'_{1}$

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

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

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

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

Ondorioz: $v\neq 0$.