Pell-en ekuazioa bateragarria da. Pellen ekuazioa : x 2 − p y 2 = 1 {\displaystyle x^{2}-py^{2}=1} , bateragarria da, p {\displaystyle p} edozein zenbaki arrunt eta ez karratu izanik. Pell-en ekuazioak beti onartzen du ebazpen neutroa: x = 1 , y = 0 {\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, p {\displaystyle 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 ] {\displaystyle [a,b]} bitartea erabiliko da, [ a , b ] ∩ Z {\displaystyle [a,b]\cap \mathbb {Z} } multzoa adierazteko, eta [ x ] {\displaystyle [x]} , berriz, x {\displaystyle x} -ren zati osoa adierazteko.
ℵ {\displaystyle \aleph } , aleph zero sinboloak infinitu kontagarria adierazten du, eta | A | {\displaystyle \left\vert A\right\vert } -k, A {\displaystyle A} multzoaren elementu kopurua.
Zenbaki arrazionalak: Q = { p q : p ∈ Z , q ∈ N } {\displaystyle \mathbb {Q} =\{{\frac {p}{q}}:p\in \mathbb {Z} ,q\in \mathbb {N} \}}
Lema α ∈ R {\displaystyle \alpha \in \mathbb {R} } emanik, ∀ n ∈ N {\displaystyle \forall n\in \mathbb {N} } -rentzat ∃ p q ∈ Q {\displaystyle \exists {\frac {p}{q}}\in \mathbb {Q} } non | α − p q | < 1 q ⋅ n {\displaystyle \left\vert \alpha -{\frac {p}{q}}\right\vert <{\frac {1}{q\cdot n}}} eta q ∈ [ 1 , n ] {\displaystyle q\in [1,n]} .
Froga. p q {\displaystyle {\frac {p}{q}}} zenbaki arrazionala, p ∈ Z {\displaystyle {p}\in \mathbb {Z} } eta q ∈ N {\displaystyle {q}\in \mathbb {N} } suposatzen da orokortasuna galdu gabe.
n ∈ N {\displaystyle {n}\in \mathbb {N} } emanik, ondorengo segida eraikiko da: x j = j α − [ j α ] {\displaystyle x_{j}=j\alpha -[j\alpha ]} non j ∈ [ 0 , n ] {\displaystyle j\in [0,n]} .
x j = j α − [ j α ] ∈ [ 0 , 1 ) = ⋃ k = 0 n − 1 [ k n , k + 1 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 )}} , ∀ j ∈ [ 0 , n ] {\displaystyle \forall j\in [0,n]} -rentzat.
Honela n + 1 {\displaystyle n+1} zenbaki , n {\displaystyle n} bitarte disjuntutan banatu dira, eta usategi printzipioa erabiliz, existitzen da bitarte bat, gutsienez bi zenbaki bere baitan dituena.
∃ r , s ∈ [ 0 , n ] {\displaystyle \exists {r},{s}\in [0,n]} , eta r < s {\displaystyle {r}<{s}} non | x r − x s | < 1 n {\displaystyle \left\vert x_{r}-x_{s}\right\vert <{\frac {1}{n}}}
| x r − x s | = | r α − s α − ( [ r α ] − [ s α ] ) | = {\displaystyle \left\vert x_{r}-x_{s}\right\vert =\left\vert r\alpha -s\alpha -([r\alpha ]-[s\alpha ])\right\vert =} ( r − s ) | α − [ r α ] − [ s α ] r − s | < 1 n {\displaystyle (r-s)\left\vert \alpha -{\frac {[r\alpha ]-[s\alpha ]}{r-s}}\right\vert <{\frac {1}{n}}}
r , s ∈ [ 0 , n ] {\displaystyle {r},{s}\in [0,n]} , eta r < s {\displaystyle {r}<{s}} ⇒ 1 ≤ r − s ≤ n {\displaystyle \Rightarrow 1\leq r-s\leq n} . Honela q = r − s {\displaystyle q=r-s} eta p = [ r α ] − [ s α ] {\displaystyle p=[r\alpha ]-[s\alpha ]} aukeratuz.
∃ p q ∈ Q {\displaystyle \exists {\frac {p}{q}}\in Q} non | α − p q | < 1 q ⋅ n {\displaystyle \left\vert \alpha -{\frac {p}{q}}\right\vert <{\frac {1}{q\cdot n}}} eta q ∈ [ 1 , n ] {\displaystyle q\in [1,n]} .
Korolarioa (Dirichleten teorema) α ∈ R − Q {\displaystyle \alpha \in \mathbb {R} -\mathbb {Q} } eta ℜ = { p q ∈ Q : | α − p q | < 1 q 2 } {\displaystyle \Re =\left\{{\frac {p}{q}}\in \mathbb {Q} :\left\vert \alpha -{\frac {p}{q}}\right\vert <{\frac {1}{q^{2}}}\right\}} ⇒ | ℜ | = ℵ {\displaystyle \Rightarrow \left\vert \Re \right\vert =\aleph }
Froga p = [ α ] , q = 1 {\displaystyle p=\left[\alpha \right],q=1} , hartuaz | α − [ α ] 1 | < 1 1 2 ⇒ [ α ] ∈ ℜ {\displaystyle \left\vert \alpha -{\frac {[\alpha ]}{1}}\right\vert <{\frac {1}{1^{2}}}\Rightarrow [\alpha ]\in \Re } . Ondorioz ℜ ≠ ∅ {\displaystyle \Re \neq \emptyset } .
Absurdura bideratuz suposa bedi, | ℜ | < ℵ {\displaystyle \left\vert \Re \right\vert <\aleph } dela (finitua).
| ℜ | < ℵ ⇒ ∃ ϵ = M i n { | α − r | : r ∈ ℜ } {\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: ∃ n ∈ N : 1 n < ϵ {\displaystyle \exists n\in \mathbb {N} :{\frac {1}{n}}<\epsilon } .
α {\displaystyle \alpha } eta n {\displaystyle n} zenbakiei, aurreko Lema aplikatuz: ∃ p q ∈ Q : | α − p q | < 1 q ⋅ n ; q ∈ [ 1 , n ] {\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, | α − p q | < 1 q ⋅ n ≤ 1 n < ϵ ⇒ p q ∉ ℜ {\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 | α − p q | < 1 q ⋅ n ≤ 1 q 2 ⇒ p q ∈ ℜ {\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 ℜ {\displaystyle \Re } finitua izatea ⇒ | ℜ | = ℵ {\displaystyle \Rightarrow \left\vert \Re \right\vert =\aleph }
Proposizioa p {\displaystyle p} zenbaki arrunt eta ez karratua bada, Pell-en ekuazioak: x 2 − p y 2 = 1 {\displaystyle x^{2}-py^{2}=1} , badu ebazpen ez neutro bat.
Froga p {\displaystyle p} zenbaki arrunt eta ez karratua bada, p {\displaystyle {\sqrt {p}}} ez da zenbaki arrazionala: p ∈ R − Q {\displaystyle {\sqrt {p}}\in \mathbb {R} -\mathbb {Q} } .
α = p ∈ R − Q {\displaystyle \alpha ={\sqrt {p}}\in \mathbb {R} -\mathbb {Q} } zenbakiari aurreko korolarioa aplikatuz, | ℜ | = ℵ {\displaystyle \left\vert \Re \right\vert =\aleph } , zeinetan:
ℜ = { x y ∈ Q : | p − x y | < 1 y 2 } {\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: ∃ m ∈ Z ; | m | < 1 + 2 p {\displaystyle \exists m\in Z;\left\vert m\right\vert <1+2{\sqrt {p}}} non | A m | = ℵ {\displaystyle \left\vert {\mathcal {A_{m}}}\right\vert =\aleph } .
Zeinetan A m = { ( x , y ) ∈ N × N : | x 2 − p y 2 | = m } {\displaystyle {\mathcal {A}}_{m}=\{(x,y)\in N\times N:\left\vert x^{2}-py^{2}\right\vert =m\}} multzoa den.
Bat : | ℜ 1 | = ℵ {\displaystyle \left\vert \Re _{1}\right\vert =\aleph } , forgatuko da, zeinetan ℜ 1 = { x y ∈ Q : | x 2 − y 2 p | < 1 + 2 p } {\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 ℜ {\displaystyle \Re } multzoko zatikiek: x y ∈ ℜ {\displaystyle {\frac {x}{y}}\in \Re } emanik: | p − x y | < 1 y 2 ⇔ | y p − x | < 1 y {\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:
| y p + x | = | 2 y p + x − y p | ≤ {\displaystyle \left\vert y{\sqrt {p}}+x\right\vert =\left\vert 2y{\sqrt {p}}+x-y{\sqrt {p}}\right\vert \leq } 2 y p + | x − y p | < 1 y + 2 y p {\displaystyle 2y{\sqrt {p}}+\left\vert x-y{\sqrt {p}}\right\vert <{\frac {1}{y}}+2y{\sqrt {p}}} .
Bi zenbakien biderketa eginez:
| x − y p | ⋅ | x + y p | = | x 2 − p y 2 | {\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 } < 1 y ⋅ ( 1 y + 2 y p ) = 1 y 2 + 2 p ≤ 1 + 2 p {\displaystyle <{\frac {1}{y}}\cdot ({\frac {1}{y}}+2y{\sqrt {p}})={\frac {1}{y^{2}}}+2{\sqrt {p}}\leq 1+2{\sqrt {p}}} . Honela: x y ∈ ℜ ⇒ x y ∈ ℜ 1 {\displaystyle {\frac {x}{y}}\in \Re \Rightarrow {\frac {x}{y}}\in \Re _{1}} . Eta emaitza frogatzen da: ℜ ⊂ ℜ 1 ; | ℜ | = ℵ ⇒ | ℜ 1 | = ℵ {\displaystyle \Re \subset \Re _{1};|\Re |=\aleph \Rightarrow |\Re _{1}|=\aleph } .
Bi | A | = ℵ {\displaystyle \left\vert {\mathcal {A}}\right\vert =\aleph } zeinetan A = { ( x , y ) ∈ N × N : | x 2 − p y 2 | < 1 + 2 p } {\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: f : A → ℜ 1 {\displaystyle f:{\mathcal {A}}\rightarrow \Re _{1}} , zeinetan f ( x , y ) = x y {\displaystyle f(x,y)={\frac {x}{y}}} .
Erraz frogatzen da ondo definitutako aplikazioa dela, eta supraiektiboa dela.
f {\displaystyle f} supraiektiboa ⇒ | ℜ 1 | ≤ | A | {\displaystyle \Rightarrow \left\vert \Re _{1}\right\vert \leq \left\vert {\mathcal {A}}\right\vert } .
ℜ 1 {\displaystyle \Re _{1}} infinitua denez, A {\displaystyle {\mathcal {A}}} ere infinitua da: | ℜ 1 | = ℵ ⇒ | A | = ℵ {\displaystyle \left\vert \Re _{1}\right\vert =\aleph \Rightarrow |{\mathcal {A}}|=\aleph } .
Hiru : ∃ m ∈ Z ; | m | < 1 + 2 p {\displaystyle \exists m\in Z;\left\vert m\right\vert <1+2{\sqrt {p}}} non | A m | = ℵ {\displaystyle \left\vert {\mathcal {A}}_{m}\right\vert =\aleph } .
Zeinetan A m = { ( x , y ) ∈ N × N : | x 2 − p y 2 | = m } {\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:
A = ⋃ | m | < 1 + 2 p A m {\displaystyle {\mathcal {A}}=\bigcup _{|m|<1+2{\sqrt {p}}}{\mathcal {A}}_{m}} .
Absurdura bideratuz ∀ m ∈ Z ; | m | < 1 + 2 p ⇒ | A m | < ℵ {\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.
| A | = | ⋃ | m | < 1 + 2 p A m | ≤ ∑ | m | < 1 + 2 p | A m | < ℵ {\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, | A | < ℵ {\displaystyle \left\vert {\mathcal {A}}\right\vert <\aleph } ondorioztatu da, zeinak | A | = ℵ {\displaystyle \left\vert {\mathcal {A}}\right\vert =\aleph } , ukatzen duen: absurdua.
Existitzen da beraz A m {\displaystyle {\mathcal {A}}_{m}} multzoren bat infinitu elementu dituena.
Behin m ∈ Z {\displaystyle m\in Z} aukeratu dugularik ( | m | < 1 + 2 p {\displaystyle \left\vert m\right\vert <1+2{\sqrt {p}}} ) eta | A m | = ℵ {\displaystyle \left\vert {\mathcal {A}}_{m}\right\vert =\aleph } , A m {\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:
∃ ( x 1 , y 1 ) , ( x 2 , y 2 ) ∈ A m {\displaystyle \exists (x_{1},y_{1}),(x_{2},y_{2})\in {\mathcal {A}}_{m}} : ( x 1 , y 1 ) ≠ ( x 2 , y 2 ) {\displaystyle (x_{1},y_{1})\neq (x_{2},y_{2})} , x 1 ≡ x 2 ( mod | m | ) ; y 1 ≡ y 2 ( mod | m | ) {\displaystyle x_{1}\equiv x_{2}{\pmod {|m|}};y_{1}\equiv y_{2}{\pmod {|m|}}} .
f : A m → Z | m | × Z | m | {\displaystyle f:{\mathcal {A}}_{m}\rightarrow Z_{|m|}\times Z_{|m|}} , aplikazioa eraikiko da zeinetan f ( x , y ) = ( x + | m | Z , y + | m | Z ) {\displaystyle f(x,y)=(x+\left\vert m\right\vert Z,y+\left\vert m\right\vert Z)} , x + | m | Z ∈ Z | m | = { 0 ¯ , 1 ¯ , . . . , m − 1 ¯ } {\displaystyle x+\left\vert m\right\vert Z\in Z_{|m|}=\{{{\bar {0}},{\bar {1}},...,{\overline {m-1}}}\}} eraztuneko elementuak izanik.
A m {\displaystyle {\mathcal {A}}_{m}} Multzoa infinitua izateagatik eta Z | m | × Z | m | {\displaystyle Z_{|m|}\times Z_{|m|}} , multzoa berriz finitua, irudi berdineko bi elementu desberdin existitzen direla ondoriozta daiteke ( zentzu zorrotzean infinitu ere exititu arren). | A m | = ℵ {\displaystyle \left\vert {\mathcal {A}}_{m}\right\vert =\aleph } , eta | Z | m | × Z | m | | = m 2 ⇒ {\displaystyle \left\vert Z_{|m|}\times Z_{|m|}\right\vert =m^{2}\Rightarrow } ∃ ( x 1 , y 1 ) , ( x 2 , y 2 ) ∈ A m {\displaystyle \exists (x_{1},y_{1}),(x_{2},y_{2})\in {\mathcal {A}}_{m}} .
f ( x 1 , y 1 ) = f ( x 2 , y 2 ) {\displaystyle f(x_{1},y_{1})=f(x_{2},y_{2})} eta ( x 1 , y 1 ) ≠ ( x 2 , y 2 ) {\displaystyle (x_{1},y_{1})\neq (x_{2},y_{2})} . ∃ ( x 1 , y 1 ) , ( x 2 , y 2 ) ∈ A m {\displaystyle \exists (x_{1},y_{1}),(x_{2},y_{2})\in {\mathcal {A}}_{m}} non
( x 1 + | m | Z , y 1 + | m | Z ) = ( x 2 + | m | Z , y 2 + | m | Z ) {\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 ∃ ( x 1 , y 1 ) , ( x 2 , y 2 ) ∈ A m {\displaystyle \exists (x_{1},y_{1}),(x_{2},y_{2})\in {\mathcal {A}}_{m}} :
( x 1 , y 1 ) ≠ ( x 2 , y 2 ) {\displaystyle (x_{1},y_{1})\neq (x_{2},y_{2})} eta x 1 ≡ x 2 ( mod | m | ) ; y 1 ≡ y 2 ( mod | m | ) {\displaystyle x_{1}\equiv x_{2}{\pmod {|m|}};y_{1}\equiv y_{2}{\pmod {|m|}}} .
Bi : Ondorengo erlazioa frogatuko da: z k h ( x 1 , y 1 ) = z k h ( x 2 , y 2 ) = d {\displaystyle {zkh}(x_{1},y_{1})={zkh}(x_{2},y_{2})=d} .
d / z k h ( x 1 , y 1 ) ⇒ d / z k h ( x 2 , y 2 ) {\displaystyle d/{zkh}(x_{1},y_{1})\Rightarrow d/{zkh}(x_{2},y_{2})} frogatuko da, lehenik. d / z k h ( x 1 , y 1 ) ⇒ d / x 1 ; d / y 1 ⇒ { d 2 / x 1 2 ; d 2 / y 1 2 x 1 2 − p y 1 2 = m ⇒ d 2 / m {\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} { { d / x 1 ; d / m x 1 ≡ x 2 ( mod | m | ) ⇒ d / x 2 { d / y 1 ; d / m y 1 ≡ y 2 ( mod | m | ) ⇒ d / y 2 ⇒ d / z k t ( x 2 , y 2 ) {\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: d / z k h ( x 2 , y 2 ) ⇒ d / z k h ( x 1 , y 1 ) {\displaystyle d/{zkh}(x_{2},y_{2})\Rightarrow d/{zkh}(x_{1},y_{1})} .
Ondorioz: z k h ( x 1 , y 1 ) = z k h ( x 2 , y 2 ) = d {\displaystyle {zkh}(x_{1},y_{1})={zkh}(x_{2},y_{2})=d} .
Hiru : ( x 1 x 2 − p y 1 y 2 , x 1 y 2 − x 2 y 1 ) ∈ A m {\displaystyle (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 ) ∈ A m ⇒ ( x 1 x 2 − p y 1 y 2 , x 1 y 2 − x 2 y 1 ) ∈ A m {\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}} .
( x 1 2 − p y 1 2 ) ( x 2 2 − p y 2 2 ) = ( x 1 x 2 − p y 1 y 2 ) 2 − p ( x 1 y 2 − x 2 y 1 ) 2 = m 2 {\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 : x 1 y 2 − x 2 y 1 ≡ 0 ( mod | m | ) {\displaystyle x_{1}y_{2}-x_{2}y_{1}\equiv 0{\pmod {|m|}}} eta x 1 x 2 − p y 1 y 2 ≡ 0 ( mod | m | ) {\displaystyle x_{1}x_{2}-py_{1}y_{2}\equiv 0{\pmod {|m|}}} frogatuko da.
{ x 2 − x 1 ≡ 0 ( mod | m | ) ⇒ y 2 ( x 2 − x 1 ) ≡ 0 ( mod | m | ) y 2 − y 1 ≡ 0 ( mod | m | ) ⇒ x 2 ( y 2 − y 1 ) ≡ 0 ( mod | m | ) {\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: x 1 y 2 − x 2 y 1 ≡ 0 ( mod | m | ) {\displaystyle x_{1}y_{2}-x_{2}y_{1}\equiv 0{\pmod {|m|}}}
x 1 y 2 − x 2 y 1 ≡ 0 ( mod | m | ) ⇒ ( x 1 y 2 − x 2 y 1 ) 2 ≡ 0 ( mod m 2 ) {\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}}}} { ( x 1 x 2 − p y 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 ≡ 0 ( mod m 2 ) ⇒ ( x 1 x 2 − p y 1 y 2 ) 2 ≡ 0 ( mod 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}}}} ( x 1 x 2 − p y 1 y 2 ) 2 ≡ 0 ( mod m 2 ) ⇒ x 1 x 2 − p y 1 y 2 ≡ 0 ( mod | m | ) {\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|}}} .
x 1 x 2 − p y 1 y 2 ≡ 0 ( mod | m | ) {\displaystyle x_{1}x_{2}-py_{1}y_{2}\equiv 0{\pmod {|m|}}} .
Bost : Pellen ekuazioaren ebazpen bat existitzen dela frogatuko da.
∃ ( u , v ) ∈ Z × Z : u 2 − p v 2 = 1 {\displaystyle \exists (u,v)\in \mathbb {Z} \times \mathbb {Z} :u^{2}-pv^{2}=1}
{ x 1 y 2 − x 2 y 1 ≡ 0 ( mod | m | ) x 1 x 2 − p y 1 y 2 ≡ 0 ( mod | m | ) ⇒ ∃ u , v ∈ Z {\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 { x 1 y 2 − x 2 y 1 = v m x 1 x 2 − p y 1 y 2 = u m {\displaystyle {\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 − p y 1 2 ) ( x 2 2 − p y 2 2 ) = ( x 1 x 2 − p y 1 y 2 ) 2 − p ( x 1 y 2 − x 2 y 1 ) 2 = m 2 {\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}} .
( x 1 x 2 − p y 1 y 2 ) 2 − p ( x 1 y 2 − x 2 y 1 ) 2 = m 2 ⇔ ( u m ) 2 − p ( v m ) 2 = m 2 ⇔ u 2 − p v 2 = 1 {\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: x = u , y = v {\displaystyle x=u,y=v} .
Sei: Ebazpena ez dela neutroa frogatuko da: v ≠ 0 {\displaystyle v\neq 0} .
Absurdura bideratuz, v = 0 {\displaystyle v=0} baldin bada: x 1 y 2 = x 2 y 1 {\displaystyle x_{1}y_{2}=x_{2}y_{1}} .
z k t ( x 1 , y 1 ) = z k t ( x 2 , y 2 ) = d {\displaystyle {zkt}(x_{1},y_{1})={zkt}(x_{2},y_{2})=d} denez ondorengo zenbakiak sortuko dira:
x 1 = d x 1 ′ , x 2 = d x 2 ′ , y 1 = d y 1 ′ , y 2 = d y 2 ′ {\displaystyle x_{1}=dx'_{1},x_{2}=dx'_{2},y_{1}=dy'_{1},y_{2}=dy'_{2}} .
Zenbaki hauen zatitzaile komunetako handiena:
z k t ( x 1 , y 1 ) = z k t ( x 2 , y 2 ) = d ⇒ z k t ( x 1 ′ , y 1 ′ ) = z k t ( x 2 ′ , y 2 ′ ) = 1 {\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: x 1 y 2 = x 2 y 1 ⇒ x 1 ′ y 2 ′ = x 2 ′ y 1 ′ {\displaystyle x_{1}y_{2}=x_{2}y_{1}\Rightarrow x'_{1}y'_{2}=x'_{2}y'_{1}}
{ { x 1 ′ y 2 ′ = x 2 ′ y 1 ′ z k t ( x 1 ′ , y 1 ′ ) = 1 ⇒ y 2 ′ y 1 ′ = x 2 ′ x 1 ′ ∈ Z { x 1 ′ y 2 ′ = x 2 ′ y 1 ′ z k t ( x 2 ′ , y 2 ′ ) = 1 ⇒ x 1 ′ x 2 ′ = y 1 ′ y 2 ′ ∈ Z {\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.
x 2 ′ x 1 ′ = − 1 ⇒ x 2 = − x 1 {\displaystyle {\frac {x'_{2}}{x'_{1}}}=-1\Rightarrow x_{2}=-x_{1}} bada, bietako bat negatiboa da, eta aukeraketa x + | m | Z ∈ Z | m | = { 0 ¯ , 1 ¯ , . . . , m − 1 ¯ } {\displaystyle x+\left\vert m\right\vert Z\in Z_{|m|}=\{{{\bar {0}},{\bar {1}},...,{\overline {m-1}}}\}} multzotik egin da ezinezkoa.
x 2 ′ x 1 ′ = 1 {\displaystyle {\frac {x'_{2}}{x'_{1}}}=1} , bada ( x 1 ′ , y 1 ′ ) = ( x 2 ′ , y 2 ′ ) ⇒ ( x 1 , y 1 ) = ( x 2 , y 2 ) {\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: v ≠ 0 {\displaystyle v\neq 0} .