Commit inicial, primer cuatrimestre.

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+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass book
+\use_default_options true
+\maintain_unincluded_children false
+\language spanish
+\language_package default
+\inputencoding auto
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
+\font_typewriter "default" "default"
+\font_math "auto" "auto"
+\font_default_family default
+\use_non_tex_fonts false
+\font_sc false
+\font_osf false
+\font_sf_scale 100 100
+\font_tt_scale 100 100
+\use_microtype false
+\use_dash_ligatures true
+\graphics default
+\default_output_format default
+\output_sync 0
+\bibtex_command default
+\index_command default
+\paperfontsize default
+\spacing single
+\use_hyperref false
+\papersize default
+\use_geometry false
+\use_package amsmath 1
+\use_package amssymb 1
+\use_package cancel 1
+\use_package esint 1
+\use_package mathdots 1
+\use_package mathtools 1
+\use_package mhchem 1
+\use_package stackrel 1
+\use_package stmaryrd 1
+\use_package undertilde 1
+\cite_engine basic
+\cite_engine_type default
+\biblio_style plain
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style swiss
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\begin_body
+
+\begin_layout Section
+Aritmética de los enteros
+\end_layout
+
+\begin_layout Standard
+Propiedades de 
+\begin_inset Formula $\mathbb{Z}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Unicidad de los neutros:
+\series default
+ 
+\begin_inset Formula $\exists!0\in\mathbb{Z}:\forall a\in\mathbb{Z},0+a=a$
+\end_inset
+
+; 
+\begin_inset Formula $\exists!1\in\mathbb{Z}:\forall a\in\mathbb{Z},1a=a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Unicidad de los opuestos:
+\series default
+ 
+\begin_inset Formula $\forall a\in\mathbb{Z},\exists!(-a)\in\mathbb{Z}:a+(-a)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Cancelación en sumas:
+\series default
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{Z},(a+b=a+c\implies b=c)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Multiplicación por cero:
+\series default
+ 
+\begin_inset Formula $\forall a\in\mathbb{Z},a0=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Reglas de signos:
+\series default
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{Z},(-(-a)=a\land a(-b)=(-a)b=-(ab)\land(-a)(-b)=ab)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+
+\series bold
+Cancelación en productos:
+\series default
+ 
+\begin_inset Formula $\forall a,b,c\in\mathbb{Z},a\neq0,(ab=ac\implies b=c)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de la división entera:
+\series default
+ 
+\begin_inset Formula $\forall a,b\in\mathbb{Z},\exists!q,r\in\mathbb{Z}:(a=bq+r\land0\leq r<|b|)$
+\end_inset
+
+.
+ Llamamos a 
+\begin_inset Formula $q$
+\end_inset
+
+ el 
+\series bold
+cociente
+\series default
+ de la división y a 
+\begin_inset Formula $r$
+\end_inset
+
+ el 
+\series bold
+resto
+\series default
+\SpecialChar endofsentence
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $a,b>0$
+\end_inset
+
+ y 
+\begin_inset Formula $R=\{x\in\mathbb{Z}|x\geq0\land\exists n\in\mathbb{Z}:x=a-bn\}\subseteq\mathbb{N}$
+\end_inset
+
+.
+ Sabemos que 
+\begin_inset Formula $R\neq\emptyset$
+\end_inset
+
+ porque 
+\begin_inset Formula $a=a-b\cdot0\in R$
+\end_inset
+
+.
+ Por tanto tiene primer elemento 
+\begin_inset Formula $r=a-bq\in R$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $r\geq b$
+\end_inset
+
+ entonces 
+\begin_inset Formula $0\leq r-b\in R\#$
+\end_inset
+
+, luego 
+\begin_inset Formula $r<b$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a<0$
+\end_inset
+
+ y 
+\begin_inset Formula $b>0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $-a>0$
+\end_inset
+
+ y 
+\begin_inset Formula $-a=bq+r$
+\end_inset
+
+ con 
+\begin_inset Formula $0\leq r<b$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $r=0$
+\end_inset
+
+, 
+\begin_inset Formula $a=b(-q)+0$
+\end_inset
+
+, y si 
+\begin_inset Formula $r\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $a=b(-q)-r=b(-q-1)+(b-r)$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $b<0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $-b>0$
+\end_inset
+
+ y 
+\begin_inset Formula $a=(-b)q+r$
+\end_inset
+
+ con 
+\begin_inset Formula $0\leq r<-b=|b|$
+\end_inset
+
+, luego 
+\begin_inset Formula $a=b(-q)+r$
+\end_inset
+
+ con 
+\begin_inset Formula $0\leq r<|b|$
+\end_inset
+
+.
+ Finalmente, si 
+\begin_inset Formula $a=0$
+\end_inset
+
+ entonces 
+\begin_inset Formula $0=b\cdot0+0$
+\end_inset
+
+.
+ Para la unicidad de 
+\begin_inset Formula $q$
+\end_inset
+
+ y 
+\begin_inset Formula $r$
+\end_inset
+
+, supongamos 
+\begin_inset Formula $a=bq+r=bq'+r'$
+\end_inset
+
+ con 
+\begin_inset Formula $0\leq r,r'<|b|$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $b(q-q')=r-r'$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $|b||q-q'|=|r-r'|$
+\end_inset
+
+, pero como 
+\begin_inset Formula $0\leq r,r'<|b|$
+\end_inset
+
+, entonces 
+\begin_inset Formula $q-q'=0$
+\end_inset
+
+ y 
+\begin_inset Formula $r-r'=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Decimos que 
+\series bold
+
+\begin_inset Formula $b$
+\end_inset
+
+ divide a 
+\begin_inset Formula $a$
+\end_inset
+
+
+\series default
+ o que 
+\series bold
+
+\begin_inset Formula $a$
+\end_inset
+
+ es múltiplo de 
+\begin_inset Formula $b$
+\end_inset
+
+
+\series default
+ (
+\begin_inset Formula $b|a$
+\end_inset
+
+) si 
+\begin_inset Formula $\exists c\in\mathbb{Z}:a=bc$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+, también decimos que 
+\series bold
+
+\begin_inset Formula $b$
+\end_inset
+
+ es divisor de 
+\begin_inset Formula $a$
+\end_inset
+
+
+\series default
+.
+ Para 
+\begin_inset Formula $b\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $b|a$
+\end_inset
+
+ equivale a que la división entera de 
+\begin_inset Formula $a$
+\end_inset
+
+ entre 
+\begin_inset Formula $b$
+\end_inset
+
+ dé resto 0.
+\end_layout
+
+\begin_layout Enumerate
+La divisibilidad es reflexiva y transitiva.
+\end_layout
+
+\begin_layout Enumerate
+No es antisimétrica, pero 
+\begin_inset Formula $a|b\land b|a\implies|a|=|b|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a|b\iff a|-b$
+\end_inset
+
+, con lo que si 
+\begin_inset Formula $b\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $b$
+\end_inset
+
+ y 
+\begin_inset Formula $-b$
+\end_inset
+
+ tienen los mismos divisores.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a|b\iff-a|b$
+\end_inset
+
+, luego 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $-a$
+\end_inset
+
+ tienen los mismos múltiplos.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c|a\land c|b\implies c|ra+sb$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a|b\land c|d\implies ac|bd$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a|b\implies ca|cb$
+\end_inset
+
+.
+ El recíproco es cierto si 
+\begin_inset Formula $c\neq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $b\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $a|b\implies|a|\leq|b|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $a,b\in\mathbb{Z}$
+\end_inset
+
+, su 
+\series bold
+máximo común divisor
+\series default
+ es 
+\begin_inset Formula $\text{mcd}(a,b)=\max\{d\in\mathbb{Z}:d|a\land d|b\}$
+\end_inset
+
+ (excepción: 
+\begin_inset Formula $\text{mcd}(0,0)=0$
+\end_inset
+
+).
+ Este existe porque el conjunto de divisores comunes es no vacío (contiene
+ al 1) y finito, luego tiene máximo.
+ Propiedades:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{mcd}(a,b)=\text{mcd}(a,|b|)=\text{mcd}(|a|,|b|)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{mcd}(a,0)=|a|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{mcd}(a,b)=0\iff a=b=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $a,b\in\mathbb{Z}$
+\end_inset
+
+ con alguno distinto de 0, 
+\begin_inset Formula $\text{mcd}(a,b)=\min\{ra+sb>0|r,s\in\mathbb{Z}\}$
+\end_inset
+
+, y todo divisor común de 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ lo es de 
+\begin_inset Formula $\text{mcd}(a,b)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Dado 
+\begin_inset Formula $\emptyset\neq D=\{ra+sb>0|r,s\in\mathbb{Z}\}\subseteq\mathbb{Z}^{+}$
+\end_inset
+
+, existe 
+\begin_inset Formula $\delta=\min D$
+\end_inset
+
+.
+ Existen entonces 
+\begin_inset Formula $\alpha,\beta\in\mathbb{Z}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $\delta=\alpha a+\beta b$
+\end_inset
+
+.
+ Por el algoritmo de la división, 
+\begin_inset Formula $a=\delta q+r$
+\end_inset
+
+ con 
+\begin_inset Formula $0\leq r<\delta$
+\end_inset
+
+, luego 
+\begin_inset Formula $r=(1-\alpha q)a+(-q\beta)b$
+\end_inset
+
+, luego 
+\begin_inset Formula $r$
+\end_inset
+
+ es combinación lineal y entonces 
+\begin_inset Formula $r\in D$
+\end_inset
+
+ o 
+\begin_inset Formula $r=0$
+\end_inset
+
+.
+ Lo primero es imposible porque 
+\begin_inset Formula $r<\delta=\min D$
+\end_inset
+
+, luego 
+\begin_inset Formula $r=0$
+\end_inset
+
+ y 
+\begin_inset Formula $\delta|a$
+\end_inset
+
+.
+ Análogamente 
+\begin_inset Formula $\delta|b$
+\end_inset
+
+.
+ Que sea máximo, y que todo divisor común de 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ lo sean de 
+\begin_inset Formula $\delta$
+\end_inset
+
+, se desprende de que 
+\begin_inset Formula $c|a\land c|b\implies c|\alpha a+\beta b=\delta$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+De aquí que para todo 
+\begin_inset Formula $a,b\in\mathbb{Z}$
+\end_inset
+
+ existen 
+\begin_inset Formula $r,s\in\mathbb{Z}$
+\end_inset
+
+ tales que 
+\begin_inset Formula $\text{mcd}(a,b)=ra+sb$
+\end_inset
+
+.
+ Una expresión de la forma 
+\begin_inset Formula $d=ra+sb$
+\end_inset
+
+ es una 
+\series bold
+identidad de Bézout
+\series default
+.
+ En particular, si 
+\begin_inset Formula $a=da'$
+\end_inset
+
+ y 
+\begin_inset Formula $b=db'$
+\end_inset
+
+ con 
+\begin_inset Formula $d=\text{mcd}(a,b)$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\text{mcd}(a',b')=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $d=\text{mcd}(a,b)$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $d|a$
+\end_inset
+
+, 
+\begin_inset Formula $d|b$
+\end_inset
+
+, 
+\begin_inset Formula $c|a\land c|b\implies c|d$
+\end_inset
+
+ y 
+\begin_inset Formula $d\geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Las propiedades (1) y (3) son por definición, y la (2) la acabamos de demostrar.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Si 
+\begin_inset Formula $a\neq0$
+\end_inset
+
+ o 
+\begin_inset Formula $b\neq0$
+\end_inset
+
+, 
+\begin_inset Formula $d$
+\end_inset
+
+ es el mayor entero que divide a 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a=b=0$
+\end_inset
+
+, como 
+\begin_inset Formula $0|a,b$
+\end_inset
+
+, entonces 
+\begin_inset Formula $0|d$
+\end_inset
+
+, luego 
+\begin_inset Formula $d=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El máximo común divisor de 
+\begin_inset Formula $a_{1},\dots,a_{n}$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{mcd}(a_{1},\dots,a_{n})=\max\{d\in\mathbb{Z}:\forall i,d|a_{i}\}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $\text{mcd}(a_{1},\dots,a_{n})=\text{mcd}(\text{mcd}(a_{1},a_{2}),a_{3},\dots,a_{n})$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Sea 
+\begin_inset Formula $d:=\text{mcd}(a_{1},\dots,a_{n})$
+\end_inset
+
+, como 
+\begin_inset Formula $d||a_{1},\dots,a_{n}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $d|(f:=\text{mcd}(a_{1},a_{2})),a_{3},\dots,a_{n}|e:=\text{mcd}(\text{mcd}(a_{1},a_{2}),a_{3},\dots,a_{n})$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $d|e$
+\end_inset
+
+.
+ Pero 
+\begin_inset Formula $e|f,a_{3},\dots,a_{n}$
+\end_inset
+
+, luego 
+\begin_inset Formula $e|a_{1},\dots,a_{n}$
+\end_inset
+
+ y 
+\begin_inset Formula $e|d$
+\end_inset
+
+, y como 
+\begin_inset Formula $d,e\geq0$
+\end_inset
+
+, 
+\begin_inset Formula $d=e$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema:
+\series default
+ Dados 
+\begin_inset Formula $a_{1},\dots,a_{n}\in\mathbb{Z}^{*}$
+\end_inset
+
+, 
+\begin_inset Formula $\text{mcd}(a_{1},\dots,a_{n})=\min\left\{ \sum_{i=1}^{n}r_{i}a_{i}>0|r_{i}\in\mathbb{Z}\right\} $
+\end_inset
+
+.
+ Además, 
+\begin_inset Formula $d=\text{mcd}(a_{1},\dots,a_{n})$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $d|a_{1},\dots,a_{n}$
+\end_inset
+
+, 
+\begin_inset Formula $c|a_{1},\dots,a_{n}\implies c|d$
+\end_inset
+
+ y 
+\begin_inset Formula $d\geq0$
+\end_inset
+
+.
+ 
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula $a,b\in\mathbb{Z}$
+\end_inset
+
+ son 
+\series bold
+coprimos
+\series default
+ o 
+\series bold
+primos entre sí
+\series default
+ si 
+\begin_inset Formula $\text{mcd}(a,b)=1$
+\end_inset
+
+, es decir, si 
+\begin_inset Formula $\exists\alpha,\beta\in\mathbb{Z}:\alpha a+\beta b=1$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ son coprimos:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a|bc\implies a|c$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Sea 
+\begin_inset Formula $1=\alpha a+\beta b$
+\end_inset
+
+, multiplicando por 
+\begin_inset Formula $c$
+\end_inset
+
+, 
+\begin_inset Formula $c=\alpha ac+\beta bc$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $a|bc$
+\end_inset
+
+, 
+\begin_inset Formula $c=\alpha ca+\beta na$
+\end_inset
+
+ y 
+\begin_inset Formula $a|c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a|c\land b|c\implies ab|c$
+\end_inset
+
+.
+\begin_inset Formula 
+\begin{multline*}
+\begin{array}{c}
+1=ra+sb\\
+\frac{c}{a},\frac{c}{b}\in\mathbb{Z}
+\end{array}\implies\frac{c}{a}=\frac{c}{a}ra+\frac{c}{a}sb=\frac{c}{b}rb+\frac{c}{a}sb=b\left(\frac{c}{b}r+\frac{c}{a}s\right)\implies\\
+\implies c=ab\left(\frac{c}{b}r+\frac{c}{a}s\right)\implies ab|c
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Se tiene que 
+\begin_inset Formula $\text{mcd}(a,b)=\text{mcd}(a-sb,b)=\text{mcd}(a,b-sa)$
+\end_inset
+
+, y en particular, si 
+\begin_inset Formula $b\neq0$
+\end_inset
+
+ y 
+\begin_inset Formula $a=bq+r$
+\end_inset
+
+ con 
+\begin_inset Formula $0\leq r<b$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\text{mcd}(a,b)=\text{mcd}(b,r)$
+\end_inset
+
+.
+ La aplicación repetida de lo anterior se conoce como 
+\series bold
+algoritmo de Euclides
+\series default
+.
+ También permite obtener identidades de Bézout.
+ Si llamamos 
+\begin_inset Formula $(a,b)=\text{mcd}(a,b)$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\begin{eqnarray*}
+a=bq_{1}+r_{1} & (a,b)=(b,r_{1}) & r_{1}<b\\
+b=r_{1}q_{2}+r_{2} & (b,r_{1})=(r_{1},r_{2}) & r_{2}<r_{1}\\
+ & \vdots\\
+r_{n-2}=r_{n-1}q_{n}+0 & (r_{n-2},r_{n-1})=(r_{n-1},0)=r_{n-1} & 0<r_{n-1}
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Como 
+\begin_inset Formula $b>r_{1}>\dots\geq0$
+\end_inset
+
+, el algoritmo acaba en un número finito de pasos.
+ Además, cada dos pasos del algoritmo, el resto se reduce a la mitad.
+ 
+\series bold
+Demostración:
+\series default
+ Sean 
+\begin_inset Formula $a=bq+r$
+\end_inset
+
+, 
+\begin_inset Formula $b=rq'+s$
+\end_inset
+
+ y 
+\begin_inset Formula $r=sq''+t$
+\end_inset
+
+, si 
+\begin_inset Formula $s\leq\frac{1}{2}r$
+\end_inset
+
+ entonces 
+\begin_inset Formula $t<s\leq\frac{1}{2}r$
+\end_inset
+
+ y hemos terminado, y si 
+\begin_inset Formula $s>\frac{1}{2}r$
+\end_inset
+
+, entonces 
+\begin_inset Formula $q''=1$
+\end_inset
+
+ y 
+\begin_inset Formula $t=r-s<r-\frac{1}{2}r=\frac{1}{2}r$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $a,b\in\mathbb{Z}^{*}$
+\end_inset
+
+, su 
+\series bold
+mínimo común múltiplo
+\series default
+ es 
+\begin_inset Formula $\text{mcm}(a,b)=\min\{m\in\mathbb{Z}^{+}:a|m\land b|m\}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a$
+\end_inset
+
+ o 
+\begin_inset Formula $b$
+\end_inset
+
+ son 0, entonces 
+\begin_inset Formula $\text{mcm}(a,b)=0$
+\end_inset
+
+.
+ Propiedades:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{mcm}(a,b)=\text{mcm}(a,|b|)=\text{mcm}(|a|,|b|)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{mcm}(a,b)=0\iff a=0\lor b=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{mcm}(a,ab)=|ab|$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Teorema:
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\text{mcm}(a,b)\text{mcd}(a,b)=|ab|$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Para 
+\begin_inset Formula $a,b>0$
+\end_inset
+
+, sea 
+\begin_inset Formula $d=\text{mcd}(a,b)$
+\end_inset
+
+ con 
+\begin_inset Formula $a=da'$
+\end_inset
+
+ y 
+\begin_inset Formula $b=db'$
+\end_inset
+
+, sea 
+\begin_inset Formula $m=a'b'd$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $a|m$
+\end_inset
+
+ y 
+\begin_inset Formula $b|m$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $c=\alpha a=\beta b$
+\end_inset
+
+ con 
+\begin_inset Formula $\alpha,\beta\in\mathbb{Z}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\alpha da'=\beta db'$
+\end_inset
+
+, luego 
+\begin_inset Formula $\alpha a'=\beta b'$
+\end_inset
+
+, y como 
+\begin_inset Formula $a'$
+\end_inset
+
+ y 
+\begin_inset Formula $b'$
+\end_inset
+
+ son coprimos, 
+\begin_inset Formula $a'|\beta$
+\end_inset
+
+ y 
+\begin_inset Formula $\beta=\gamma a'$
+\end_inset
+
+ con 
+\begin_inset Formula $\gamma\in\mathbb{Z}$
+\end_inset
+
+.
+ Sustituyendo, 
+\begin_inset Formula $c=\gamma a'b=\gamma a'db'=\gamma m\geq m$
+\end_inset
+
+, luego 
+\begin_inset Formula $m=\text{mcm}(a,b)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a|c\land b|c\implies\text{mcm}(a,b)|c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+El mínimo común múltiplo de 
+\begin_inset Formula $a_{1},\dots,a_{n}$
+\end_inset
+
+ es 
+\begin_inset Formula $\text{mcm}(a_{1},\dots,a_{n})=\min\{m\in\mathbb{Z}^{+}:\forall i,a_{i}|m\}$
+\end_inset
+
+.
+ Así, 
+\begin_inset Formula $m=\text{mcm}(a_{1},\dots,a_{n})$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $a_{1},\dots,a_{n}|m$
+\end_inset
+
+, 
+\begin_inset Formula $a_{1},\dots,a_{n}|c\implies m|c$
+\end_inset
+
+ y 
+\begin_inset Formula $m\geq0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Una ecuación del tipo 
+\begin_inset Formula $ax+by=c$
+\end_inset
+
+ en la que se buscan soluciones enteras es una 
+\series bold
+ecuación diofántica lineal
+\series default
+, en este caso de dos variables.
+ Tiene solución si y sólo si 
+\begin_inset Formula $d=\text{mcd}(a,b)|c$
+\end_inset
+
+, y entonces estas son de la forma
+\begin_inset Formula 
+\[
+\left\{ \begin{array}{ccc}
+x & = & x_{0}+x'\\
+y & = & y_{0}+y'
+\end{array}\right.
+\]
+
+\end_inset
+
+donde 
+\begin_inset Formula $x_{0},y_{0}$
+\end_inset
+
+ es una solución particular y 
+\begin_inset Formula $x',y'$
+\end_inset
+
+ es una solución de la 
+\series bold
+ecuación homogénea asociada
+\series default
+, 
+\begin_inset Formula $ax+by=0$
+\end_inset
+
+.
+ En particular, si 
+\begin_inset Formula $\alpha a+\beta b=d$
+\end_inset
+
+ y 
+\begin_inset Formula $c=c'd$
+\end_inset
+
+, entonces 
+\begin_inset Formula $x_{0}=c'\alpha$
+\end_inset
+
+ e 
+\begin_inset Formula $y_{0}=c'\beta$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\implies]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Sean 
+\begin_inset Formula $x,y\in\mathbb{Z}$
+\end_inset
+
+ con 
+\begin_inset Formula $ax+by=c$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $d|ax+by=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\impliedby]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+Multiplicando la identidad de Bézout, 
+\begin_inset Formula $(c'\alpha)a+(c'\beta)b=c'd=c$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si 
+\begin_inset Formula $d=\text{mcd}(a,b)$
+\end_inset
+
+, 
+\begin_inset Formula $a=a'd$
+\end_inset
+
+ y 
+\begin_inset Formula $b=b'd$
+\end_inset
+
+, las soluciones de 
+\begin_inset Formula $ax+by=0$
+\end_inset
+
+ son
+\begin_inset Formula 
+\[
+\left\{ \begin{array}{ccc}
+x & = & -b't\\
+y & = & a't
+\end{array}\right.
+\]
+
+\end_inset
+
+para cualquier 
+\begin_inset Formula $t\in\mathbb{Z}$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $ax=-by$
+\end_inset
+
+, luego 
+\begin_inset Formula $a'x=-b'y$
+\end_inset
+
+.
+ Como 
+\begin_inset Formula $a'$
+\end_inset
+
+ y 
+\begin_inset Formula $b'$
+\end_inset
+
+ son coprimos y 
+\begin_inset Formula $a'|-b'y$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a'|y$
+\end_inset
+
+, luego existe 
+\begin_inset Formula $t\in\mathbb{Z}$
+\end_inset
+
+ con 
+\begin_inset Formula $y=a't$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $a'x=-b'a't$
+\end_inset
+
+ y 
+\begin_inset Formula $x=-b't$
+\end_inset
+
+.
+ Multiplicando, todos los enteros de esta forma son solución.
+\end_layout
+
+\begin_layout Standard
+Un entero 
+\begin_inset Formula $p\neq1,-1$
+\end_inset
+
+ es 
+\series bold
+primo
+\series default
+ si sus únicos divisores son 1, 
+\begin_inset Formula $-1$
+\end_inset
+
+, 
+\begin_inset Formula $p$
+\end_inset
+
+ y 
+\begin_inset Formula $-p$
+\end_inset
+
+.
+ Así,
+\begin_inset Formula 
+\[
+p\text{ es primo}\iff(p|ab\implies p|a\lor p|b)\iff(p|a_{1}\cdots a_{n}\implies\exists i:p|a_{i})
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Si 
+\begin_inset Formula $p|a$
+\end_inset
+
+ ya está.
+ Si no, 
+\begin_inset Formula $\text{mcd}(p,a)=1$
+\end_inset
+
+ y 
+\begin_inset Formula $p|b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $2\implies3]$
+\end_inset
+
+ Por inducción con 
+\begin_inset Formula $a_{1}\cdots a_{n}=a_{1}(a_{2}\cdots a_{n})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Description
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ Si 
+\begin_inset Formula $a|p$
+\end_inset
+
+ entonces 
+\begin_inset Formula $p=ab$
+\end_inset
+
+ para cierto 
+\begin_inset Formula $b$
+\end_inset
+
+, y bien 
+\begin_inset Formula $p|a$
+\end_inset
+
+ (con lo que 
+\begin_inset Formula $a=p$
+\end_inset
+
+ o 
+\begin_inset Formula $a=-p$
+\end_inset
+
+) o 
+\begin_inset Formula $p|b$
+\end_inset
+
+ (con lo que 
+\begin_inset Formula $a=1$
+\end_inset
+
+ o 
+\begin_inset Formula $a=-1$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema Fundamental de la Aritmética:
+\series default
+ Todo entero distinto de 
+\begin_inset Formula $0$
+\end_inset
+
+ y 
+\begin_inset Formula $\pm1$
+\end_inset
+
+ puede escribirse como producto de primos, y la factorización es única salvo
+ signo y orden.
+ 
+\series bold
+Demostración:
+\series default
+ Consideremos el conjunto de todos los positivos distintos de 1 que no se
+ factorizan en primos y, si este no es vacío, sea 
+\begin_inset Formula $a\in\mathbb{Z}$
+\end_inset
+
+ su mínimo.
+ 
+\begin_inset Formula $a$
+\end_inset
+
+ no es primo, luego 
+\begin_inset Formula $a=bc$
+\end_inset
+
+ con 
+\begin_inset Formula $b,c\in\mathbb{Z}^{+}\backslash\{1\}$
+\end_inset
+
+.
+ Pero como 
+\begin_inset Formula $a$
+\end_inset
+
+ es mínimo, entonces 
+\begin_inset Formula $b$
+\end_inset
+
+ y 
+\begin_inset Formula $c$
+\end_inset
+
+ sí se factorizan en primos, luego 
+\begin_inset Formula $a$
+\end_inset
+
+ también.
+\begin_inset Formula $\#$
+\end_inset
+
+ Ahora sea 
+\begin_inset Formula $a=p_{1}\cdots p_{n}=q_{1}\cdots q_{m}$
+\end_inset
+
+ con 
+\begin_inset Formula $p_{1}\cdots p_{n},q_{1}\cdots q_{m}$
+\end_inset
+
+ primos y supongamos 
+\begin_inset Formula $n\leq m$
+\end_inset
+
+.
+ Procedemos por inducción sobre 
+\begin_inset Formula $n$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $n=1$
+\end_inset
+
+, 
+\begin_inset Formula $a=p_{1}=q_{1}\cdots q_{m}$
+\end_inset
+
+, y como 
+\begin_inset Formula $p_{1}$
+\end_inset
+
+ no tiene más divisores primos que 
+\begin_inset Formula $-p_{1}$
+\end_inset
+
+ y 
+\begin_inset Formula $p_{1}$
+\end_inset
+
+, debe ser 
+\begin_inset Formula $m=1$
+\end_inset
+
+ y 
+\begin_inset Formula $q_{1}=p_{1}$
+\end_inset
+
+.
+ Si suponemos el resultado válido para 
+\begin_inset Formula $n-1$
+\end_inset
+
+, entonces 
+\begin_inset Formula $p_{n}$
+\end_inset
+
+ divide a 
+\begin_inset Formula $a=q_{1}\cdots q_{n}$
+\end_inset
+
+ y por tanto divide a algún 
+\begin_inset Formula $i\in\{1,\dots,m\}$
+\end_inset
+
+.
+ Reordenamos los factores para obtener 
+\begin_inset Formula $i=m$
+\end_inset
+
+, es decir 
+\begin_inset Formula $p_{n}|q_{m}$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $q_{m}=\pm p_{n}$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $p_{1}\cdots p_{n-1}p_{n}=q_{1}\cdots q_{m-1}(\pm p_{n})$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $p_{1}\cdots p_{n-1}=\pm q_{1}\cdots q_{m-1}$
+\end_inset
+
+ y 
+\begin_inset Formula $n-1=m-1$
+\end_inset
+
+, luego 
+\begin_inset Formula $n=m$
+\end_inset
+
+ y además, después de ordenar si hiciera falta, 
+\begin_inset Formula $q_{i}=\pm p_{i}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así, para 
+\begin_inset Formula $a\in\mathbb{Z},a\neq0,\pm1$
+\end_inset
+
+, 
+\begin_inset Formula $a=\pm p_{1}^{n_{1}}\cdots p_{s}^{n_{s}}$
+\end_inset
+
+ y estos primos y sus exponentes son únicos (salvo orden).
+ Entonces podemos calcular el 
+\begin_inset Formula $\text{mcd}(a,b)$
+\end_inset
+
+ tomando el producto de primos comunes a 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ elevados a la mínima potencia y el 
+\begin_inset Formula $\text{mcm}(a,b)$
+\end_inset
+
+ tomando el producto de primos entre ambos elevados a la máxima potencia.
+\end_layout
+
+\begin_layout Standard
+Como 
+\series bold
+teorema
+\series default
+, el conjunto de los números primos es infinito.
+ Si no lo fuera, y fuera 
+\begin_inset Formula $\{p_{1},\dots,p_{n}\}$
+\end_inset
+
+, el número 
+\begin_inset Formula $N:=p_{1}\cdots p_{n}+1$
+\end_inset
+
+ también lo es.
+\begin_inset Formula $\#$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Section
+Congruencias
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $x,y\in\mathbb{Z},m\in\mathbb{Z}^{+}$
+\end_inset
+
+, 
+\begin_inset Formula $x$
+\end_inset
+
+ e 
+\begin_inset Formula $y$
+\end_inset
+
+ son 
+\series bold
+congruentes módulo 
+\begin_inset Formula $m$
+\end_inset
+
+
+\series default
+, 
+\begin_inset Formula $x\equiv y\mod m$
+\end_inset
+
+ ó 
+\begin_inset Formula $x\equiv y\,(m)$
+\end_inset
+
+, si 
+\begin_inset Formula $m|x-y$
+\end_inset
+
+.
+ Esta relación es de equivalencia.
+ Propiedades:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $r$
+\end_inset
+
+ es el resto de 
+\begin_inset Formula $a/m$
+\end_inset
+
+ entonces 
+\begin_inset Formula $a\equiv r\,(m)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a\equiv b\,(m)\land0\leq a,b<m\implies a=b$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a\equiv b\,(m)$
+\end_inset
+
+ si y sólo si 
+\begin_inset Formula $a$
+\end_inset
+
+ y 
+\begin_inset Formula $b$
+\end_inset
+
+ dan el mismo resto entre 
+\begin_inset Formula $m$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a\equiv a'\,(m)\land b\equiv b'\,(m)\implies a+b\equiv a'+b'\,(m)$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula $a-a'=\lambda m$
+\end_inset
+
+ y 
+\begin_inset Formula $b-b'=\mu m$
+\end_inset
+
+ para ciertos 
+\begin_inset Formula $\lambda,\mu\in\mathbb{Z}$
+\end_inset
+
+, luego 
+\begin_inset Formula $(a+b)-(a'+b')=(\lambda+\mu)m$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a\equiv a'\,(m)\land b\equiv b'\,(m)\implies ab\equiv a'b'\,(m)$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula 
+\[
+ab-a'b'=(a'+\lambda m)(b'+\mu m)-a'b'=a'b'+(a'\mu+b'\lambda+\lambda\mu m)m-a'b'\equiv0\,(m)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $a\equiv b\,(m)\implies ac\equiv bc\,(m)$
+\end_inset
+
+.
+ El recíproco es cierto si 
+\begin_inset Formula $c$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+ son coprimos.
+\begin_inset Newline newline
+\end_inset
+
+La primera parte se sigue de lo anterior.
+ Para la segunda, 
+\begin_inset Formula $m|ac-bc=(a-b)c\implies m|a-b\implies a\equiv b\,(m)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $c\neq0\implies(a\equiv b\,(m)\iff ac\equiv bc\,(mc))$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Formula $a-b=\lambda m\iff ac-bc=\lambda mc$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Denotamos la clase de equivalencia (llamada 
+\series bold
+clase de congruencia módulo 
+\begin_inset Formula $m$
+\end_inset
+
+
+\series default
+) de 
+\begin_inset Formula $a\in\mathbb{Z}$
+\end_inset
+
+ por 
+\begin_inset Formula $\overline{a}$
+\end_inset
+
+, y su 
+\series bold
+representante canónico
+\series default
+ es el resto de 
+\begin_inset Formula $a/m$
+\end_inset
+
+.
+ Llamamos entonces 
+\begin_inset Formula $\mathbb{Z}/(m)$
+\end_inset
+
+ o 
+\begin_inset Formula $\mathbb{Z}_{m}$
+\end_inset
+
+ al conjunto cociente, que tiene exactamente 
+\begin_inset Formula $m$
+\end_inset
+
+ elementos.
+ Así, para 
+\begin_inset Formula $\overline{a},\overline{b}\in\mathbb{Z}_{m}$
+\end_inset
+
+, definimos 
+\begin_inset Formula $\overline{a}+\overline{b}=\overline{a+b}$
+\end_inset
+
+ y 
+\begin_inset Formula $\overline{a}\cdot\overline{b}=\overline{a\cdot b}$
+\end_inset
+
+, y vemos que están bien definidas y que 
+\begin_inset Formula $\mathbb{Z}_{m}$
+\end_inset
+
+ es un anillo conmutativo.
+ Dado 
+\begin_inset Formula $\overline{a}\in\mathbb{Z}_{m}$
+\end_inset
+
+:
+\begin_inset Formula 
+\begin{multline*}
+\overline{a}\text{ tiene inverso (}\exists\overline{b}\in\mathbb{Z}_{m}:\overline{a}\cdot\overline{b}=1\text{)}\iff\overline{a}\text{ es cancelable (}\overline{a}\cdot\overline{x}=\overline{a}\cdot\overline{y}\implies\overline{x}=\overline{y}\text{)}\iff\\
+\iff\text{mcd}(a,m)=1
+\end{multline*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $1\implies2]$
+\end_inset
+
+ Multiplicando por 
+\begin_inset Formula $\overline{a}^{-1}$
+\end_inset
+
+ en ambos lados de 
+\begin_inset Formula $\overline{a}\cdot\overline{x}=\overline{a}\cdot\overline{y}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $2\implies3]$
+\end_inset
+
+ Si 
+\begin_inset Formula $\text{mcd}(a,m)=d>1$
+\end_inset
+
+, sean 
+\begin_inset Formula $a=a'd$
+\end_inset
+
+ y 
+\begin_inset Formula $m=m'd$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\overline{a}\cdot\overline{m'}=\overline{a'}\cdot\overline{d}\cdot\overline{m'}=\overline{a'}\cdot\overline{m}=\overline{0}=\overline{a}\cdot\overline{0}$
+\end_inset
+
+, pero 
+\begin_inset Formula $\overline{m'}\neq\overline{0}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $3\implies1]$
+\end_inset
+
+ Existen 
+\begin_inset Formula $r,s\in\mathbb{Z}$
+\end_inset
+
+ con 
+\begin_inset Formula $ra+sm=1$
+\end_inset
+
+, luego 
+\begin_inset Formula $\overline{r}\cdot\overline{a}=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Así, 
+\begin_inset Formula $\mathbb{Z}_{m}$
+\end_inset
+
+ es un cuerpo si y sólo si 
+\begin_inset Formula $m$
+\end_inset
+
+ es primo, pues entonces todos los elementos tienen inverso.
+\end_layout
+
+\begin_layout Standard
+Un entero es divisible por 3 si y sólo si la suma de sus cifras lo es.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $3|m\iff m\equiv0\,(3)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $10\equiv1\,(3)$
+\end_inset
+
+, luego 
+\begin_inset Formula $10^{s}\equiv1\,(3)$
+\end_inset
+
+ para todo 
+\begin_inset Formula $s$
+\end_inset
+
+ y si 
+\begin_inset Formula $m$
+\end_inset
+
+ se escribe como 
+\begin_inset Formula $a_{n}\cdots a_{0}$
+\end_inset
+
+, entonces 
+\begin_inset Formula $m=a_{n}10^{n}+\dots+a_{0}\equiv a_{n}+\dots+a_{0}\,(3)$
+\end_inset
+
+.
+ De forma parecida se pueden sacar reglas para el 9 y el 11.
+\end_layout
+
+\begin_layout Standard
+Dados 
+\begin_inset Formula $a,b,t\in\mathbb{Z}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Standard
+\begin_inset Formula 
+\[
+\overline{t}\text{ es sol. de }\overline{a}x=\overline{b}\in\mathbb{Z}_{m}\iff t\text{ es sol. de }ax\equiv b\,(m)\iff\exists s\in\mathbb{Z}:(t,s)\text{ es sol. de }ax-my=b
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+La ecuación 
+\begin_inset Formula $ax\equiv b\,(m)$
+\end_inset
+
+ tiene solución si y sólo si 
+\begin_inset Formula $d:=\text{mcd}(a,m)|b$
+\end_inset
+
+, y las soluciones son todos los enteros 
+\begin_inset Formula $x=x_{0}+\lambda\frac{m}{d}$
+\end_inset
+
+ con 
+\begin_inset Formula $\lambda\in\mathbb{Z}$
+\end_inset
+
+, donde 
+\begin_inset Formula $x_{0}$
+\end_inset
+
+ es una solución particular, de modo que la ecuación tiene 
+\begin_inset Formula $d$
+\end_inset
+
+ soluciones distintas módulo 
+\begin_inset Formula $m$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ 
+\begin_inset Formula $ax\equiv b$
+\end_inset
+
+ equivale a la ecuación diofántica 
+\begin_inset Formula $ax-my=b$
+\end_inset
+
+, que tiene solución si y sólo si 
+\begin_inset Formula $d|b$
+\end_inset
+
+.
+ Sean pues 
+\begin_inset Formula $b=db'$
+\end_inset
+
+, 
+\begin_inset Formula $a=da'$
+\end_inset
+
+ y 
+\begin_inset Formula $m=dm'$
+\end_inset
+
+, entonces 
+\begin_inset Formula $ax-my=b$
+\end_inset
+
+ equivale a 
+\begin_inset Formula $a'x-m'y=b'$
+\end_inset
+
+ y las soluciones son
+\begin_inset Formula 
+\[
+\left\{ \begin{array}{ccc}
+x & = & x_{0}+m'\lambda\\
+y & = & y_{0}+a'\lambda
+\end{array}\right.
+\]
+
+\end_inset
+
+Entonces 
+\begin_inset Formula $x_{0}+\lambda m'\equiv x_{0}+\mu m'\,(m)\iff\lambda m'\equiv\mu m'\,(dm')\iff\lambda\equiv\mu\,(d)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Teorema Chino de los Restos
+\end_layout
+
+\begin_layout Standard
+Sean 
+\begin_inset Formula $b_{1},\dots,b_{k}\in\mathbb{Z}$
+\end_inset
+
+ arbitrarios y 
+\begin_inset Formula $m_{1},\dots,m_{k}\in\mathbb{Z}^{+}$
+\end_inset
+
+ coprimos dos a dos, el sistema de congruencias
+\begin_inset Formula 
+\[
+\left\{ \begin{array}{cc}
+x\equiv b_{1} & (m_{1})\\
+\vdots\\
+x\equiv b_{k} & (m_{k})
+\end{array}\right.
+\]
+
+\end_inset
+
+tiene solución única módulo 
+\begin_inset Formula $M:=m_{1}\cdots m_{k}$
+\end_inset
+
+.
+ En particular, esta es 
+\begin_inset Formula $b_{1}M_{1}N_{1}+\dots+b_{k}M_{k}N_{k}$
+\end_inset
+
+, donde 
+\begin_inset Formula $M_{i}=\frac{M}{M_{i}}$
+\end_inset
+
+ y 
+\begin_inset Formula $N_{i}$
+\end_inset
+
+ es tal que 
+\begin_inset Formula $M_{i}N_{i}\equiv1\,(m_{i})$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Demostración:
+\series default
+ Si 
+\begin_inset Formula $p$
+\end_inset
+
+ es un número primo que divide a 
+\begin_inset Formula $M_{i}$
+\end_inset
+
+ y 
+\begin_inset Formula $m_{i}$
+\end_inset
+
+, entonces divide a algún 
+\begin_inset Formula $m_{j}$
+\end_inset
+
+ con 
+\begin_inset Formula $j\neq i$
+\end_inset
+
+, lo cual contradice que 
+\begin_inset Formula $\text{mcd}(m_{i},m_{j})=1$
+\end_inset
+
+, luego 
+\begin_inset Formula $M_{i}$
+\end_inset
+
+ y 
+\begin_inset Formula $m_{i}$
+\end_inset
+
+ son coprimos y 
+\begin_inset Formula $M_{i}$
+\end_inset
+
+ tiene inverso 
+\begin_inset Formula $N_{i}$
+\end_inset
+
+ módulo 
+\begin_inset Formula $m_{i}$
+\end_inset
+
+, teniendo en cuenta que 
+\begin_inset Formula $M_{i}N_{i}\equiv0\,(m_{j})$
+\end_inset
+
+ para 
+\begin_inset Formula $j\neq i$
+\end_inset
+
+.
+ Entonces 
+\begin_inset Formula $x_{0}=b_{1}M_{1}N_{1}+\dots+b_{k}M_{k}N_{k}$
+\end_inset
+
+ es solución del sistema.
+ Ahora bien, si 
+\begin_inset Formula $x$
+\end_inset
+
+ e 
+\begin_inset Formula $y$
+\end_inset
+
+ son soluciones del sistema, 
+\begin_inset Formula $x,y\equiv b_{i}\,(m_{i})$
+\end_inset
+
+, luego 
+\begin_inset Formula $x\equiv y\,(m_{i})$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $x-y$
+\end_inset
+
+ es múltiplo de todos los 
+\begin_inset Formula $m_{i}$
+\end_inset
+
+ y por tanto 
+\begin_inset Formula $x\equiv y\,(M)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Si los módulos no son coprimos, intentamos simplificar cada ecuación dividiéndol
+a entre un número, pues 
+\begin_inset Formula $a'dx\equiv b'd\,(m'd)\iff a'x\equiv b'\,(m')$
+\end_inset
+
+.
+ Si esto no es posible, resolvemos una ecuación y sustituimos en el resto.
+\end_layout
+
+\begin_layout Section
+Teoremas de Euler y Fermat
+\end_layout
+
+\begin_layout Standard
+Denotamos 
+\begin_inset Formula $\mathbb{Z}_{m}^{*}=\{x\in\mathbb{Z}_{m}|x\text{ es invertible}\}$
+\end_inset
+
+, y definimos la 
+\series bold
+función 
+\begin_inset Formula $\phi$
+\end_inset
+
+ de Euler
+\series default
+ como 
+\begin_inset Formula $\phi:\mathbb{N}\rightarrow\mathbb{N}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $\phi(m)=|\{x\in\mathbb{N}|1\leq x\leq m\land\text{mcd}(x,m)=1\}|=|\mathbb{Z}_{m}^{*}|$
+\end_inset
+
+.
+ Así:
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $p$
+\end_inset
+
+ es primo, 
+\begin_inset Formula $\phi(p)=p-1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $p$
+\end_inset
+
+ es primo, 
+\begin_inset Formula $\phi(p^{n})=p^{n-1}(p-1)$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Los no-coprimos con 
+\begin_inset Formula $p^{n}$
+\end_inset
+
+ son precisamente los múltiplos de 
+\begin_inset Formula $p$
+\end_inset
+
+, por lo que estos son 
+\begin_inset Formula $\frac{p^{n}}{p}=p^{n-1}$
+\end_inset
+
+ y 
+\begin_inset Formula $\phi(p^{n})=p^{n}-p^{n-1}=p^{n}(p-1)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $\text{mcd}(n,m)=1$
+\end_inset
+
+, entonces 
+\begin_inset Formula $\phi(nm)=\phi(n)\phi(m)$
+\end_inset
+
+.
+\begin_inset Newline newline
+\end_inset
+
+Definimos 
+\begin_inset Formula $f:\mathbb{Z}_{nm}^{*}\rightarrow\mathbb{Z}_{n}^{*}\times\mathbb{Z}_{m}^{*}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f(x)=(x_{n},x_{m})$
+\end_inset
+
+, donde 
+\begin_inset Formula $x_{n}$
+\end_inset
+
+ y 
+\begin_inset Formula $x_{m}$
+\end_inset
+
+ son los restos de dividir 
+\begin_inset Formula $x$
+\end_inset
+
+ entre 
+\begin_inset Formula $n$
+\end_inset
+
+ y 
+\begin_inset Formula $m$
+\end_inset
+
+, respectivamente.
+ Para abreviar asumimos que está bien definida, y pasamos a ver que es biyectiva.
+ Si 
+\begin_inset Formula $f(x)=(x_{n},x_{m})=f(y)$
+\end_inset
+
+ entonces 
+\begin_inset Formula $x\equiv y\,(m)$
+\end_inset
+
+ y 
+\begin_inset Formula $x\equiv y\,(n)$
+\end_inset
+
+, luego 
+\begin_inset Formula $nm|(x-y)$
+\end_inset
+
+ y en 
+\begin_inset Formula $\mathbb{Z}_{nm}^{*}$
+\end_inset
+
+ es inyectiva.
+ Para ver que es suprayectiva, consideramos 
+\begin_inset Formula $(a,b)\in\mathbb{Z}_{n}^{*}\times\mathbb{Z}_{m}^{*}$
+\end_inset
+
+.
+ Al existir una identidad de Bézout 
+\begin_inset Formula $1=rn+sm$
+\end_inset
+
+, podemos hacer 
+\begin_inset Formula $x=brn+asm$
+\end_inset
+
+, con lo que 
+\begin_inset Formula $x\equiv a\,(n)$
+\end_inset
+
+ y 
+\begin_inset Formula $x\equiv b\,(m)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Enumerate
+Si 
+\begin_inset Formula $m=p_{1}^{n_{1}}\cdots p_{s}^{n_{s}}$
+\end_inset
+
+ es la descomposición de 
+\begin_inset Formula $m$
+\end_inset
+
+ en factores primos, entonces
+\begin_inset Formula 
+\[
+\phi(m)=\prod_{i=1}^{s}p_{i}^{n_{i}-1}(p_{i}-1)=m\left(1-\frac{1}{p_{1}}\right)\cdots\left(1-\frac{1}{p_{s}}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema de Euler:
+\series default
+ Sea 
+\begin_inset Formula $1<m\in\mathbb{Z}$
+\end_inset
+
+.
+ Si 
+\begin_inset Formula $a$
+\end_inset
+
+ es coprimo con 
+\begin_inset Formula $m$
+\end_inset
+
+ entonces 
+\begin_inset Formula $a^{\phi(m)}\equiv1\,(m)$
+\end_inset
+
+.
+ 
+\series bold
+Demostración:
+\series default
+ Esto equivale a que 
+\begin_inset Formula $\overline{a}^{\phi(m)}=\overline{1}\in\mathbb{Z}_{m}$
+\end_inset
+
+.
+ Sea 
+\begin_inset Formula $\mathbb{Z}_{m}^{*}=\{\overline{x_{1}},\dots,\overline{x_{\phi(m)}}\}$
+\end_inset
+
+ y 
+\begin_inset Formula $\overline{a}\cdot\mathbb{Z}_{m}^{*}=\{\overline{a}\overline{x}|\overline{x}\in\mathbb{Z}_{m}^{*}\}$
+\end_inset
+
+.
+ Demostramos que 
+\begin_inset Formula $\overline{a}\cdot\mathbb{Z}_{m}^{*}=\mathbb{Z}_{m}^{*}$
+\end_inset
+
+:
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\subseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $x=\overline{ax_{i}}\in\overline{a}\cdot\mathbb{Z}_{m}^{*}\implies x\overline{a}^{-1}\overline{x_{i}}^{-1}=\overline{1}\implies x\in\mathbb{Z}_{m}^{*}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Argument item:1
+status open
+
+\begin_layout Plain Layout
+\begin_inset Formula $\supseteq]$
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula $\overline{x_{i}}\in\mathbb{Z}_{m}^{*}\implies\overline{x_{i}}=\overline{a}\,\overline{a}^{-1}\overline{x_{i}}\in\overline{a}\cdot\mathbb{Z}_{m}^{*}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Entonces 
+\begin_inset Formula $\prod_{i=1}^{\phi(m)}\overline{x}_{i}=\prod_{i=1}^{\phi(m)}\overline{ax_{i}}=\overline{a}^{\phi(m)}\prod_{i=1}^{\phi(m)}\overline{x_{i}}$
+\end_inset
+
+, y dividiendo entre 
+\begin_inset Formula $\prod_{i=1}^{\phi(m)}\overline{x_{i}}$
+\end_inset
+
+, porque es invertible, se obtiene el resultado.
+\end_layout
+
+\begin_layout Standard
+
+\series bold
+Teorema Pequeño de Fermat:
+\series default
+ Si 
+\begin_inset Formula $a\in\mathbb{Z}$
+\end_inset
+
+ y 
+\begin_inset Formula $p$
+\end_inset
+
+ es un número primo que no divide a 
+\begin_inset Formula $a$
+\end_inset
+
+, entonces 
+\begin_inset Formula $a^{p-1}\equiv1\,(p)$
+\end_inset
+
+, y para todo 
+\begin_inset Formula $x\in\mathbb{Z}$
+\end_inset
+
+, 
+\begin_inset Formula $x^{p}\equiv x\,(p)$
+\end_inset
+
+.
+ Esto se deriva del teorema de Euler y de que 
+\begin_inset Formula $\phi(p)=p-1$
+\end_inset
+
+.
+\end_layout
+
+\end_body
+\end_document