Gebruiker:Wim Coenen/Kladblok

Poging drie

$p$ is een priemgetal. $p$ is groter dan twee.

$A\neq ap$

$B\neq ap$

Is $A^{p-1}-1$ deelbaar door $p$?

Elk getal $C$ kunnen we schrijven als $C=(C-1)+1$

${\frac {C^{p-1}-1}{p}}={\frac {((C-1)+1)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(C-1)}^{k}1^{p-1-k}-1}{p}}$

${\frac {C^{p-1}-1}{p}}={\frac {1+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(C-1)^{k}p^{k}1^{p-1-k}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(C-1)^{k}p^{k}}{p}}$

${\frac {C^{p-1}-1}{p}}=\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{{\frac {(C-1)^{k}}{p^{k}}}p^{k-1}}$

${\displaystyle {\frac {C^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(C-1)^{k}}$

$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(C-1)}^{k}$ moet een p-voud zijn.

Inductie stap een.

$\displaystyle {\frac {C^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(2-1)^{k}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

De som van de binomiaalcoëfficiënten. 

{${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!}{(p-1-k)!k!}}+{\frac {(p-1)!}{(p-1-k-1)!(k+1)!}}$

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!(k+1)}{(p-1-k)!k!(k+1)}}+{\frac {(p-1)!(p-1-k)}{(p-1-k-1)!(k+1)!(p-1-k)}}$

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!(k+1+p-1-k)}{(p-(k+1))!(k+1)!}}={\binom {p}{k+1}}$

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {p(p-1)!}{(p-(k+1))!(k+1)!}}=pq$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1!}}=q$

${\frac {C^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}=q$

Inductie stap twee.

Voor een zeker getal $B$ geldt:

${\frac {B^{p-1}-1}{p}}=Q$

$Q\in \mathbb {N}$

Inductie stap drie.

Het getal $A$ is gelijk aan: $A=p+q$

Het getal $B$ is gelijk aan: $B=p-q$

Dan volgt daaruit: $A+B=2p\Longrightarrow p={\frac {1}{2}}(A+B)$

${\frac {A^{p-1}-1}{p}}-{\frac {B^{p-1}-1}{p}}=2{\frac {A^{2k}-B^{2k}}{A+B}}=2\textstyle \sum _{n=1}^{2k}\displaystyle {(-1)^{n-1}}A^{2k-n}B^{n-1}=Z$

${\frac {A^{p-1}-1}{p}}=2\textstyle \sum _{n=1}^{2k}\displaystyle {(-1)^{n-1}}A^{2k-n}B^{n-1}+{\frac {B^{p-1}-1}{p}}=R$

${\frac {A^{p-1}-1}{p}}=Z+Q=R\Longrightarrow R\in \mathbb {N}$

Een andere weg.

Elk getal $A$ kunnen we schrijven als $A=ap+1$

${\frac {A^{p-1}-1}{p}}={\frac {(ap+1)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(ap)}^{k}1^{p-1-k}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {1+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}a^{k}p^{k}1^{p-1-k}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}a^{k}p^{k}}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}a^{k}p^{k}}{p}}=\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}a^{k}p^{k-1}$

${\frac {A^{p-1}-1}{p}}=\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{{\frac {(A-1)^{k}}{p^{k}}}p^{k-1}}$

${\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}}$

$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}$ moet een p-voud zijn.

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(2-1)^{k}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!}{(p-1-k)!k!}}+{\frac {(p-1)!}{(p-1-k-1)!(k+1)!}}$

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!(k+1)}{(p-1-k)!k!(k+1)}}+{\frac {(p-1)!(p-1-k)}{(p-1-k-1)!(k+1)!(p-1-k)}}$

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!(k+1+p-1-k)}{(p-(k+1))!(k+1)!}}={\binom {p}{k+1}}$

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=^{2}\uparrow \sum _{k=1}^{p-1}{\binom {p}{k+1}}=p(^{2}\uparrow \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-(k+1))!(k+1!}})=pq$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=^{2}\uparrow \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-(k+1))!(k+1!}}=q$

${\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}}$

Een kortere weg.

$A\neq ap$

Elk getal $A$ kan worden geschreven als $A=(A-1)+1$

${\frac {A^{p-1}-1}{p}}={\frac {((A-1)+1)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}1^{p-1-k}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {1+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(A-1)^{k}1^{p-1-k}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(A-1)^{k}}{p}}$

${\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}}$

$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}$ moet een p-voud zijn.

Inductiestap een.

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(2-1)^{k}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

De som van de binomiaalcoëfficiënten. 

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!}{(p-1-k)!k!}}+{\frac {(p-1)!}{(p-1-k-1)!(k+1)!}}$

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!(k+1)}{(p-1-k)!k!(k+1)}}+{\frac {(p-1)!(p-1-k)}{(p-1-k-1)!(k+1)!(p-1-k)}}$

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!(k+1+p-1-k)}{(p-(k+1))!(k+1)!}}={\binom {p}{k+1}}$

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {p(p-1)!}{(p-(k+1))!(k+1)!}}=pq$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}=q$

${\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}=q$

$q\in \mathbb {N}$

Voor $A=2$ is de stelling waar.

Inductiestap twee

We nemen de waarheid van de stelling voor een zeker getal $A$ aan.

Dus:

${\frac {A^{p-1}-1}{p}}=R$

$R\in \mathbb {N}$

Inductiestap drie.

${\frac {(A+p)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}{\binom {p-1}{k}}A^{p-1-k}p^{k}-1\displaystyle }{p}}$

${\frac {(A+p)^{p-1}-1}{p}}={\frac {\textstyle A^{p-1}+\sum _{k=1}^{p-1}{\binom {p-1}{k}}A^{p-1-k}p^{k}-1\displaystyle }{p}}$

${\frac {(A+p)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}{\binom {p-1}{k}}A^{p-1-k}p^{k}\displaystyle }{p}}+{\frac {A^{p-1}-1}{p}}$

${\frac {(A+p)^{p-1}-1}{p}}=\textstyle \sum _{k=1}^{p-1}{\binom {p-1}{k}}A^{p-1-k}p^{k-1}\displaystyle +R=Q$

$Q\in \mathbb {N}$

De stelling is dus juist.

Een alternatief.

$p=A-B$

${\frac {A^{p-1}-1}{p}}-{\frac {B^{p-1}-1}{p}}={\frac {A^{2k}-B^{2k}}{A-B}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle$

${\frac {B^{p-1}-1}{p}}={\frac {A^{p-1}-1}{p}}-\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle$

${\frac {B^{p-1}-1}{p}}=R-\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle =Z$

$Z\in \mathbb {N}$

De stelling is dus juist.

De definitieve versie

$A\neq ap$

Elk getal $A$ kan worden geschreven als $A=(A-1)+1$

${\frac {A^{p-1}-1}{p}}={\frac {((A-1)+1)^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}1^{p-1-k}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {1+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(A-1)^{k}1^{p-1-k}-1}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}(A-1)^{k}}{p}}$

${\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}}$

$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(A-1)}^{k}$ moet een p-voud zijn.

Stap een.

$\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(2-1)^{k}}={\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}$

De som van de binomiaalcoëfficiënten. 

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!}{(p-1-k)!k!}}+{\frac {(p-1)!}{(p-1-k-1)!(k+1)!}}$

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!(k+1)}{(p-1-k)!k!(k+1)}}+{\frac {(p-1)!(p-1-k)}{(p-1-k-1)!(k+1)!(p-1-k)}}$

${\binom {p-1}{k}}+{\binom {p-1}{k+1}}={\frac {(p-1)!(k+1+p-1-k)}{(p-(k+1))!(k+1)!}}={\binom {p}{k+1}}$

$\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {p(p-1)!}{(p-(k+1))!(k+1)!}}=pq$

${\frac {1}{p}}\sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}=\sum _{k=1,3,5\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}=q$

${\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}=q$

$q\in \mathbb {N}$

Voor $A=2$ is de stelling waar.

Stap twee

${\frac {b^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}={\frac {1}{p}}\sum _{k=1,3,5,\ldots }^{p-2}{\binom {p}{k+1}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}=q$

$A=a+p\Longrightarrow p=A-a$

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}={\frac {A^{2k}-a^{2k}}{A-a}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle$

${\frac {A^{p-1}-1}{p}}={\frac {a^{p-1}-1}{p}}+\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle$

Stel dat: $a=2$

${\frac {A^{p-1}-1}{p}}={\frac {2^{p-1}-1}{p}}+\textstyle \sum _{n=1}^{2k}A^{2k-n}2^{n-1}\displaystyle$

${\frac {A^{p-1}-1}{p}}=\sum _{k=1,3,5,\ldots }^{p-2}{\frac {(p-1)!}{(p-(k+1))!(k+1)!}}+\textstyle \sum _{n=1}^{2k}A^{2k-n}2^{n-1}=Q\displaystyle$

$Q\in \mathbb {N}$

Een alternatieve route.

$B=A+1$

$A+1=(A+p)-(p-1)=(A+p)-c$

${\displaystyle {\frac {((A+p)-c)^{p-1}-1}{p}}={\frac {1}{p}}\textstyle (\sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(-1)^{k}(A+p)^{p-1-k}c^{k}}-1)$

${\displaystyle {\frac {((A+p)-(p-1))^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=0}^{p-2}\displaystyle {\tbinom {p-1}{k}}}{(-1)^{k}(A+p)^{p-1-k}c^{k}}+{\frac {{\binom {p-1}{0}}(-1)^{p-1}(A+p)^{p-1}c^{0}}{p}}-{\frac {1}{p}}$

${\displaystyle {\frac {((A+p)-c)^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(-1)^{k}(A+p)^{p-1-k}c^{k}}+{\frac {(A+p)^{p-1}}{p}}-{\frac {1}{p}}$

$p=A-B$

${\frac {A^{p-1}-1}{p}}-{\frac {B^{p-1}-1}{p}}={\frac {A^{2k}-B^{2k}}{A-B}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle$

${\frac {B^{p-1}-1}{p}}={\frac {A^{p-1}-1}{p}}-\textstyle \sum _{n=1}^{2k}A^{2k-n}B^{n-1}\displaystyle$

Slot

${\displaystyle {\frac {A^{p-1}-1}{p}}={\frac {1}{p}}\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}}=R$

$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}=pR$

$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}=\textstyle \sum _{k=1,3,5\ldots }^{p-2}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}+\textstyle \sum _{k=2,4,6\ldots }^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}$$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}=\textstyle \sum _{k=1,3,5\ldots }^{p-2}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}+\textstyle \sum _{k=1,3,5\ldots }^{p-2}\displaystyle {\tbinom {p-1}{k+1}}}{(A-1)^{k}+\textstyle \sum _{k=2,4,6\ldots }^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}-\textstyle \sum _{k=1,3,5\ldots }^{p-2}\displaystyle {\tbinom {p-1}{k+1}}}{(A-1)^{k}$$\textstyle \sum _{k=1,3,5\ldots }^{p-2}\displaystyle {\tbinom {p-1}{k+1}}}{(A-1)^{k}=\textstyle \sum _{k=2,4,6\ldots }^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k-1}$$\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}=\textstyle \sum _{k=1,3,5\ldots }^{p-2}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}+\textstyle \sum _{k=1,3,5\ldots }^{p-2}\displaystyle {\tbinom {p-1}{k+1}}}{(A-1)^{k}+\textstyle \sum _{k=2,4,6\ldots }^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k}-\textstyle \sum _{k=2,4,6\ldots }^{p-1}\displaystyle {\tbinom {p-1}{k}}}{(A-1)^{k-1}$$\textstyle \sum _{k=1,3,5\ldots }^{p-2}\displaystyle {\tbinom {p}{k+1}}}{(A-1)^{k}+\textstyle \sum _{k=2,4,6\ldots }^{p-1}\displaystyle {\tbinom {p-1}{k}}}[{(A-1)^{k}-(A-1)^{k-1}]$

${\frac {A^{p-1}-1}{p}}={\frac {({\frac {a}{b}})^{p-1}-1}{p}}={\frac {{\frac {a^{p-1}}{b^{p-1}}}-{\frac {b^{p-1}}{b^{p-1}}}}{p}}={\frac {\frac {a^{p-1}-b^{p-1}}{b^{p-1}}}{p}}={\frac {a^{p-1}-b^{p-1}}{pb^{p-1}}}$

De kleine stelling van Fermat 1

$A\neq cp$

$A=a+p\Longrightarrow p=A-a$

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}={\frac {A^{p-1}-a^{p-1}}{A-a}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle$

We gaan er vanuit dat $A^{p-1}-1$ en $a^{p-1}-1$ beiden niet deelbaar zijn door $p$.

$B$ en $b$ zijn gehele getallen. $r_{1}$ en $r_{2}$ zijn resten na deling door $p$.

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}=B+{\frac {r_{1}}{p}}-\left(b+{\frac {r_{2}}{p}}\right)=B-b+{\frac {r_{1}-r_{2}}{p}}=\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle \Longrightarrow$

${\frac {r_{1}-r_{2}}{p}}=0\Longrightarrow r_{1}=r_{2}=r$

${\frac {A^{p-1}-1}{p}}-{\frac {a^{p-1}-1}{p}}=B+{\frac {r}{p}}-\left(b+{\frac {r}{p}}\right)=B-b=\textstyle \sum _{n=1}^{2k}A^{2k-n}a^{n-1}\displaystyle$

Een stap verder.

${\frac {A^{p}-A}{p}}-{\frac {a^{p}-a}{p}}={\frac {A^{p}-a^{p}}{A-a}}-{\frac {A-a}{A-a}}={\frac {A^{2k+1}-a^{2k+1}}{A-a}}-1=\textstyle \sum _{n=1}^{2k+1}A^{2k+1-n}a^{n-1}-1\displaystyle$

$A{\frac {A^{p-1}-1}{p}}-a{\frac {a^{p-1}-1}{p}}=\textstyle \sum _{n=1}^{2k+1}A^{2k+1-n}a^{n-1}-1\displaystyle$

Wederom gaan we er vanuit dat $A^{p-1}-1$ en $a^{p-1}-1$ beiden niet deelbaar zijn door $p$.

$B$ en $b$ zijn gehele getallen. $r$ is de rest na deling door $p$.

$A{\frac {A^{p-1}-1}{p}}-a{\frac {a^{p-1}-1}{p}}=A\left(B+{\frac {r}{p}}\right)-a\left(b+{\frac {r}{p}}\right)=AB-ab+(A-a){\frac {r}{A-a}}$

$A{\frac {A^{p-1}-1}{p}}-a{\frac {a^{p-1}-1}{p}}=AB-ab+r$

$AB-ab+r=\textstyle \sum _{n=1}^{2k+1}A^{2k+1-n}a^{n-1}-1\displaystyle \Longrightarrow r=0$

$r$ is nul, dus zijn $A^{p-1}-1$ en $a^{p-1}-1$ beiden deelbaar door $p$.

De kleine stelling van Fermat

$A\neq ap$

$A=ap+1$

${\displaystyle {\frac {A^{p-1}-1}{p}}=\displaystyle {\frac {(ap+1)^{p-1}-1}{p}}}$

${\frac {A^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-1}\displaystyle {\tbinom {p-1}{k}}{(ap)}^{p-1-k}1^{k}-1}{p}}={\frac {1+\textstyle \sum _{k=0}^{p-2}\displaystyle {\tbinom {p-1}{k}}{(ap)}^{p-1-k}1^{k}-1}{p}}$

${\frac {A^{p-1}-1}{p}}={\frac {\textstyle \sum _{k=0}^{p-2}\displaystyle {\tbinom {p-1}{k}}{(ap)}^{p-1-k}1^{k}}{p}}=\textstyle \sum _{k=0}^{p-2}\displaystyle {\tbinom {p-1}{k}}{a}^{p-1-k}{p}^{p-2-k}1^{k}=z$

${\frac {A^{p-1}-1}{p}}=z\Longrightarrow A{\frac {A^{p-1}-1}{p}}={\frac {A^{p}-A}{p}}=Az=u$

${\frac {(A+1)^{p}-(A+1)}{p}}={\frac {\textstyle \sum _{k=0}^{p}\displaystyle {\tbinom {p}{k}}{A}^{p-k}1^{k}-A-1}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}={\frac {{A}^{p}-A+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p}{k}}{A}^{p-k}1^{k}+1-1}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}={\frac {{A}^{p}-A+\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p}{k}}{A}^{p-k}1^{k}}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}-{\frac {A^{p}-A}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p}{k}}{A}^{p-k}1^{k}}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}-{\frac {pu}{p}}={\frac {p\textstyle \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}{A}^{p-k}}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}={p}{\frac {u+\textstyle \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}{A}^{p-k}}{p}}$

${\frac {(A+1)^{p}-(A+1)}{p}}=u+\textstyle \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}{A}^{p-k}=q$

${\frac {(A+1)^{p-1}-{1}}{p}}={\frac {u+\textstyle \sum _{k=1}^{p-1}{\frac {(p-1)!}{(p-k)!k!}}{A}^{p-k}}{A+1}}={\frac {q}{A+1}}$

Binomiale ontwikkeling

Bewijs dat: $(a+b)^{n}=\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}$

Inductiestap een:

$(a+b)^{0}=\textstyle \sum _{k=0}^{n=0}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}=\displaystyle {\binom {0}{0}}a^{0}b^{0}=1$

Voor $n=0$ is de bewering waar.

Inductiestap twee:

We nemen aan dat de bewering voor $n$ waar is:

Dus: $(a+b)^{n}=\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}$

Inductiestap drie:

We onderzoeken of de bewering voor $n+1$ waar is.

$(a+b)^{n+1}=(a+b)\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}$

$(a+b)^{n+1}=a\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}+b\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}$

$(a+b)^{n+1}=\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n+1-k}b^{k}+\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k+1}$

$\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k+1}=\textstyle \sum _{k=1}^{n+1}\displaystyle {\binom {n}{k-1}}a^{n+1-k}b^{k}$

$(a+b)^{n+1}=a^{n+1}+\textstyle \sum _{k=1}^{n}\displaystyle {\binom {n}{k}}a^{n+1-k}b^{k}+\textstyle \sum _{k=1}^{n}\displaystyle {\binom {n}{k-1}}a^{n+1-k}b^{k}+b^{n+1}$

$(a+b)^{n+1}=a^{n+1}+\textstyle \sum _{k=1}^{n}{\biggl (}\displaystyle {\binom {n}{k}}+{\binom {n}{k-1}}{\biggr )}a^{n+1-k}b^{k}+b^{n+1}$

De som van de binomiaalcoëfficiënten. 

${\binom {n}{k}}+{\binom {n}{k-1}}={\frac {n!}{(n-k)!k!}}+{\frac {n!}{(n-k+1)!(k-1)!}}$
${\binom {n}{k}}+{\binom {n}{k-1}}={\frac {n!(n-k+1)}{(n-k)!k!(n-k+1)}}+{\frac {n!k}{(n-k+1)!(k-1)!k}}$

${\binom {n}{k}}+{\binom {n}{k-1}}={\frac {{(n+1)}!}{(n+1-k)!k!}}={\binom {n+1}{k}}$

$(a+b)^{n+1}=a^{n+1}+\textstyle \sum _{k=1}^{n}\displaystyle {\binom {n+1}{k}}a^{n+1-k}b^{k}+b^{n+1}$

$(a+b)^{n+1}=\textstyle \sum _{k=0}^{n+1}\displaystyle {\binom {n+1}{k}}a^{n+1-k}b^{k}$

De bewering is waar voor $n+1$.

Dus de bewering $(a+b)^{n}=\textstyle \sum _{k=0}^{n}\displaystyle {\binom {n}{k}}a^{n-k}b^{k}$ is waar.

QED.

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Pythagorese drietallen

Voor het genereren van Pythagorese drietallen zijn er een zestal formules.

Voor de even waarden $2p$:

$(2p)^{2}=4p^{2}=2p^{2}-(-2p^{2})$

$4p^{2}=p^{4}+2p^{2}+1-(p^{4}-2p^{2}+1)$

$(2p)^{2}=(p^{2}+1)^{2}-(p^{2}-1)^{2}$

$x=2p$ $y=p^{2}-1$ $z=p^{2}+1$

Vult men de waarde $p=2$ in, dan vindt men $x=4$, $y=3$, $z=5$

Voor de oneven waarden $2p+1$:

$(2p+1)^{2}=4p^{2}+4p+1^{2}=4p(p+1)+1^{2}$

$(2p+1)^{2}=(2p(p+1))^{2}+2.2p(p+1)+1^{2}-(2p(p+1))^{2}$

$(2p+1)^{2}=(2p(p+1)+1)^{2}-(2p(p+1))^{2}$

$x=2p+1$ $y=2p(p+1)$ $z=2p(p+1)+1$

Vult men de waarde $p=1$ in, dan vindt men $x=3$, $y=4$, $z=5$