User:Wim Coenen/sandbox

Is A^(p-1)-B^(p-1) divisible by p?

$A,B,p,q$ are whole numbers.

$p$ is a prime number.

$A\neq qp$

You choose a random number $B$.

Then there is always a number $A=B+p$

$A=B+p$

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

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

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {\textstyle \sum _{k=1}^{p-1}\displaystyle {\tbinom {p-1}{k}}{B^{p-1-k}p^{k}+B^{p1}-B^{p-1}}}{p}}$

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

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

$A^{P-1}-B^{p-1}$ is divisible by $p$.

Is A^(p-1)-B^(p-1) divisible by p. A different path.

$A,B,p,q$ are whole numbers.

$p$ is a prime number.

$A\neq qp$

There is a number $A=p+q$ and there is a number $B=p-q$.

$A=p+q$

$B=p-q$

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {(p+q)^{p-1}-(p-q)^{p-1}}{p}}$

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

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

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

$A^{P-1}-B^{p-1}$ is divisible by $p$.

Fermat's little theorem unravelled.

$A,B,p,q,r$ are whole numbers.

p is a prime number.

Here is the proof that $A^{p-1}-1$ is divisible by $p$. ${\frac {A^{p-1}-1}{p}}=?$

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

${\frac {A^{p-1}-B^{p-1}}{p}}={\frac {(p+q)^{p-1}-(p+r)^{p-1}}{p}}$

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

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

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

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

The following is worth mentioning..

${\binom {p-1}{p-1}}p^{p-2-(p-1)}\{q^{p-1}-(-p+1)^{p-1}\}={\frac {{\binom {p-1}{p-1}}\{q^{p-1}-(-p+1)^{p-1}\}}{p}}$

$\{q^{p-1}-(-p+1)^{p-1}\}$ is divisible by $p$

Analytical proof: Binomial development

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

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

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

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

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

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

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

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

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

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

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