Is A^(p-1)-B^(p-1) divisible by p?[edit]
are whole numbers.
is a prime number.
You choose a random number
.
Then there is always a number
is divisible by
.
Is A^(p-1)-B^(p-1) divisible by p. A different path.[edit]
are whole numbers.
is a prime number.
There is a number
and there is a number
.
is divisible by
.
Fermat's little theorem unravelled.[edit]
are whole numbers.
p is a prime number.
Here is the proof that
is divisible by
.
The following is worth mentioning..
is divisible by
Analytical proof: Binomial development[edit]
![{\displaystyle {\binom {n}{k}}+{\binom {n}{k-1}}={\frac {n!}{(n-k)!k!}}+{\frac {n!}{(n-k+1)!(k-1)!}}}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/8b95cd360f86db06086ecc17cc8ccfc6e3cbaa01)
![{\displaystyle {\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}}}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/1058edbf8a53641e199779c21e9caf12ddc65cdd)