who wants to solve chinese cartoon math?

>>8104199
Fuck off weaboo, this is trivial.

Isn't that from madoka

let p = 3, n = 2,

(1+3)^3 - 2^3 - 1 = 55
55 is not divisible by 3.

Little Fermat

>>8104272
[math](1+n)^p - n^p -1[/math]
you can see it in the moonrunes section.

>>8104272
3 times 18 is 54 so it is preeeeeetty close.

It still holds up.

>>8104199
1+p mod p = 1
1^p = 1 mod p
n^p= n mod p

unless n mod p = 0 it isn't

For the original Japanese
(1+n)^p = 1+n mod p
so n+1-1-n = 0 mod p QED

>>8104199
It's obviously mistranslated, it's supposed to be (1+n)^p-n^p-1.

>>8104199
Literally trivial.
Expand (n+1)^p with binmial theorem.
The first term, n^p and the last term, +1, will be cancelled out.
All remaining coefficients will have a p in the numerator due to the nCp.
This is literally immediate.

>>8104481
>binomial theorem

Fermat's little theorem/Euler's theorem/Lagrange's theorem is so much easier.

>>8104504
>>8104481
Doesn't one of the most common proofs of Fermat's little employ the binomial theorem, literally in the way mentioned (...coefficients will have p in the numerator due to the choose...)

>>8104199
You just apply Fermat's little theorem to both exponent terms and everything cancels

>>8104504
>use elementary theorem T
>no! use stronger theorem Z that uses theorem T and more things to be proved, it's easier

???

>>8104606

Only for CS plebs too afraid of group theory