Fermat theorem, Euler's function

Prove an interesting theorem before proving these theorems.

For 0 mod m,n mod m, 2n mod m, 3n mod m, 4n mod m ... (m-1) n mod m

The corresponding solution set sequence must have M/D part 0 D 2d 3d. M-d. (not necessarily in order) such a solution. where d = gcd (n,m)

Specific point: n=8,m=12. D = gcd (n,m) =4

For 0 mod 12,8 mod 12,16 mod 12,24 mod 12,32 mod 12,40 mod 12...88 mod 12.

Corresponding Solutions 0 8 4 0 8 4 ..... 4

The obvious conclusion should be right.

Prove:

Fermat theorem:

N^p-1≡1 (mod p), n^p-1 mod p = 1. where n and P are mutually vegetarian.

Prove:

We can construct a p-1 n mod p,2n mod p,3n mod p ... (p-1) n mod p

According to the congruence operation (to turn the remainder into the same Yu Shizi):

Then there is (P-1)! n^ (p-1) ≡ (p-1)! (mod p)

WHERE (p-1)! P is not divisible so the formula becomes n^ (p-1) ≡ (mod p), proof.

You can also multiply n on both sides. N^p≡n (mod p)

Here's a little thought episode:

Consider the following proof:

Due to n and P-biotin. So there is n≡1 (mod p). According to the congruence theorem n^ (p-1) ≡1^ (p-1) (mod p), n^ (p-1) ≡ (mod p), the certificate.

Simple and clear. But it's wrong. Because the premise is wrong. Due to n and P reciprocity. Not necessarily n≡1 (mod p)

Can only be said that due to N and P reciprocity, exists kn≡1 (mod p) (k|1,2,3..p-1) (according to the above theorem)

Example n=2 p=5. There is 3n mod p = 1. and n mod p = 2.

Euler functions:

function: Enter n, do you want to know how many numbers in 1~n and n are mutually-vegetarian?

This is a function of Euler's function. It also allows the extended Fermat theorem to be expanded. The sum of the qualitative factors

The Euler function is a φ function, which means φ (1) = 1,φ (2) =1,φ (3) =2 ...

for φ (n).

If n is a prime number, that is, φ (n) = n-1.

If n is composite, that is, φ (n) < n-1.

Obviously this is not enough for the application. Of course we want to be able to enter N to get concrete results, not just scope.

For this problem: we can solve the problem in the next proof and find something new.

If M is a prime power p^k (such as 2^3), φ (p^k) can still be calculated.

How many and p^k are there in 1~p^k-1?

Then we can find out what the number is without P factor.

Then we can find the number of P-factor. Then the total number minus and p^k the number of non-reciprocity.

Fermat theorem, Euler's function