The cost of the study of number theory and Euler

Review of number theory and the cost of the horse and Euler

QB_UDG years One Month 8 Day 10:16:18

1. Fermat theorem Fermat theory

if P is prime , and a with the P coprime, i.e. gcd (a,p) =1 so (a^p-1) ≡1 (mod p)

application : solving multiplication inverse element

multiplication Inverse element : (x*x ') ≡ 1 (mod p) called X ' is the multiplication inverse of the X-mode P (note that it must be 1)

Inverse: (b/a) (mod n) = (b * x) (mod n). x represents the inverse of a. and a*x≡1 (mod n) Note: The inverse is only present when A and n coprime

Note that the inverse can also be understood, to find a minimum positive integer x (inverse), so that a Times x to M of the remainder equal to 1 m of the remainder, so m=1, the inverse is 1

Because (a^p-1) ≡1 (mod p) then (a*a^p-2) ≡1 (mod p)
So a^p-2 is a multiplicative inverse of modulo p .

Use of inverse element:

When solving the division modulo problem (A/b)%m, we can convert to (a% (b?m))/b,
However, if B is large, there will be a burst accuracy problem, so we avoid using division to calculate directly.
You can convert a division to a multiplication by using an inverse element:
Suppose B has a multiplicative inverse, that is, with M coprime (sufficient and necessary). Set C is the inverse of B, i.e. b?c≡1 (mod m), then there is a/b= (A/b)?1<=> (A/b)? B?cóa?c (mod m)
That is, dividing by a number modulo equals the inverse modulo of the number.

So (A/b) mod p is converted to (a*b ') mod p = ((a mod p) * (b^p-2 modp)) mod p

Applies to p as prime number

2. Euler functions
for a positive integer x , less than x and and the x coprime of the positive integers of the number , remember to do: φ (x)
where φ (1) is defined as 1, but does not have any real meaning

General Formula 1:φ (x) =x* (1-1/P1)(1-1/p2) (1-1/p3) ..... (1-1/PN)
Where P1,P2,P3......PN is all the mass factor of X

General Formula 2: if x is the k Power of prime number P , i.e. x=p^k, with φ (x) = p^k-p^ (k-1) = (p-1) *p^ (k-1)
Because the quality factor of X is only p, so in addition to multiples of p, the other numbers are coprime with X. (In the program we generally write φ (x) as Phi (X)).

Properties:

1. If x, y coprime (i.e. gcd (x, y) =1), then φ (x*y) =φ (x) *φ (y)
2. If x is odd φ (2x) =φ (x)
3. If x is prime number φ (x) =x-1
4. if x>2,φ (x) is even

The above four can be launched according to Euler function! And obviously!

3. Euler's theorem

If the a,n is a positive integer and a,n coprime (that is, gcd (a,n) =1), then there are:

A^φ (n) ≡1 (mod n)

According to this theorem, the Fermat theorem is obviously, because the Phi (p) of prime number p is equal to P-1

Application:

Power down!

A and n coprime, when B is large, the Euler theorem can be used to power down

(a^b) mod n = (a^ (Bmod Phi (n))) mod n

Finally, how do you find Phi (p)?

See Euler screens :

The following properties are required ( p is prime ):
Phi (P) =p-1
Because the prime number p except 1 is only p, so the integer 1 to p is only p and p not coprime

If I mod p = 0, then phi (i * p) =p * PHI (i)
If the integer n is not with I coprime, n+i remains with I not coprime
If the integer n with i coprime, n+i and I remain coprime, proving that:

If I modp≠0, then Phi (i * p) =phi (i) * (p-1)
I mod p is not 0 and P is prime, so I and P coprime, then according to the Euler function of the product phi (I * p) =phi (i) * PHI (P) where Phi (p) =p-1 is the first property

#include<iostream>
#include<cstdio>
#DefineN 40000
using namespaceStd
intN
intPhi[n+10],prime[n+10],tot,ans;
BOOLMARK[N+10];
void Getphi()
{
intI,j;
Phi[1]=1;
for(i=2;i<=n;i++)//equivalent to decomposition of the inverse process of the mass-type
{
if(!mark[i])
{
prime[++tot]=i;//the number of sieve primes first determines whether I is a prime number.
phi[i]=i-1;//When I is a prime number phi[i]=i-1
}
for(j=1;j<=tot;j++)
{
if(i*prime[j]>n) Break;
mark[i*prime[j]]=1;//OK I*prime[j] not prime
if(i%prime[j]==0)//Then we will see if PRIME[J] is an approximate of I
{
PHI[I*PRIME[J]]=PHI[I]*PRIME[J]; Break;
}
Elsephi[i*prime[j]]=phi[i]* (prime[j]-1);//actually here prime[j]-1 is phi[prime[j]], using the product of Euler function
}
}
}
int Main()
{
Getphi ();
}

To summarize:
Plainly, dividing by a number modulo, the inverse modulo effect is the same as multiplying the number.
The inverse is actually the a*x = 1 (mod C) set up X, where a is the divisor of division (seemingly also pull to expand,,)

To solve the problem (A/b)%m, we can turn to (a*b ') mod m to resolve, where B ' is the inverse of B

Power Down,
A and n coprime, when B is large, the Euler theorem can be used to power down

(a^b) mod n = (a^ (Bmod Phi (n))) mod n

The cost of the study of number theory and Euler