11 Basic Theory of knowledge



1. Number of primitive hooks:

Concept: A ternary group (A,B,C), where a,b,c is not common factor and satisfies: a^2+b^2=c^2

First, the number of this primitive tick is infinite, and the conditions of construction are satisfied:

a=s*t,b= (s^2-t^2)/2,c= (s^2+t^2)/2

Where S>t>=1 is an odd number with no common factor.

From the above concept can be exported any one of the primitive hook array.

2, Prime number count (prime theorem)

The number of primes (x) from 1 to X

The highest number-theory achievement of the 19th century is the following:

Lim (x->∞) {π (x)/(X/LN (x))}=1

Number-theoretic highest achievement, highest achievement ... There are wood ...

3, Goldbach conjecture (+)

A large even number (>=4) must be divided into two prime numbers and, although no one has been able to prove theoretically, but I based on the scientists use the results of computer operations, if there is an even can not be split, then the even at least a few hundred numbers.

So in the ACM World (data volume is often only 2^63 below) Goldbach conjecture is established. So the split program will be able to achieve

4. Popularization of Goldbach conjecture

An integer of any >=8 must be able to be split into four primes and

Prove:

First of all, 8=2+2+2+2, (four minimum prime number and) can no longer find than 2 small prime, so when n is less than 8, it is impossible to split into four prime numbers and.

Then when n is greater than or equal to 8, it can be discussed in some cases:

(1) n&1==0 (n is even), then n must be split into two even-numbered and

So according to Goldbach conjecture, even can be split into two prime numbers and, so, n must be split into four primes and

(2) n&1==1 (n is odd), n must be split into two even +1

Since there is an even number of primes, 2, then the odd number must have the following split: 2+3+ prime number + prime number

Evidence.

5. Euler's function (Euler's formula)

The Euler function ph (n) means the number of all numbers less than N and N coprime

For example ph (12) =4,[1,5,7,11 and 12 coprime]

Euler formula

A^ph (m) =1 (mod m)

6. Fermat theorem

Fermat theorem is a special case of Euler's formula

Because when P is prime, Ph (p) =p-1 this is obvious.

Then the Fermat theorem is obtained with the Euler formula.

a^ (p-1) =1 (mod p)

P is prime number (prime)

7. Drawer principle

Drawer principle is actually nonsense, the key is to use

The nonsense is that if there are 3 apples now, put in 2 drawers, then at least one drawer will have two apples, this is nonsense.

8. Application of drawer principle

The drawer principle itself is just a piece of crap, but his use is very powerful

Now suppose there is a sequence of positive integers a1,a2,a3,a4.....an, try to prove that we can find a continuous sequence and that this is a multiple of n, the proof of this proposition uses the drawer principle

We can first construct a sequence si=a1+a2+...ai

Then for the SI to take the mold, if there is a sk%n==0, then the A1+a2+...+ak must be a multiple of n (this kind of case is proof)

The following is the opposite of the previous case, that is, any SK is not 0 for the remainder of n

For this situation, we can consider the following because si%n!=0

Then the scope of si%n is bound to be in the n-1, so the original sequence SI produced n range in the remainder of the sum (n-1), so the drawer principle came, n number into n-1 box inside, must have at least two remainder will repeat, then these two sk1,sk2 difference is necessarily a multiple of N,

And Sk1-sk2 is a sequential sequence, then the original proposition has been proved

9, judge whether n! can be divisible by M

The calculation method is to decompose m into mass factor, and to see if each of the mass factors of M can be found in n!;

How many x are in the middle of the n! (x is any number, but in general we are only talking about X as prime), the answer to this question is:
n/x+n/(x^2) +n/(x^3) ..... [has been added to the exponent of x not more than n], the proof of this theorem is very simple, here is not to repeat the

According to the above view, it is possible to calculate whether each of M's mass factors is fully contained, and if one is not included, then it cannot be divisible.

10, Factor and calculation method

The god horse is called factor and: a number of so factors and is called factor and ...

Well, here's an example: 12 of the factors and: 1+2+3+4+6+12

The method of calculation is to decompose 12 into the expression of the mass factor 2^2*3

Then his factor is: (1+2+2^2) * (1+3)

proved to be more troublesome to write, the general idea is Newton's two-term ...

11. Judging the parity of combinatorial number C (n,m)

There's a way I don't know how to prove it.

When N&m==m is an odd number, the inverse is an even number

Transferred from: http://blog.csdn.net/xieshimao/article/details/6425099