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Alledged greatness[edit]

Untitled[edit]

The article claims Chen is one of the greats of the 20th century, but no citation is given. This claim sounds dubious to me. Someone removed the same claim from Grothendieck's article.Aliotra (talk) 15:25, 11 December 2010 (UTC)[reply]

There is nothing to be gained from a discussion of the relative greatness of mathematicians. This is so, with and without citations. — Preceding unsigned comment added by 81.95.123.216 (talk) 07:19, 19 May 2014 (UTC)[reply]

Goldbach guess any greater than 2 even, can be expressed as the sum of two prime numbers[edit]

When all the integer]] $N>3$ , is bound to the existence of $X$ ,

$N+X$ and $N-X$ are the prime - Goldbach's conjecture.

because even 2N= (N+X) + (N-X).

Framework[edit]

 In 300 BC in ancient Greece, Ella Torsten Ni created a sieve method, can produce all the prime numbers within an arbitrary large number of:
to get all theprime numbers less than a natural number   $n$  , as long as in multiples of   $2-n$      will not be more than ${\sqrt {n}}$ 
primes can be all off.。

The sieve method can be summarized as

1, if n is a composite number, then it is a factor of D satisfying 1<d≤ ${\sqrt {n}}$ 。

2, if the natural number n cannot be less than ${\sqrt {n}}$ for any prime numberdivisible, then $n$ is a prime number.

The 2 Chinese characters into English letters indicate the equivalence of content

   $n=p_{1}m_{1}+a_{1}=p_{2}m_{2}+a_{2}=\dots =p_{k}m_{k}+a_{k}.$ ....(1)

The . $p_{1},p_{2},\dots ,p_{k}$ ,order prime 2，3，5，....。 $a$ ≠0。 If $n<P_{k+1}^{2}$ , then n is a prime number。

We can think of (1) content and equivalent transformation of congruence grouprepresentation：

  $n\equiv a_{1}{\pmod {p_{1}}},n\equiv a_{2}{\pmod {p_{2}}},\dots ,n\equiv a_{k}{\pmod {p_{k}}}$ 。...（2）

Because of (2) $p_{1}$ , $p_{2}$ ,..., $p_{k}$ . two two coprime, based on the Chinese Remainder Theorem (Chinese remainder theorem) that, for a given $a_{1}$ , $a_{2}$ ,..., $a_{k}$ ， (2) has a unique solution in the range of $p_{1}$ $p_{2}$ ... $p_{k}$ 。

Example[edit]

For example: k=1, $n=2m_{1}+1$ ， solutionn=3 ，5，7， The （3， $3^{2}$ ） all prime number interval.。

K=2,， $n=2m_{1}+1=3m_{2}+1$ ， solution, n=7 13, 19；

$n=2m_{1}+1=3m_{2}+2$ ，Solution n=5, 11, 17， 23。

The（5， $5^{2}$ ） all prime number interval.

k=3	$5m_{3}+1$	$5m_{3}+2$	$5m_{3}+3$	$5m_{3}+4$
$n=2m_{1}+1=3m_{2}+1=$	31	7,37	13,43	19
$n=2m_{1}+1=3m_{2}+2=$	11,41	17,47	23	29

  |} The （7， $7^{2}$ ）all prime number interval。

Copy this be all prime numbers within a not leak to the given number.。

 For all the possible  $a_{1},a_{2}\cdot ,a_{k}$ ，(1)and(2)In the $p_{1}$  $p_{2}$ ... $p_{k}$ Within the scope of，

（ $p_{1}-1$ ）（ $p_{2}-1$ ）( $p_{3}-1$ )...（ $p_{k}-1$ ） Solution。

Reasonable framework of Goldbach conjecture[edit]

How to make the two natural number addition and subtraction are primes, N+X become the prime, N-X is also a prime number. According to the division theorem:"given a positive integer a and b, b ≠ 0, has a unique integer q and r (0 ≤ r < b),the a=bq+r". Then according to the congruence theorem: "every integer exactly with the 0,1,2,3,..., m-1 congruence (mod m)". So, given a natural number N(N>4), can only be said:

$N=p_{1}m_{1}+e_{1}=p_{2}m_{2}+e_{2}=\dots =p_{k}m_{k}+e_{k}.$ (3)

The . $p_{1},p_{2},\dots ,p_{k}$ order prime $e_{i}=0,1,2,...,P_{i}-1$ 。

${\frac {p_{k}^{2}}{2}}$ < N < ${\frac {p_{k+1}^{2}}{2}}$

Now is the time to ask, whether there is X

$X=p_{1}h_{1}+f_{1}=p_{2}h_{2}+f_{2}=\dots =p_{k}h_{k}+f_{k}.$ （4）

$f_{i}$ ≠ $e_{i}$ ，

$f_{i}$ ≠ $p_{i}-e_{i}$ 。

If X<N-2,， then N+X and N-X are prime, because they correspond to (1) (2) type 。

Example[edit]

Let N=20， $20=2m_{1}+0=3m_{2}+2=5m_{5}+0$ ；

${\frac {5^{2}}{2}}$ < 20 < ${\frac {7^{2}}{2}}$

 $e_{1}=0$ , $e_{2}=2$ , $e_{3}=0$ .

Structure x x	$5h_{3}+1$	$5h_{3}+2$	$5h_{3}+3$	$5h_{3}+4$
$X=2h_{1}+1=3h_{2}+0=$	21	27	3	9
$f_{i}$ ≠ $e_{i}$ ， $f_{i}$ ≠ $p_{i}-e_{i}$	$f_{1}=1$ ， $f_{2}=0$ ， $f_{3}=1$ .	$f_{1}=1$ ， $f_{2}=0$ ， $f_{3}=2$ .	$f_{1}=1$ ， $f_{2}=0$ ， $f_{3}=3$ .	$f_{1}=1$ ， $f_{2}=0$ ， $f_{3}=4$ .

The four solutions are：21，27，3，9。 Less than N-2 X， 3 and 9。, we learn that, 20+3 and 20-3 is a pair of prime numbers; 20+9 and 20-9 is a pair of prime numbers. This is the use of primality test: minimum residual is not zero。And $N+X<P_{k+1}^{2}$ , N+X and N-X is a pair of prime numbers. Because (N+X) +(N-X) =2N。 This is the famous Goldbach conjecture conjecture,The problem is transferred to the elementary number range.

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