Introducing two New Sieves for Factorization Natural Odd Numbers

For each non-prime odd number as FF = pppp, if we consider mm/nn as an approximation for pp/pp and choose kk = mmnn, then by proving some lemmas and theorems, we can compute the values of mm and nn. Finally, by using Fermat’s factorization method for FF aannaa 4kkFF as difference of two non-consecutive natural numbers, we should be able to find the values of pp and pp. Then we introduce two new and powerful sieves for separating composite numbers from prime numbers.


Introduction
There are many features for identification of prime numbers from non-prime numbers. In this paper, we

Development of Fermat's factorization method
We know each non-prime odd number as = (3 ≤ < ), can be written as a difference of squares of two nonconsecutive natural numbers as following: It is clear that by assuming = , we have = 2 .
Therefore, for each natural number bigger than 1 as , we can write 4 as difference of squares of two even numbers or odd numbers, in which is a non-prime odd number: If we assume = , in this case, we can write: Definition: In this paper, the expressions ��√4 � + � If is a square number and assume = , it means that: Since and are natural numbers, so will be divided on .This means that is a non-prime number. By assuming = , we will have: When is a non-prime number, both and are natural numbers bigger than 1.

Lemma 1:
Whenever is a natural even number such that > ( −2) 2 8 and is a natural number, in this case, we will have: Proof: From the basic algebra, we have: If + 2 − 1 isn't the biggest integer number smaller than √ 2 + , we should have at least: Because � + 2 − 1� 2 < 2 + , so we will have > ( −2) 2 8 .

Lemma 2:
Suppose is a natural number and is a natural even number such that > ( +2) 2 8 . Then, we will have: Proof: From basic algebra, we have: If − +2 2 isn't the biggest integer component smaller than √ 2 + , then we should have: is the biggest integer number smaller than √ 2 − and from bracket function definition, we will have: And so: 2 8 .
In this paper, the approximate value of a number as will be shown in the form of ~( ) or �.

Theorem 2:
In a non-prime odd number as then the result of (1) will be square and we will have: Proof: Since = − is even, then from lemma (2) we have: In this case, according to lemma (1), we have:

Lemma 3:
If is a natural odd number and is a natural number such that > ( −1 2 ) 2 , then we will have: Proof: From the basic algebra, we will have: is not the biggest integer number smaller than √ 2 + , then, we should have at least: According to lemma (1), we will have: Based on lemma (1), we have: and according to the relation (1), we can write According to lemma (1), we have: According to lemma (1), we should have: Based on the relation (2), we will have:

Theorem 4:
For each natural odd number as = , (1 ≤ < ), the value of ( , 1) will be square when lies in interval ( ) except in a case that = .
Proof (1): If k > , in this case 2 > 2 and because the result of = 2 − 2 is even, then from lemma (1), we will have: According to lemma (1), we should have: If < , in this case 2 < 2 and because the result of = 2 − 2 is even, then from lemma (1), we will get: Based on the lemma (1), we should have: Now, from (1) and (2), we will have , thus if = , we can write:

Proof (2):
According to theorem(3), when we assume = � = and = 1, we will have: and shown the difference of by then for every integer number which is lied in that interval, the result of ( , 1) will be square. Therefore, We will have the Maximum value of when = 1. In this case, we have = 4√ .
If the number of natural numbers located in interval ( , ) are demonstrated by , so we will When the value of is maximum, then the value of will be maximum too. , ℎ , then the value of ( , 1) will be square.
Proof: For proof in the first case, we propose > Based on lemma (1) , we should have: Therefore we should have − < 2� + 1 .
In the case < , the proof is similar as above.
Theorem 6: For each non-prime odd number as = (1 ≤ < ), if we assume = �(� − � ) 2 � + 1 and doesn't be square, then the value of ( , ) will be square so that At that rate for any k value, we can write: In the sequel we have: For example when = 89 × 911 = 81,079 and = 7 , we can calculate k values as bellow: We can observe when = 6 , we can't find any value for k.

Note 2: For any value of k as a natural number, exist a nonnegative integer number as
For example, when = 17 × 23 = 391 and propose m=5 and n=2, then we should have: have � =~( / ) = / and = , then the value of ( , 1) would be square whenever: Proof: From theorem (3), we have: According to = � and = � as well as assuming = difference of will be shown by , so we will get: ℎ by choosing = 10 and = 1 … ����������� , we should have: If we show the number of natural numbers that is located in interval ( , ) by , so for any we should have: In other way ⌊ ⌋ = ⌊ ⌋ = − 1, therefore for each value of and , we can conclude: For any value of S , we define ( ): as below Then for any ( ): we have: We can calculate for F as below: If we propose i as a natural number from zero to , then define s value as = − .

For any we have:
For any we have: In this case from = √ 16 we should have: Therefore: For calculated the value of ( ):1 ,we should have: Therefore: Therefore in ( , 1) test we can do as bellow: We can observe in any s zone for a natural number as F, when the value of F increases, then the difference between √ and ( ):1 increases, because :

3.Introducing −
Since for each odd composite number we have: Therefore, if we propose S=x+y+1 and R=xy, then we will have: By assuming = 2 − so that becomes a square number, we will to have: In general case, for each value of S, we can obtain its corresponding p as follows: Therefore by assuming = 2 − = 2 , we will have = − = + .
For any F=p q we can conclude that = � � and for any of the two different values of P, we have: It is seen that for one definite number as F and a constant value of ∆ , with increase in values of 1 and 2 , the ∆ S is decreased. In other words, with decrease in p values, the values, increase. When the difference between and √ is represented by n, we will have: In this case we have: By considering the points mentioned, we can recommend a sieve to identify an odd composite number is prime or composite. This is based on the premise that if we are able to find values of S in a way that the value of becomes square, therefore F will be a composite number.
By considering the relation K=2R+S-1 we will have: Therefore, for different cases, we have: It is noteworthy to mention that in this sieve by finding the first response point, the compositeness of the For proving, by assume = (… 3 2 1 ���������� ) 2 we should have: and in a way that is even, we have: ( is a natural number)

Proof: Since any natural number can be represented by 2 + in a way that if the number is even, then = 0 and if it is odd, then = 1 and that is a non-negative integer number; therefore, by
assuming that ∆ is a perfect square number, we will have: We will continue the operation above until the value of in 2 + equal to 1.
When we assume:
Since we have: Therefore the number of the tests that we can do in a definite will be as below: Thus we will have: Therefore: For the case that is even, we have 1 = 0 and its proof is like the previous one; then for this case we will have: Therefore, the number of tests, when ∆ is even, will be as below: Thus we have: Then in general case we will have: Since we have ∆ = (2 1 − 1) − (−1) = 2 1 = 2 therefore for two value of A with difference 2 −1 so that (2 − 1) = 2 × 2 −2 − and (2 − 1) = 2 × 2 −1 − (when i is a odd number and located in interval from 1 to 2 −1 − 1 ) then we have: Therefore the number of tests can be calculated by 2 = 2 −2 .
In an ascending arrangement of the A values, for both of its values which lie in a symmetrical position, we will have: Since the result of (2 −2 − 2 − 1)is positive, then N must be a number greater than 3. Therefore the total number of the remaining values of A decreases from 2 −2 to 2 −3 ,which is represented by .
Thus in general case when is odd, for every test we will have: Then the total number of tests equal to = 2 −3 .
For the case in which is even, the proof process is completely similar and for each test we have: In a way that the total number of each test is calculated like that of the previous case, thus we will have = 2 −3 , therefore the proof is complete. that a ́= ± , and r can have one of the digits from 1 to 9, then we will have: It means ́− 2 count by 2 and the proof is complete. In this sieve we can use theorems 7 and 8 in tests as below.
It means the number 2 − − − 2 . In other words − test is positive and the proof is complete.  Proof: It means that the 2 − 2 and the proof is complete.

In tests for a definite S, only for a one value of x, the test result is positive. It is because if we assume
for the two different values 1 and 2 the test result is positive, then it means 1 and 2 will have the same outcome. If in the first (1) all the tests of are negative, it means that the odd number under the test is a prime number.
We must do the tests only in (1) If we assume that the number of + tests in (1) is equal to w, then the total number of tests in − sieve can be calculated by − + = � − 10 � × 10 .
Note 8: If = −1 … 3 2 1 ���������������������� will be a prefect square number, then we must to have∶ If we represent the odd values of 1 for 1 by 1 and the even values of 1 for 1 by 1 ,thus we will have: 3,5,7,9)  When the result of ( , 1) is a square number, we will have: Therefore, when we have = + , the value of ( , ) will be a square number and the desired S value, will be = + . Considering that the values of m and n are odd or even, through k=mn we can determine if the values of S are odd or even.
In order to use this sieve between two values of , we will do as follows: By calculating the minimum value of S in the sieve interval of ∆, we will have: To calculate the maximum value of S in the sieve interval of ∆, we will do as follows: In the best case, If we consider 1 and 2 as consecutive integer number and 1 = 2 = , then we will have: If one is located in the sieve interval of ∆, we represented it by ∆ .For each in a sieve interval of ∆ , we have: In orders that the sieve should be easier, between the two consecutive integer values of , especially for big values of F, it is better to divide the distance into y equal parts in which each part is an independent sieve zone.
∆ − represents the it h of sieve zone related to K. In a particular case when the distance between the two integer values of consecutive is selected as one sieve zone, we will do as follows: In zones of ∆ − , the length of the sieve intervals is decreased by increasing the i. In other words: In order to utilize the − sieve in the zone of = 0, it will only suffice to use the integer values of , located in this zone, as the k in ( , 1) tests.
If the value of �