CHAPTER 4
On the history of the problem of recognizing prime numbers by two sided theorem in particular Wilson’s theorem and its consequences
In order to a nonone natural number "m>1" be prime, the necessary and sufficient condition is that "m" aliquot (m1)_{}
Proof. At first, we prove theorem inverse. We suppose that
_{} (1)
If "m" is not prime number, a natural number like "d" is one of numbers_{}. d (m1)! . In other words, according to the thesis _{}. So, according to (1), it is impossible. Now, we conclude Wilson’s theorem from Dirichlet’s theorem.
Suppose that "p" is odd prime number and_{}. If a number like "b" exists so that _{} then, congruence _{} has two non congruent answers whit module "p" exactly.
Proof. At least two non congruent answers exists for congruence. Because:
_{}, _{}
For _{}we show that congruence hasn’t another answers except these two answers. For this purpose suppose that_{}. Therefore _{}. So _{}then _{} or _{}. In first case is _{} and in second case is:
_{}.
Suppose that "_{}" be prime number and _{}If congruence _{}hasn’t answered, then _{}and if has answer_{}.
Proof. If "_{}", then assertion is obvious. So suppose "_{}" is prime number. According to theorem, if _{}, then there will be an individual number "n", so that 1_{} and _{}. Now, we call number "m" and "n" as corresponding. If _{}hasn’t answer, then for corresponding numbers "m" and "n" we will have _{}. Therefore, we can classify the integer numbers in the interval "1" up to _{}to_{} couples of corresponding numbers. Multiplication of all these corresponding numbers is congruent with _{} to module "p" and obviously it is also congruent with_{}. Therefore _{}. If the congruence _{} is solvable so there is an answer like "b" that "_{}".
Another answer of this congruence is_{}. Now we consider _{} as residual number. It is obvious that these _{} number, aren’t answers for congruence. Therefore we can convert them to corresponding pairs.
Multiplication of these pairs is congruent to module "p" with _{}and_{}, so:
_{}
We see that grand result of Dirichlet's theorem is Wilson’s theorem. In 1771, Lagrange presented a very complicated proof of Wilson’s theorem. But Wilson’s theorem is obtained simply from theorem (4.1.2).
If "p" be prime number, then "_{}"
Proof. In (4.1.2) theorem we set_{}, especially _{}has answer. So:
_{} _{}
Therefore, Wilson’s theorem is a test for primality of considered number.
Although this test is unusable, because _{} is a very big number.
Consider that "p" be prime and_{}. So congruence _{} is (is not) solvable
if:
_{} _{}
Proof. Consider that congruence has answer. Therefore, according to Wilson’s theorem:
_{} ; _{}
Now, factorize result is obtained by multiplication of both side in "–1". If congruence hasn’t answer, proof it the same.
I. First time, Werring in his book "thoughts in algebra" (1770) concerned theorem to sir john Wilson (17411793), so this theorem is known as Wilson’s theorem Although Leibniz had said an equivalent assertion. The inverse of theorem was propounded by Lagrange (1771) who also proved.
II. Wilson's theorem gives a simple criterion for investigation the investigating the primality of nonone natural number "m". It is enough to identify that _{}is divisible by "m" or not. But this criterion has no practical advantage. Because _{} in crease with "m", rapidly.
III. According to Wilson's theorem, for each prime number "p" number _{}is a natural number. Every prime number like "p" so that _{}is a natural number, has called as a Wilson’s prime number. By testing, it’s clear that only 5, 13, 563 are Wilson’s prime numbers among prime numbers:
_{}
If "p" is a prime number and _{}, then:
_{}
Proof. Congruence (1) is established for _{} by Wilson’s theorem and it is provocations for p=2.
We suppose that _{}and congruent are in module "p". Since_{}, so:
_{}.
According to _{}and Wilson’s theorem:
_{}.
Suppose that "p" is an odd prime number. Order (1) in last example results for _{}:
_{} (2)
So, if "p" be _{} or _{} then:
_{} (_{}) (3)
_{} (_{}) (4)
For example:
_{} (_{})
If and only if odd prime number "p" to be in form of "_{}" then_{}.
The number of natural numbers like "n" that _{} is composite for them is infinite.
If _{} and _{}, then "m" is composite. Conversely, if "m" is composite and distinguished from 4, it adapts in (*).
If _{}be prime number and greater than "5", for every natural number of _{}:
_{} (5)
The condition of necessary and sufficient in order that natural number of _{} be prime:
_{} (6)
If multiply expression of _{} in the two side relation (6), then to obtain Wilson's theorem:
_{} ; _{} (7)
Here, we present some determinate function formula by Wilson's theorem:
_{}
_{}
According to Wilson’s theorem, supposing that "k" is an integer number, for every _{} and _{}:
_{} (5)
It is necessary to mention that we can build a lot of identification functions by using the Wilson’s theorem, that for example, we present one of them:
[1]_{} (6)
(_{}: Number integer part)
By Wilson’s theorem even can not testing number "100001" limit of _{} defines in below form:
_{} (11)
Of course, we sow important results of this theorem. Very mathematicians presented a lot of theorems and testing for characteristic of prime numbers that we will attempt which showing on the history of obtained results.
Here, with present other formula for characteristic of numbers that very interesting and it express by Euler’s arithmetic function namely _{}we follow subject:
_{}
We can determine number _{} by below determinant:
_{}
In these functions number of_{}exist. It is obvious that working with it is impossible and even for a number near to_{}, it is also unfaultable. For this reason, these identification functions can’t be useful in real and in fact we can use them to present pure theories, because for example the number" 69!" is greater than _{} (a 99 digits number).
The congruence _{}is establish for every prime value of "n" and no natural composite number adapts in that , except of 4, 6, and 22.
If_{}, then "n" is a prime number, if and only if :
_{}
If "p" is a prime number then for every natural number "n":
_{}
If and only if natural number _{} to be prime number then for every natural number "n", so that _{}:
_{}
If and only if natural number _{} to be prime number then for every natural number "n":
_{}
At first we express this interesting criterion with language which its discoverers said and then by mathematics language as a theorem.
In Pascal mathematical triangle, we transfer every row two cell toward right, so that the coefficients in "n" the row, namely the numbers:
_{}
Start from "2n" column and finish in "3n" column. Then mentioned triangle changes some part of it is shown below:
k n 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 
0 
1 























1 


1 
1 




















2 




1 
2 
1 

















3 






1 
3 
3 
1 














4 








1 
4 
6 
4 
1 











5 










1 
5 
10 
10 
5 
1 








6 












1 
6 
15 
20 
15 
6 
1 





7 














1 
7 
21 
35 
35 
21 
7 
1 


8 
















1 
8 
28 
56 
70 
56 
28 
8 
9 


















1 
9 
36 
84 
126 
126 
10 




















1 
10 
45 
120 
11 






















1 
11 
In this new table, in "n" row, entries which are divisible by "n", has been printed in bold letters. Like "1" and "1" in row "1" and "2" in row "2", "3" and "3" in row "3" and etc.
If and only if the numbers of column "1" in above table to be prime then all coefficients this column, become printed in bold letters. According to the structure of the table, only and only an entry of "n" row is in "k" column that_{}. In other word only and only an entry of "n" row is in a specific "k" column when _{}.
If and only if for every nonone natural number "k" to be prime number, then for every "n" (_{}):
_{}_{}
If "N" is odd and greater than 9 and if and only if "N" is prime number then below numbers:
_{}
aren’t perfect square.
Because, it is obvious:
_{}.
Some numbers are representable as sum of two perfect squires in two different ways. For example:
_{}
representations which their difference is just in arranging of factors aren’t distinct. We can factorization these numbers to multiplication of two factors. Fernekle and Mersenne discovered this thought but widely. Apparently we express solution of factorization problem we say some primary observation.
I. necessary condition for an odd number to be sum of two squares is that it be in "_{}" form.
II. We consider that
_{} (1)
If _{} then _{} repeating the above reasoning relation (1) will result a relation like _{}and_{}.
III. If equality (1) is established and _{} but_{}, then_{}, _{} so: _{}.
Every natural number is composite if it has two different representations of sum of two squares.
Also: _{}
_{}
So: _{}
_{}
The mentioned decomposition method in above theorem is Euler’s method. For using in decomposition of an odd number like "N", at first we must sure that this number is in the form of "_{}". (For example:_{}).
Because If "_{}", then "_{}".
For example:
_{}
_{}.
_{}
_{}
_{}
Attention. Prime numbers which are in form of"_{}" aren't present in sum of two square.
All of numbers which are in form of "_{}" are composite except of 5. Because _{}
So, "_{}" may be prime number only when one of below equalities is established:
I) _{} , II) _{}
It is resulted from (1) that _{} and in this case _{} is equal to "5" as a prime number. And it is resulted from (2) that _{}and _{} that is a composite number.
For every two natural numbers "a" and "b" and every prime number "p", has just one representation as:
_{}.
The number 16000001 which is equal to 1040^{2}+3860^{2} is decomposable to multiplication of two factors or multiplication of three factors, because:
_{}
If we suppose that _{} so that "a", "b", "c" and "d" are natural numbers and _{} then it is possible to prove that If "_{}", then "_{}".
According to _{}and two mentioned factorization be distinct, so we can prove:
I)_{}
II)_{}
It is clear that multiplication of two numbers in "_{}" form is: _{}
Number "N" is convertible to different forms namely "_{}" and "_{}" that_{}, so we can prove:
_{}
For example, we can convert the number 4361 to "_{}" and we can factorization it.
Now, dispense with expression another ways and ways can be useful for identification "N", comparatively greater number.
Now we express certain solution of identification number problem by identification formula_{}.