The distribution of prime numbers is a difficult problem in number theory. The proof of prime theorem attracts many scholars and puzzles many wise people. Mathematicians have proved thousands of theorems on the premise of Riemann conjecture.
In 1737, the Swiss mathematician Leonhard Euler published a formula:
Here (1.1) is called Euler product formula. Where n is an integer, p is a prime and s is a real number.
Euler studied this formula. Let s = 1. For large x, Euler obtains an asymptotic formula
From 1792 to 1793, Gauss, a German mathematician, studied the number of primes in 1000 adjacent integers around x. he found that for large x, the “average distribution density” of primes should be 1/logx, which is shown as follows     :
where, the number of primes not exceeding x is π(x). General expression
Here (1.3) is Gauss’s guess. Represents the number of primes in r adjacent integers near x.
Let x = 1238, r = 238, by (1.3) get
According to (1.3), we can divide x into n parts. In this way, we can get π(x).
Let integern, x = nr, by (1.3) can get
Set up , Substituting (1.4) get
Let x = ar, get
Here (1.5) is a prime theorem without remainder. It is equal to the prime theorem π(x) ~ Lix.
Let x = 64, r = 16, a = x/r = 4, by (1.6) calculation,
In this paper, we improve the Riemann ladder function and prove a strong prime number theorem by using the metens theorem .
Before proving (1.6), we use a simple method to prove (1.3). Obviously, if we prove (1.3), we will prove (1.5).
2. The Elementary Proof of Prime Number Theorem
It is difficult to prove the prime theorem. It is especially difficult to prove the prime theorem in a simple way. We prove the prime theorem (1.3) by using the simple method according to Euler’s asymptotic formula
Let the integer r, prime p, x − r ≤ p ≤ x obviously 
Let x = 16, r = 12, x − r = 4, 4 ≤ p ≤ 16, π(16) − π(4) = 4, by (2.1) get
By (2.1) can ge
Let , get
By (2.2) and (2.3) can get:
According to Euler’s asymptotic Formula (1.2), get
From this, we get
Take main item , get:
Take the main item again , From this, we get
By (2.5) Substituting (2.4),
This proves (1.3).
Now, let’s look at the mean of the prime p, by (x − r) ≤ p ≤ x can get
According to (2.4), we change x into an average (x − r/2) get
For example x = 271, r = 34, 237 ≤ p ≤ 271, prime number 241, 251, 257, 263,
By (2.6) get
actual π(271) − π(237) = 4.
We divide x into n parts, n is a positive integer.
Let x = 2n, r = 2, can get (x − r/2) = (2n − 1), by (2.6) we get
Therefore, we can improve the Riemann ladder function.
3. Improved Riemann Ladder Function
In 1859, German mathematician Riemann published a formula :
Here (3.1) is Riemann’s prime distribution formula . Where J(x1/n) is called Riemannian ladder function . The calculation is very complicated.
According to the definition of Riemann ladder function:
A new ladder function is obtained by improving the ladder function
Let x ≥ 2, Integer n, prime p, get 
Here (3.2) is an improved Riemannian prime distribution formula. It is strictly equal to π(x).
For example x = 16, a = 16/2 = 8, n = 1, 2, 3, 4, 5, 6, 7, 8.
According to (3.2), a prime number is crossed each time, and the calculation is as follows:
In fact, π(16) = 6, it is now prove that (3.2).
If there is a prime p between 2n − 2 and 2n, it must be
Set up , by (3.2) and (3.3) get,
According to (3.4), we can get the
By (3.5) get
Confirm (3.2) certification.
4. Merdens’ Theorem
In 1874, the mathematician Merdens proved that :
Here (4.1) is called: merdens theorem . Where the mertensian constant
Let the coefficient c (x) be obtained from (4.1)
For the convenience of using (4.2), we transform (3.2).
For example x = 8, a = 8/2 = 4, by (3.2) can get
From the above
Replace (4.3) with (4.2), where
It is obtained by substituting (4.3) above
Let a = x/2, s(x) denote the logarithmic part, which is obtained by substituting (3.2) above
Here (4.6) is strictly equal to π(x).
For example x = 8, x/2 = 4, by (4.6) get
By (4.2) calculation coefficient
Actual π(8) = 4.
By (4.6) we can prove the prime theorem with remainder.
5. Prime Theorem with Remainder
The remainder estimation of prime number theorem is very complicated. The key is to use the mertensian constant to calculate the coefficient.
Let’s look at the remainder of (4.6). Among
Let even numbers y, x > y, according to (3.2) a = x/2, can get,
From this we get
where s(x) is the main term followed by the remainder. Now let’s look at the remainder of (5.1). Among
Here (5.2) is the key remainder. Let’s see c(x) and c(y + 2n).
In order to let c(y + 2n) approach M, let y = 2[x1/2]. According to (4.1) and (4.2), x tends to infinity, and we obtain
By (5.2) get
Let’s look at a very small x.
For example x = 16, y = 4, (x − y)/2 = 6, Substitute (5.3) and calculate according to (4.2)
By (5.3) get
1.3074861 near 0.929918.
x The bigger, the closer (y + 1)M.
Now, let’s look at the scope of the remainder.
By (5.3) Substitute (5.1),
For the y of sufficient size, it is obvious that the estimation is
Substitute (5.4) get
Let’s look at the coefficient,
From (4.1) and (4.2) Confirm M/2 < c(2n) < 1, obviously
By (5.6) Substitute (5.5) get
Let y = 2[x1/2], by (5.7) can get
By (5.8) Confirm, The prime theorem (1.6) is proved.
Previously, we discussed the Euler asymptotic formula
From (6.2), we prove the prime theorem in a simple way
Here (6.2) is a prime theorem without remainder.
We also discuss the distribution formula of prime number of Riemann
We make a great improvement on the distribution formula of Riemannian prime numbers
This is a function strictly equal to π(x). In this way, we prove that Riemann’s conjecture is correct in principle.
According to the (6.3) theorem, we prove the strong prime number theorem
Here (6.6) is a prime theorem with remainder. Where s(x) and Lix are equal.
 Atiyah, M. and Singer, I.M. (1963) The Index of Elliptic Operators on Compact Manifolds. Bulletin of the American Mathematical Society, 69, 422-433.