Any even number greater than 4 can be expressed as the sum of two prime numbers.

P2x( 1, 1) and its supremum are derived by the method of satisfying the expectation of prime numbers, so as to prove 2x ≡ P 1+P2, (x > 2).

It is explained that π( 1)≠0, π (1) = 1.

Lemma 1. Establish the density function of prime number distribution: y = xπ (x)/x, and get

(x/㏒ x) 1a)。 ⑴

Certificate. Establish a function: y = x π (x)/x, then π (x) = (x/㏒ x) ㏒ y.

∫limπ(x)/x = lim 1/㏒x，(x→∞)。 [ 1]

We have lim x π (x)/x = lim x1㏒ x, (x→∞).

∵ x 1/㏒ x= e，lim xπ(x)/x=e= ymin，(x→∞)。 ㏒ ymin= 1。

When x > a, ymin < y ≤ ymax.

∴ (1) hold. Prove lemma 1.

Lemma 2. Let p2x( 1, 1) be the number of prime numbers p 1 or P2 be 2x = P 1+P2. When X is a constant, (P 1, the number of groups of P2). X is greater than

The natural number of 2, 2 < p 1 ≤ p2.

p2x( 1, 1)≥[((2x-3)/㏒(2x-3)-((x- 1)/㏒(x- 1))㏒ymax)(x/㏒x-π(2))/((x- 1)/2)]+ 1

=[k(x)]+ 1，(a