P^2+Σ2P

Every odd number greater than 5 can be expressed as the sum of three primes. (A prime may be used more than once in the same sum.) This is called Goldbach's weak conjecture. I think that it is almost the same as Goldbach's conjecture . a＞2 2a+1=P1+P2+P3 2a=(P1+P2+P3)-1 P is prime numbers. a＞2 2a+1=O1+O2+O3 2a=(O1+O2+O3)-1 O is odd numbers. For example, 2a=6 (a=3) 6=(3+2+2)-1 This is prime numbers. 6=(3+3+1)-1=(5+1+1)-1 This is odd numbers. 2a=8 (a=4) 8=(5+2+2)-1=(3+3+3)-1 This is prime numbers. 8=(7+1+1)-1=(5+3+1)-1=(3+3+3)-1 This is odd numbers. 2a=10 (a=5) 10=(7+2+2)-1=(5+3+3)-1 This is prime numbers. 10=(5+5+1)-1=(7+3+1)-1=(9+1+1)-1=(5+3+3)-1 This is odd numbers. ∴ (P1+P2+P3)-1⊂(O1+O2+O3)-1⊂2a Therefore, Goldbach's weak conjecture is correct .

Every even integer greater than 2 can be expressed as the sum of two primes. This is called Goldbach's conjecture. 2a=P1+P2 when a is 2, 4=2+2. In the case of a>2, P1 and P2 must be odd numbers. P is prime numbers. O is odd numbers. ∴ P1+P2⊂O1+O2⊂2a I have found the pattern of prime numbers which is based on odd numbers. P1+P2⊂O1+O2⊂2a is never broken. It is quite natural that you can keep finding P1. P2=2a-P1 Therefore, Goldbach's conjecture must be correct.

When you keep adding integers infinitely, O (not Zero) is usable to express this difficulty. It is called Landau symbol. O(n), O(n^2), O(nlogn) This is expansion anyway. Traveling salesman problem is well known as Landau symbol. In this case, there are 4 places. The salesman starts from A and comes back to the same place. He must choose the fastest. This is easy one. However, when you keep adding places, the salesman would be confused. It is said that there are (n-1)!/2 choices. If you have 10 places, (10-1)!/2=(9*8*7*6*5*4*3*2*1)/2=181440. There are 181440 choices. If you have 14 places, (14-1)!/2=3113510400. This is expansion. Max=T(1) k=2 T(k)＞T(1) Max=T(k) k＜n and T(k+1)=T(n)＞T(k) Max=T(k+1)=T(n) ∴

Twin prime seems to be related to Galois theory. Galois described field extension and intermediate field. F=coefficient, K=field extension, M=intermediate field. G=Galois group. Therefore F⊂M⊂K G(K/M) G(K/F) ∴ G(K/M)⊂G(K/F) There is cubic equation. x^3+ax+b=(x-α)(x-β)(x-γ)=0 α+β+γ=0, αβ+βγ+αγ=a, αβγ=-b G e f1 f2 g1 g2 g3 α α β γ β α γ β β γ α α γ β γ γ α β γ β α e⊂H⊂G H1=(e,g1) H2=(e,g2) H3=(e,g3) H=(e,f1,f2) H e f1 f2 α α β γ β β γ α γ γ α β I define F⊂M⊂K and G(K/M)⊂G(K/F) Therefore, G=F and H=M. K⊂M⊂F This is upside-down. ∴ K(G(M))=M, G(K(H))=H

I describe symmetry which is based on Galois theory. I think that it must be related to prime numbers. Quadratic equation is like a mirror. There is the formula. f(x)=0 have two solutions, α and β. ∴α+β=-a As you know, f(α)=f(β)=0. If f(α)=β and f(β)=α, f(f(α))=α and f(β)=α. I define α+β=-a. Therefore f(α+β)=f(-a) f(α)+f(β)=β+α=-a β+f(β)=-a ∴f(β)=-a-β=α+β-β=α. There is symmetry.

When you count prime numbers in each 1000000, there are less prime numbers. This is known as power law . I have found the pattern of prime numbers . Шn+1=Шn^2+2(д-1)Шn This is absolute convergence . Therefore, it would be replaced as f(z). I can pick up all prime numbers infinitely, so, in this case, f(z) must be the sum of prime numbers in each 100. Z=Z(T)=x(T)+iy(T) T is a parameter. f(Z)=f(Z(T)) C is the curve of this graph, and you can divide it by N. T0=Ta,T1,T2,.....,TN=Tb Z0=Za,Z1,Z2,.....,ZN=Zb ΔZn≡Zn-Zn-1, ΔTn≡Tn-Tn-1 Z=Z(T) and ΔZn≡Zn-Zn-1 ∴

This figure is piling squares which are prime numbers. The pattern includes complex plane , so I think that I can describe Riemann surface. θ=2nπ 　(n=0,±1,±2........) and arg Z=θ+2nπ (-π＜θ≦π). Therefore, arg Z change according to n. n=0 →　-π＜arg Z≦π n=1 →　π＜arg Z≦3π n=5 →　9π＜arg Z≦11π n=-5 → -11π＜arg Z≦-9π ∴(2n-1)π＜arg Z≦(2n+1)π Complex planes Z depend on n, and these are piled infinitely.