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.
0 件のコメント:
コメントを投稿