Two prime numbers{3,5}.
{3,5}→{3+3= 6，3+5=8，5+5=10} →{6,8,10}.
{{10}→(5+5=10 = 3 + 7) → 7}Increased by 7 →{3, 5, 7}.
Note changes:{3,5}→{ 6,8,10}→ {3, 5, 7}.
{3,5,7}→{ 3+3= 6, 3+5=8, 5+5=10, 5+7=12, 7+7=14}→{ 6,8,10,12,14}.
{{14}→(7+7=14 = 3 + 11) → 11} Increased by 11 →{3, 5, 7, 11}.
Note changes: {3,5,7}→{ 6,8,10,12,14}→ {3, 5, 7,11}.
The same logic would be:{3,5,7,11}→{ 6,8,10,12,14,16}→ {3, 5, 7,11,13}.
The same logic would be:
{3,5,7,11,13}→{ 6,8,10,12,14,16,18,20}→ {3, 5, 7,11,13,17}.
【Note: if a prime number is added, n consecutive even numbers (n≥1) can be obtained.】
Sequential continuation, ......
If you expand infinitely in the above specified mode: {3, 5, 7, 11, 13, 17, 19, 23,...
Get: { 6,8,10,12,14,16,18,20,22,...
The above is: the continuity of prime numbers can lead to even continuity.
Get: Goldbach conjecture holds.
If it is mandatory: Authenticity stops at an even number 2n.
{{3, 5, 7, 11, 13, 17, 19, 23,...,p1}→{ 6,8,10,12,14,16,18,20,22,...,2n}.
{3, 5, 7, 11, 13, 17, 19, 23,...,p1} ↛(2n+2).
∀p+∀p≠2n+2 } It can be proved that: It violates the "Bertrand Chebyshev" theorem.
∴ {3, 5, 7, 11, 13, 17, 19, 23,...
→{ 6,8,10,12,14,16,18,20,22,...
Get: Goldbach conjecture holds.