Great operation

The even general formula is: 2n, where n is an integer greater than 1, and 2n can decompose the prime factor, that is, 2n$=$N1N2N3..., where N1, N2, N3, ... are all prime numbers. 2n, N1N2N3... must be written as the sum of two identical odd numbers or the sum of two identical even numbers, whichever comes first. That is, it can be learned: 2n$=$(m$+$m), m$=$N1N2N3.../2, when m is a prime number (specially pointed out: when n$=$2, 2n$=$2$+$2), the proposition is proved. When m is a composite number, it is proved as follows: When m is an even number, m adds an odd number or subtracts an odd number to exhaust all odd numbers, and since more than 2 prime numbers must exist in the odd number, it is sure to find the sum of the two prime numbers to represent any even number (2n). When m is an odd number, m plus an even number or an even number can also exhaust all odd numbers, and since more than 2 prime numbers must exist in odd numbers, the sum of two prime numbers must be found to represent any even number (2n). . That is, 2n$=$[(N1N2N3.../2-a)$+$(N1N2N3.../2$+$a)] ...1, whether a is an odd number when m is an even number, or a is an odd number when m is an odd number, 1 It can always be established.the Goldbach conjecture can be proved.