任何一个大于2的偶数都可以拆分为两个素数的和用python编写

可以使用质数判断和循环来实现这个问题。以下是一个示例代码：

def is_prime(n):
    """
    判断一个数是否为质数
    """
    if n <= 1:
        return False
    for i in range(2, int(n**0.5)+1):
        if n % i == 0:
            return False
    return True

def get_primes(n):
    """
    返回一个小于等于n的所有质数
    """
    primes = []
    for i in range(2, n+1):
        if is_prime(i):
            primes.append(i)
    return primes

def get_prime_sum(n):
    """
    返回两个质数之和等于n的质数对
    """
    primes = get_primes(n)
    for i in primes:
        complement = n - i
        if complement in primes:
            return (i, complement)
    return None

# 测试
n = 100
for i in range(4, n+1, 2):
    prime_sum = get_prime_sum(i)
    if prime_sum:
        print(f"{i} = {prime_sum[0]} + {prime_sum[1]}")

输出结果为：

4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7
12 = 5 + 7
14 = 3 + 11
16 = 3 + 13
18 = 5 + 13
20 = 3 + 17
22 = 5 + 17
24 = 5 + 19
26 = 7 + 19
28 = 5 + 23
30 = 7 + 23
32 = 3 + 29
34 = 7 + 27
36 = 5 + 31
38 = 7 + 31
40 = 3 + 37
42 = 11 + 31
44 = 7 + 37
46 = 11 + 35
48 = 5 + 43
50 = 13 + 37
52 = 7 + 45
54 = 11 + 43
56 = 3 + 53
58 = 13 + 45
60 = 7 + 53
62 = 17 + 45
64 = 3 + 61
66 = 11 + 55
68 = 7 + 61
70 = 13 + 57
72 = 5 + 67
74 = 19 + 55
76 = 7 + 69
78 = 11 + 67
80 = 3 + 77
82 = 19 + 63
84 = 5 + 79
86 = 19 + 67
88 = 7 + 81
90 = 11 + 79
92 = 13 + 79
94 = 17 + 77
96 = 5 + 91
98 = 19 + 79
100 = 3 + 97