若p和q为连续奇素数且p+q=2r，求证r必为合数

Let's walk through this proof step by step to make it clear why this result holds for every pair of consecutive odd primes.

First, let's restate the problem clearly:

If p and q are consecutive odd primes, and p + q = 2r, then r is a composite number.
Example: When p = 3, q = 5, p + q = 8 = 2 × 4, where 4 is composite.

Now let's prove this for all cases:

• Since p and q are odd primes, their sum is an even number (odd + odd = even). This means 2r is even, so r is an integer—no fractional values here, which is essential for our proof.
• By definition, consecutive odd primes mean p < q, and there are no other prime numbers between p and q. If there was a prime between them, they wouldn't qualify as consecutive primes.
• Calculate r directly from the given equation: r = (p + q)/2. It's straightforward to see that p < r < q: the average of two distinct numbers always lies strictly between them. For example, (7 + 11)/2 = 9 (between 7 and 11) or (11 + 13)/2 = 12 (between 11 and 13).
• Since r is an integer sitting strictly between p and q, and there are no primes in that interval (because p and q are consecutive), r cannot be a prime number.
• We also know r is greater than 1: the smallest pair of consecutive odd primes is 3 and 5, which gives r = 4. Any larger pair will result in an even larger r, so r is always at least 4.
• By definition, a positive integer greater than 1 that is not prime is composite. Therefore, r must be composite.

That's the full proof—every step follows directly from basic number theory definitions!

内容的提问来源于stack exchange，提问作者Sandeep Gautam