Show that the sum of two consecutive primes is never twice a prime. Show that the sum of two consecutive primes is never twice a prime.
My first language is not English, and would just like to understand the problem. Does someone could give me a concrete example? P.S. I can remove the question later if you find that the question is not relevant to the website.
 A: Let $a, b \in \mathbb{N}$, $a < b$. Then $a < (a + b) / 2 < b$.
Let $p_n$ denote the $n$th prime number. Following the same rule,
$$ p_n < \frac{p_n + p_{n+1}}{2} < p_{n+1}$$
Since $p_n, p_{n+1}$ are consecutive primes, $(p_n + p_{n+1})/2$ is not prime.
A: Example: $2 + 3 = 5 \neq 2p$ for any prime p. 
Beyond 2, the sum of two consecutive primes is at least even. Where $(p_n)_{n \ge 0}$ enumerates the primes, the question amounts to:
$$
\text{For $n > 0$, is } \frac {p_n + p_{n+1}} 2 \text{ prime?}
$$
Examples: $(11 + 13)/2 = 6$ is not prime; $(19 + 23)/2 = 21$ is not prime.
A: Consider the odd primes modulo $4$. If $p_n \equiv 1 \pmod 4$ and $p_{n + 1} \equiv 3 \pmod 4$ (or vice-versa), then $p_n + p_{n + 1} \equiv 0 \pmod 4$ and therefore $$\frac{p_n + p_{n + 1}}{2} \equiv 0 \textrm{ or } 2 \pmod 4.$$
For example, $3 + 5 = 8$, and half that is $4$; $5 + 7 = 12$ and half of that is $6$.
What if both $p_n$ and $p_{n + 1}$ are congruent to $1$, or both to $3$? It might be productive to switch to modulo $6$. If $p_n \equiv 1 \pmod 6$ and $p_{n + 1} \equiv 3 \pmod 6$ (or vice-versa), then $p_n + p_{n + 1} \equiv 0 \pmod 6$ and therefore $$\frac{p_n + p_{n + 1}}{2} \equiv 0 \textrm{ or } 3 \pmod 6,$$ which is in either case divisible by $3$.
On second thought, maybe that was not such a good idea. But as you did not want the whole answer, I suppose I can leave it at this, and I'm not going to say anything about the special case of $2 + 3$.
