if $p$ and $q$ are successive odd primes and $p + q = 2r$ then $r$ is composite,

Question

As $p$ and $q$ are successive odd primes, for example if $p = 3$, $q = 5$ then $p + q = 8 = 2 \times 4$ here 4 is a composite number. But how to prove it generally in all conditions?

Answer 1

Are you requiring $p$ and $q$ to be twin primes? As in $p = r - 1$ and $q = r + 1$? Then $p + q = (r - 1) + (r + 1) = 2r$. This means that $r$ is even and thus composite. e.g., 5 + 7 = 12 which is twice 6.

But if you don't require $p$ and $q$ to be twin primes, but merely that every integer strictly between $p$ and $q$ be composite, then we have $p = r - k$ and $q = r + k$, where $r$ may be odd or even but certainly composite, and we're not too concerned with the value of $k$. e.g., 7 + 11 = 18, which is twice 9.

The foregoing assumes we're only talking about positive primes. But considering negative primes hardly changes things: $-2 + 2 = 0 = 2 \times 0$.

Answer 2

Hint: $\,p \lt \dfrac{p+q}{2} \lt q\,$.

Answer 3

Suppose $p < q$. If $r$ were a prime number, then we would have $p < r < q$. But that is a contradiction since $p$ and $q$ are supposed to be successive prime numbers. It follows that $q$ must be a composite number.

Answer 4

Suppose $p<q$. Then

$2p < p+q < 2q \Rightarrow 2p<2r<2q \Rightarrow p<r<q$

How $p$ and $q$ are successive primes, then $r$ can't be prime.

Answer 5

Assume $r$ is also prime. Then two cases: $$r<p \ \ \text{or} \ \ q<r.$$ First case: $$2r=r+r<p+q \ \ \text{contradiction}.$$ Second case: $$p+q<r+r=2r \ \ \text{contradiction}.$$ Hence $r$ is not prime.