Spring 2017, problem 45

Let $q=p+k$. As $k$ is even, $k=2\times k'$.Then $p+q= p+(p+2\times k')=2\times (p+k')$.

But $ p$ $<$ $ p+k' $ $<$ $p+k=q$. So $p+k'$ is a composite number ,because is between two consecutive primes,

and then is a product of at least $2$ natural numbers greater than$ 1$.

Hence $p+q$ is a product of at least $3$ (not necessarily different) natural numbers greater than $1$.

Suppose p+q cannot be written as a product of 3 integers. then p+q is either prime or a product of two primes. Clearly p+q is not prime because p+q is even. So it is a product of two primes; also one of those primes is 2. So p+q=2r where r is prime, that is r is the average of two prime numbers and is itself prime. Thus p and q cannot be consecutive primes. This proves the contrapositive

Define $n$ as an odd natural number greater than 1. First, it can be shown that the quantity

$ \frac{n + (n+2)}{4} $

is also a natural number greater than 1.

$ \frac{n + (n+2)}{4} $

$ \frac{2(n+1)}{4} $

$ \frac{1}{2}(n+1) $

But $n$ is an odd number, so $n+1$ must be even. Therefore, $ \frac{1}{2}(n+1) $ is a natural number. The smallest odd natural number greater than 1 is 3. $ \frac{1}{2}(3+1) $ = 2, so the answer must always be greater than 1. Now, consider the problem given. $q$ can be written as $p+2$ since the two are consecutive primes. As previously shown,

$ \frac{p + (p+2)}{4} $

is a natural number greater than 1. So, the sum $p+q$ can always be written as the product of at least 3 natural numbers:

$p + q = 2(2)(\frac{p + q}{4})$.

Q.E.D.