Prove two subsequent primes cannot be written as a product of two primes Suppose we have two subsequent primes, say $p$ and $p'$. Prove their sum cannot be written as a product of two primes, say $p_1$ and $p_2$.

I wanted to proof by contradiction. I started by thinking about parity of the sum. Suppose $p=2$, then $p'=3$. But this sum cannot be written as a product of two primes $p_1$ and $p_2$. So we know that $p&gt;2$; this implies that both $p$ and $p'$ are odd, so $p+p'$ is even. This means that either $p_1$ or $p_2$ must equal $2$. This results in: 
\begin{equation} p+p'=2p_1.\end{equation}

Now, how can I finish the proof?
For $p=2$ then it's immediate. For odd primes $p,p'$ you would have $p_1p_2=even$ namely $p_1=2$ and $p_2=\frac{p+p'}{2}$. But then $p_2$ is a prime that lies between two consecutive primes $p,p'$. Contradiction.
The lowest pair to check is $2$ and $3$. That sum can not be represented as the product of two primes. Therefore $p &gt; 2$. Because of that $p$ must be $odd$. So the sum must be $even$. To get an $even$ product, one of the two factors must be $even$.
So we got two facts: 


  $p &gt; 2$, therefore $p$ is $odd$


and 


  $p_1 \times p_2$ = $x$, whereas $x$ is $even$


That means either $p_1$ or $p_2$ must be $even$. That contradicts with our first conclusion that $p &gt; 2$ and $p$ must be $odd$.

That means that we have proven this problem by contradiction.