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 $p>2$; 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.

  • $\begingroup$ Ah yes! That was the missing part, thank you! $\endgroup$ – user408670 Feb 11 '17 at 21:05

The lowest pair to check is $2$ and $3$. That sum can not be represented as the product of two primes. Therefore $p > 2$. 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:

$p > 2$, therefore $p$ is $odd$


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

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

That means that we have proven this problem by contradiction.

  • 1
    $\begingroup$ I think there is a mistake in here. We checked that the lowest pair $(p,p')$ equals $(2,3)$ and indeed this sum cannot be a product of two primes. Now, we know that $p,p'>2$. We can then conclude that $p_1 \cdot p_2 = x$, where $x$ is indeed even ($x=p_1\cdot p_2)$. But we did not assume that $p_1$ or $p_2$ are greater than 2, in contrast, as I already mentioned in my OP, either $p_1$ or $p_2$ is equal to $2$ (and this leads to the contradiction, as Test123 showed). $\endgroup$ – user408670 Feb 11 '17 at 21:24
  • $\begingroup$ You just stated it again: $(p, p') > 2$. That means $(p, p') $ is an element of $]2;\infty[$. So $p$ can not be $2$. All other primes are $odd$. And $x$ is $even$. However, $odd \times odd$ is $odd$. That leads to a contradiction as well, as I stated in my answer. $\endgroup$ – manuel-hoelzl Feb 11 '17 at 21:49

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