TITLE: Given two primes $p$ and $q$, show that there always exists an $x$ such that only one of $p+x$ and $q+x$ is prime
QUESTION [5 upvotes]: Given any two primes $p$ and $q$ with $p \neq q$, show that there always exists an $x \in \mathbb{N}$, such that only one of $p+x$ and $q+x$ is prime.
How do I show this?

REPLY [6 votes]: Lemma: There exist arbitrarily long chains of consecutive composite natural numbers.
Proof: Let $n>2$ be given.  For each $2\le k\le n$, $n!+k$ is obviously divisible by $k$.  Therefore, every natural number from $n!+2$ through $n!+n$ is composite.  Since $n$ was arbitrary, you can make consecutive composite chains as long as you like.

Now, we return to the problem given above.  Without loss of generality, assume $p<q$.  Consider some chain of at least $(q-p)$ consecutive composite numbers, and let $N$ be the largest prime number less than the members of that chain.  
Then $p+(N-p)$ is prime and $q+(N-p)$ is composite, since $p+(N-p)=N$ which was given to be prime and $q+(N-p)=N+(q-p)$ is either in the chain of composite numbers that we considered or between N and the start of that chain.