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 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?
 A: 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.
A: Elementary solution 1
WLOG suppose that $p < q$. Let $q = p + k$ for some positive integer $k$.  
Now consider $x=k$, so
$$\{p+x,q+x\} = \{q,q+k\}$$
If $q+k$ is not prime then we are done, since $q$ is prime. Otherwise we get a new set of primes
$$
\{p',q'\} = \{q,q+k\}
$$
and we can repeat the process by adding $k$.  
Each time round the new $p'$ is necessarily prime, so this will only fail if we can produce $q'$ that is prime forever.  
However at the $q$-th iteration this must fail since
$$
\{p+x,q+x\} = \{p+qk,q+qk\} = \{p+qk,q(1+k)\}
$$
and $q(1+k)$ is not prime.  

Solution 2:
Let $x=dq,d\geq 1$ so that $q+x$ is never prime. Then one of $p+x$ is prime by Dirichlet's theorem on arithmetic progression.
