Can there be $N$ such that $P \equiv a \mod n$ a prime $\Rightarrow P+N$ also prime? Show there's no number which, when added to each of the primes of a particular non-trivial* modulo class, yields only primes.
* i.e. a modulo class containing more than one prime, $a \mod n$ with $gcd(a,n)=1$
Note: This only came up as something I wanted to use in answering another question. Well it's a grand generalisation of what I needed. I assume it's true, because there's no such thing as a free prime number and also for some vague distributiony reason, neither of which I could formalise.

Proposition: Let $\gcd(a,n) = 1$ and $N \neq 0$. Then there are infinitely many primes $P \equiv a \pmod{n}$ such that $P + N$ is composite.
Proof: Let $q$ a prime not dividing $n\cdot N$. By the Chinese remainder theorem, there is an integer $b$, $0 < b < \lvert n\cdot q\rvert$, such that $b \equiv a \pmod{n}$ and $b \equiv -N \pmod{q}$. By Dirichlet's theorem on primes in arithmetic progressions there are infinitely many primes $P \equiv b \pmod{n\cdot q}$. For such primes we have $P \equiv a \pmod{n}$, and $P + N \equiv 0 \pmod{q}$, so $P + N$ can be prime at most once.
Aside: You didn't rule out $N = 0$ explicitly, so the answer to the question as stated is "Yes, $N = 0$".
