primes whose difference is a multiple of $n$ Hello everyone this is my first question here so if there are any suggested edits  feel free to participate!
Is it true that for every $n$ there two prime numbers $p$ and $q$ with $n\mid p-q$?
I have verified this empirically but  am not sure if there is a proof for this
 A: Yes. This is true. There are only finitely many ($n$ to be precise) residue classes modulo $n$. There are infinitely many primes. Therefore (by the pigeonhole principle) at least one of the residues classes must contain more than a single prime (actually infinitely many, but that's besides the point). If $p$ and $q$ are two primes in the same residue class then their difference is divisible by $n$.
A: Yes, but I don't expect it's trivial.
To prove it, choose any prime q not dividing n.  Then look at the congruence $$x \equiv q \;\; mod \,n$$
By Dirichlet's theorem on Primes in Arithmetic Progressions there are infinitely many prime solutions and any of these will work.
A: By the Dirichlet's box principle, if we take the first $n+1$ primes $2=p_1,p_2,\ldots,p_{n+1}$ at least two of them must have the same remainder when divided by $n$. So there is a solution of:
$$ p_i\equiv p_j\pmod{n} $$
with $i\neq j$ and $\max(p_i,p_j)\leq C n \log n$. Obviously, large prime numbers may fall only in $\varphi(n)$ residue classes $\!\!\pmod{p}$, so the previous inequality can be improved a bit. At last, the Bombieri-Vinogradov theorem proves that primes are almost uniformly distributed over the previous residue classes. However, to provide explicit bounds for the Chebyshev's bias is an extremely difficult problem.
