# Does there exist some prime $k$ for which there will be exactly two primes of the form $n!+k$?

This is a question related to my recent question Conjecture: “For every prime $$k$$ there will be at least one prime of the form $$n!\pm k$$” true?

Using PARI/GP I searched for the number of primes of the form $$n!+k$$ possible for each prime $$k\le 2300$$. I observed that for some prime $$k$$, all numbers less than $$24$$ were found except $$2$$.

Note that when $$n \ge k$$, $$n! \pm k$$ cannot be prime as $$k$$ will be a factor of $$n! \pm k$$. This means that there are a finite number of primes of the form $$n! \pm k$$ for each $$k$$.

User Peter and I, searched for some prime $$k$$ for which $$n!+k$$ has two primes upto $$k\le 10^5$$ and found none in the range. Here is the code we used in PARI/GP if anyone wants to extend the search:

forprime(k=2,10^7,s=0;for(n=1,k,if(s<3,if(ispseudoprime(n!+k)==1,s=s+1)));if(s==2,print(k)))

## Conjecture:

For any prime $$k$$ there cannot be exactly two primes of the form $$n!+k$$.

Can prove/disprove this? Any ideas, heuristics, guesses are welcome.

Update: Upto $$k \le 10^6$$, $$k=714377$$ is the only candidate remaining for which we have only found one prime of the form $$n!+k$$ till $$n\lt 3000$$. If a second prime of the form $$n!+714377$$ is found for the range $$3000 \lt n \lt 714377$$, then the conjecture is disproved!

• Can someone explain the downvotes? – Mathphile Jul 15 '19 at 15:02
• Although I don't see a quick proof, I think it extremely unlikely. For $n\ge 3,\ 6\mid n!$. Therefore, $k=2$, the only prime $n!+k=3$ and $k=3$ the only prime is $n!+k=5$. For $k>3,\ k=6m\pm 1$. Thus $n!+k=6N+6m\pm 1=6M\pm 1$. Hard to believe that for $M=\frac{n!}{6}+m$, there is some $m$ for which only two instance of $n$ render $6M\pm1$ prime. – Keith Backman Jul 15 '19 at 17:16
• @KeithBackman hence why it's rather interesting that the PARI/GP code may suggest otherwise ;) – Mr Pie Jul 15 '19 at 18:03
• Any pair of n,m such that adding k to their factorials is prime, will knock out a value $p=m+n$ such that $p! +k^2{{p}\choose{n}}$ and $p!+k^2{{p}\choose{m}}$ are both 3-almost primes. Taking $k=1$ and choose functions that are primes, we would get to knock out primes. I know it's too stupid. – user645636 Jul 24 '19 at 15:22