TITLE: Is there any $k$ , for which we can prove that $n^n+k$ is never prime?
QUESTION [7 upvotes]: Is there any positive integer $k$, such that we can prove that $n^n+k$ is not prime for any positive integer $n$ ?

$$n^n+1805$$ has a prime factor not exceeding $43$ up to $n=1805$. However, for the multiples of $1806$ this is in general no more the case. 
I think that for no $k$, $n^n+k$ must have a "small" factor for all $n$.
Is this true, and if yes, does this destroy any hope for a proof ?

REPLY [9 votes]: The sequence consisting of smallest $n$ depending on $k$ is on OEIS as A087037. As stated on the page, for $k=(6m-1)^3$ there is no such $n$. For completeness I repeat the proof here.
Let $k=(6m-1)^3$ and consider $N=n^n+k$. For $n$ odd, $N$ is even, hence composite. If $3\mid n$, then $N$ is a sum of cubes, hence composite, since $a^3+b^3$ factors as $(a+b)(a^2-ab+b^2)$. Finally, if $n$ is even but not divisible by $3$,
$$n^n+k\equiv(\pm 1)^n+(-1)^3\equiv 1-1\equiv 0\pmod 3$$
and hence is composite.
This method gives $k=125$ as the least number for which we know $n$ doesn't exist, but as the OEIS entry notes, existence of $n$ is unknown for many smaller $k$, the least being $k=8$.