TITLE: What is the smallest prime of the form $n^n+5$?
QUESTION [10 upvotes]: This question asks about the smallest prime of the form $n^n+8$. However, I haven't be able to find a prime of the form $n^n+5$ where $n$ is natural, assuming we take $0^0=1$. 
If $n$ is odd, then $n^n+5$ is even. If $n \equiv 2 \mod 6$ or $n \equiv 4 \mod 6$ then $n$ is even and $\gcd(n,3)=1$ and hence by the Euler-Fermat theorem $n^n \equiv 1 \mod 3$, and hence $n^n+5 \equiv 0 \mod 3$. 
So only numbers in the form $n=6m$ can qualify. Of course, one also needs that $n$ is not divisible by 5. I also checked $n \leq 71$, they are not prime. 
So I wonder whether there exists a prime in the form $n^n+5$, and if so, what is the smallest?

REPLY [4 votes]: $444^{444}+5$ is prime, and is the smallest of that form.  The next is $3948^{3948}+5$.
perl -Mntheory=:all -Mbigint -E 'for (1..1e5) { say if is_prime((0+$_)**$_+5); }'
It's a little faster using -Mbigint=lib,GMP or -MMath::GMP=:constant.  A bit under 0.3 seconds to find the first, albeit this is uses a robust PRP test rather than doing a proof.
See:  factordb entry for a primality certificate.