# What is the smallest integer $n$>1 such that $n^{5000}+n^{2013}+1$ is prime?

Which is the smallest integer $n>1$, such that $$n^{5000}+n^{2013}+1$$ is prime ? Since $x^{5000}+x^{2013}+1$ is irreducible over $\mathbb{Q}$ and has value $1$ for $x=0$, there should be infinitely many such $n$, if Bunyakovsky's conjecture is true.

• Mathematica says $n^{5000}+n^{2013}+1$ is not prime for $1<n<3000$. May 21, 2013 at 1:21
• According to Maple is not prime for $1<n<5000$.
– P..
May 21, 2013 at 4:45

$n=23205$ produces the smallest prime value of the polynomial (aside from the trivial $n=1$). Interestingly, $23205=3\times 5\times 7\times 13\times 17$.
$n\in\{44579, 55754, 78120, 78515, 94154, 99045\}$ produce all the remaining primes with $n<10^5$. In all cases, OpenPFGW has been used to find the primes and prove primality using Brillhart-Lehmer-Selfridge method.
• I'd be happy to see someone confirm my findings using an independent tool for checking primality of large numbers (there are quite a few of them). As for the number of digits, the prime corresponding to $n=23205$ has 21828 digits and the one for $n=55754$ has 23732 of them. May 22, 2013 at 7:56
• All $3$ values check out using Mathematica's PrimeQ function. I'm not sure what method it uses to test for primality. May 23, 2013 at 0:05