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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.

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  • $\begingroup$ Peter, assuming you were the user who suggested the edit, see this page for help getting access to your account again. $\endgroup$ May 20, 2013 at 21:17
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    $\begingroup$ Mathematica says $n^{5000}+n^{2013}+1$ is not prime for $1<n<3000$. $\endgroup$
    – Jared
    May 21, 2013 at 1:21
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    $\begingroup$ According to Maple is not prime for $1<n<5000$. $\endgroup$
    – P..
    May 21, 2013 at 4:45

1 Answer 1

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$n=23205$ 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.

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  • $\begingroup$ I can't imagine the number of digits the number has. And I can't even be sure that it is a prime. $\endgroup$
    – Inceptio
    May 22, 2013 at 7:46
  • $\begingroup$ 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. $\endgroup$ May 22, 2013 at 7:56
  • $\begingroup$ Well, I assume an elementary approach to the problem is still open. $\endgroup$
    – Inceptio
    May 22, 2013 at 8:03
  • $\begingroup$ Certainly so; at the moment we don't even know if the search would finish at all. $\endgroup$ May 22, 2013 at 8:10
  • $\begingroup$ All $3$ values check out using Mathematica's PrimeQ function. I'm not sure what method it uses to test for primality. $\endgroup$
    – Jared
    May 23, 2013 at 0:05

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