Is there a positive integer $k$, such that $k^{2018}+2018$ is prime ? If yes, which $k$ is the smallest ?

According to my calculation, $k$ must be greater than $10^5$ and therefore such a prime must be a gigantic prime (at least $10^4$ digits). Also, I did not find a reason that there is no such prime (such as forced divisors or algebraic factors).

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    $\begingroup$ Well, obviously even numbers and multiples of 1009 need not apply. +1. $\endgroup$ – Oscar Lanzi Jan 29 '18 at 1:01
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    $\begingroup$ Obviously, I do not know how to do this, so I am going to run it on my python program. $\endgroup$ – Yash Jain Jan 29 '18 at 1:05
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    $\begingroup$ I just found a probable prime! : $\color \red {k=129\ 735}$ $\endgroup$ – Peter Jan 29 '18 at 11:52
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    $\begingroup$ @Peter $k=129735$ is the smallest, indeed. No solutions up to $10^{5}$. $\endgroup$ – Cleyton Muto Jan 30 '18 at 10:41
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    $\begingroup$ Peter, post as an answer. $\endgroup$ – Ed Pegg Mar 12 '18 at 1:22

Not sure why this question has no answer since comments contain one, but here are couple more values found by brute force search in Maple. Smallest (probable) primes of these form are given by:

$$ k=129735,145563,147165,\dots $$

Also the leading coefficient is positive, the polynomial is irreducible (by Eisenstein for example) and values it represents are coprime (by $\gcd(f(2),f(3))=1$ for example). Thus it satisfies conditions for Bunyakovsky conjecture, by which it should represent infinitely many primes.

Bonus: Since we are already in $2019$, here are couple examples for $k^{2019}+2019$:



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