# Is $k^{2018}+2018$ prime for some positive integer $k$ ? If yes, which $k$ is the smallest?

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).

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

$$k=129735,145563,147165,\dots$$
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 $$k^{2019}+2019$$:
$$k=16294,36688,42188,\dots$$