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I know that it still is not known whether the sequence $n^2+1$ contain an infinite number of prime numbers.

I guess that this is also not known for any sequence of the form $n^k+1$ where $k\geq2$ is an even natural number.

But what if we look at all these sequences together and then raise a question does the set of their values contain an infinite number of prime numbers?

What exactly do I mean by this? Well, let us define the set $P$ as $P=\bigcup_{i=2}^{\infty} \{n^i+1:n\in\mathbb N\}$.

Does $P$ contain an infinite number of prime numbers?

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It is equivalent to the first question. In fact, if $k$ is odd, then $n^k + 1$ is not prime, except for $n=1$. If $k$ is even, then $n^k$ is a square, so you can write it as $(n^s)^2+1$.

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  • $\begingroup$ it's not equiVALENT $\endgroup$
    – Asinomás
    Commented Jun 3, 2016 at 20:00
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    $\begingroup$ why not? Could you explain? $\endgroup$
    – Exodd
    Commented Jun 3, 2016 at 20:02
  • $\begingroup$ I think you needed to say if $k$ has an odd prime factor then $n^k+1$ is not prime. $\endgroup$
    – Asinomás
    Commented Jun 3, 2016 at 20:11
  • $\begingroup$ @CarryonSmiling Exodd's claim is true, though, and sufficient: suppose $\{n^i+1: i, n\in\mathbb{N}\}$ contains infinitely many primes. Then (by Exodd's claim) all but finitely many of those are of the form $n^{2j}+1$, hence of the form $(n^j)^2+1=m^2+1$. While what you say is true, it's not relevant. $\endgroup$
    – Noah Schweber
    Commented Jun 3, 2016 at 20:15
  • $\begingroup$ Oh ok, I see, thank you, my mistake. $\endgroup$
    – Asinomás
    Commented Jun 3, 2016 at 20:16

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