# Infinitely many prime numbers of the form $n^{2^k}+1$?

I am not a specialist of number theory, so please excuse my ignorance: is the following question still an open problem? Let $k \in \mathbb{N}^*$, are there infinitely many prime numbers of the form $n^{2^k}+1$?

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What's known about the problem if both n and k are allowed to vary? –  Qiaochu Yuan Feb 16 '10 at 4:42
@Qiaochu: In this problem, allowing n and k to vary among positive integers is the same as setting k=1 and letting n vary. –  Bjorn Poonen Feb 16 '10 at 5:14

Your question is still open. It is a special case of Schinzel's Hypothesis H applied to the polynomial $f(x)=x^{2^k}+1$.

As Bjorn mentions in his comment, the case of $k=1$ is a particularly famous unsolved problem. It is the fourth of Landau's problems (Edmund Landau was a famous German number theorist during the early twentieth century).

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Thank you Ben, this is useful. –  Portland Feb 16 '10 at 5:04
This may not be clear from the Wikipedia article: Bunyakovsky's conjecture is not known for any particular polynomial of degree greater than 1. –  Qiaochu Yuan Feb 16 '10 at 5:11

It took me a while to find this: http://www.pnas.org/content/94/4/1054.full

Anyway by Friedlander and Iwaniec (1997). They proved that there are infinitely many primes of the form $x^2 + y^4 .$ They mention near the end that they do not have a proof for primes of the form $x^2 + y^6$ but would like one. So there is a way to go to settle $x^2 + 1.$

FYI, what I did (not remembering title, authors, anything but the result) was write a program to give the primes $x^2 + y^4$ and put the first dozen in Sloane's sequence site search feature.

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I think that the best result thus far is Iwaniec's proof that there are infinitely many positive integers n such that n^2+1 is divisible by at most 2 primes. –  Ben Linowitz Feb 16 '10 at 5:23
oeis.org/classic/A028916 –  Charles Sep 13 '10 at 16:03