Primes of form $N^2 - 2$ where $N$ is odd and greater than $1$

Question

Let $S$ be the set of all odd numbers $N$ greater than $1$, and $f(N) =N^2 - 2$ for all $N$. Let $P$ be the subset of $S$ in which all members of $P$ are primes. Prove that $P$ has infinitely many primes.

P should be all values of f(N) that are prime! Sorry bout that. — Bradford Gile, Dec 03 '13 at 18:57
So you are simply asking "Prove that there are infinitely many primes of the form $N^2 - 2$." — fretty, Dec 03 '13 at 19:04
@Bradford, it is best to edit the main question body (see the edit link under it) instead of explaining such a crucial point here in the comments. — Jyrki Lahtonen, Dec 03 '13 at 19:05
See the following:
http://math.stackexchange.com/questions/144334/are-there-infinitely-many-primes-of-the-form-n2-d-for-any-d-not-a-square — kvmu, Dec 03 '13 at 19:09
@kvmu I would even add that as a solution! With some extra text it would be more than adequate! — mathematics2x2life, Dec 03 '13 at 19:10

score 3 · Answer 1 · answered Dec 03 '13 at 19:06

This should be an open problem! Your problem is essentially prove that $N^2-2$ is prime infinitely often. This would be a special case of Bunyakovsky conjecture. If $N^2-2$ were not prime infinitely often, this would prove a counterexample to the conjecture (roughly speaking, it does not satisfy all the necessary conditions).

Primes of form $N^2 - 2$ where $N$ is odd and greater than $1$

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