TITLE: Are there infinitely many primes of the form $n^2 - d$, for any $d$ not a square?
QUESTION [8 upvotes]: Clearly, for $d$ a square number, there is at most one prime of the form $n^2 - d$, since $n^2-d=(n+\sqrt d)(n-\sqrt d)$.
What about when $d$ is not a square number?

REPLY [7 votes]: There's a host of conjectures that assert that there an infinite number of primes of the form $n^2-d$ for fixed non-square $d$.  For example Hardy and Littlewood's Conjecture F, the Bunyakovsky Conjecture, Schinzel's Hypothesis H and the Bateman-Horn Conjecture.
As given by Shanks 1960, a special case of Hardy and Littlewood's Conjecture F, related to this question, is as follows:

Conjecture:  If $a$ is an integer which is not a
  negative square, $a \neq -k^2$, and if $P_a(N)$ is the number of
  primes of the form $n^2+a$ for $1 \leq n \leq N$, then \[P_a(N) \sim
  \frac{1}{2} h_a \int_2^N \frac{dn}{\log n}\] where $h_a$ is the
  infinite product \[h_a=\prod_{\text{prime } w \text{ does not divide
  } a}^\infty \left(1-\left(\frac{-a}{w}\right) \frac{1}{w-1}\right)\] taken over
  all odd primes, $w$, which do not divide $a$, and for which $(-a/w)$
  is the Legendre symbol.

The integral is (up to multiplicative/additive constants) the logarithmic integral function.