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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?

mike
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  • "Clearly, for $d$ a square number, there is only one prime..." - I don't see any primes for $d = 16$. – TMM May 12 '12 at 20:16
  • @TMM, updated to be "at most one prime" – mike May 12 '12 at 20:28
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    It would be surprising if this were known. Nobody knows whether there are infinitely many primes $n^2 + 1$ and there is no resolution of the problem in sight. – Will Jagy May 12 '12 at 20:38
  • See the answer to this question http://math.stackexchange.com/questions/4506/are-there-infinitely-many-primes-of-the-form-4n23 – Zander May 14 '12 at 20:31
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    @Will, everybody knows that there are infinitely many primes of the form $n^2+1$, it's just that nobody knows how to prove it. – Gerry Myerson May 19 '12 at 09:14
  • Is this not to mean that a norm from a quadratic extension is a prime? Of course there are infinitemy many, if $d$ is a negative square, hence not a square. – awllower Jun 16 '12 at 07:11
  • No, it would be easy otherwise. You are asking for the norm to be of a special form which isn't easily studied. Equivalently you are asking which primes factor in $\mathbb{Q}(\sqrt{d})$ into the special form $(n+\sqrt{d})(n-\sqrt{d})$ for some $n$. – fretty Dec 03 '13 at 19:11

1 Answers1

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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.

Douglas S. Stones
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