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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?

  • 0
    "Clearly, for $d$ a square number, there is only one prime..." - I don't see any primes for $d = 16$.2012-05-12
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    @TMM, updated to be "at most one prime"2012-05-12
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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.2012-05-12
  • 0
    See the answer to this question http://math.stackexchange.com/questions/4506/are-there-infinitely-many-primes-of-the-form-4n232012-05-14
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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.2012-05-19
  • 0
    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.2012-06-16
  • 0
    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 $n$.2013-12-03

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