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Let $\mathbb{P}_{\mathbb{C}}$ be the set of Gaussian primes and $\mathbb{P}_{\mathbb{N}}$ the set of primes in $\mathbb{N}$.

Let $\pi_{\mathbf{C}}(\sqrt{n})$ be the number of Gaussian primes with norm $\leq \sqrt{n}$ and $\pi_{\mathbf{N}}(n)$ be, as usual, the number of primes $\leq n$ in $\mathbb{N}$. Recall that norm($x+iy$)=N($x+iy$)=$x^2 +y^2$; hence, my taking of a square root above.

I am interested to know what the order of magnitude is for $$\frac{\pi_{\mathbf{N}}(n)}{\pi_{\mathbf{C}}(\sqrt{n})}$$i.e. Has the extension of the definition of primes increased/decreased the relative density of primes with respect to their set of definition? A rather quixotic question could be " Is there a general asymptotic for the number of primes in an arbitrary infinite field with the definition of being a prime as usual?"

Fact:

  1. Prime numbers of the form $4n + 3$ are also Gaussian primes.

  2. Gauss's circle problem which asks for the number of Gaussian integers with norm less than a given value is presently unresolved. I think this is tangentially related to the asymptotic I am looking for.

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    The _asymptotic_ solution to the circle problem is just $\pi r^2$. The problem itself is about the exact value.2011-03-15
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    In dillema of which answer to accept. How can I distribute the incentive I have in hand? To Akhil for exposition or to Qiaochu for mentioning the last theorem.2011-03-15
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    @Qiaochu, that is true and that was why I said it only relates to my case tangentially.2011-03-15
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    I don't understand why you say "recall that norm($x+iy$)=$x^2+y^2$"... this is not the usual definition, and as indicated by the answers, leads to some confusion. The norm, or absolute value, of a complex number $x+iy$ is $\sqrt{x^2+y^2}$.2011-03-15
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    @mjqxxxx, that is a standard definition for norm of Gaussian integers as also stated in [wikipedia](http://en.wikipedia.org/wiki/Gaussian_integer).2011-03-15

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