# primes of type norm($a+b\sqrt{-c}$) = primes of type $1$ $mod$ $c 2^{n-1}$?

Let $a,b,c,n$ be strictly positive integers where $c$ is a prime and such that the normed ring $a+b\sqrt{-c}$ is a UFD. I noticed the primes of norm($a+b\sqrt{-1}$) are $1 \bmod 4$. Also the Eisenstein primes are of type $1 \bmod 3$. So I wondered if any prime of type norm($a+b\sqrt{-c}$) = prime of type $1\bmod c 2^{n-1}$ ?

• I think there are is no prime other than $c=2$ for which the set of all $a+b\sqrt{-c}$ is a unique factorization domain. Note, for example, that when $c=3$, you need to let in things of the form $(a/2)+(b/2)\sqrt{-c}$ with $a-b$ even to get a UFD. – Gerry Myerson Jan 31 '13 at 23:14
• Gauss conjectured that $$-1, -2, -3, -7, -11, -19, -43, -67, -163$$ all have unique factorization, and this turned out true. and there are lots of real one too. . en.wikipedia.org/wiki/Class_number_problem – user58512 Jan 31 '13 at 23:18
• @user58512 No, Gauss knew that those fields have unique factorization. What he conjectured was that the list was complete (for imaginary quadratic fields). Gerry's point is that the stated ring is not the ring of integers in $\mathbb Q(\sqrt{-c})$ and need not be a UFD. For instance, $(1+\sqrt{-3})(1-\sqrt{-3}) = 2\cdot 2$, and those factors are all irreducible in $\mathbb Z[\sqrt{-3}]$. – Erick Wong Feb 1 '13 at 1:17
• thats what I meant – user58512 Feb 1 '13 at 12:31