This is an extension of my question asked in "Prime Ideals in Imaginary Quadratic Fields", but I feel this deserves a new section. I noticed that when $p (> 2) \in \mathbb{Z}$ is prime and remains prime in $\mathbb{Z[\frac{1 + \sqrt{-47}}{2}]}$, there does not exist $m,n,k \in \mathbb{N}$ such that $2^kp = m^2
+ 47n^2$. I proved this by contradiction (essentially since p is prime in the new ring we see that p divides m and n, so p divides $2^k$ in the integers). However, this statement appears to be true if p is not prime in the ring $\mathbb{Z[\frac{1 + \sqrt{-47}}{2}]}$.
For instance we see that 3 is not prime in $\mathbb{Z}[\frac{1 + \sqrt{-47}}{2}]$ because $x^2 – x + 12$ is reducible in $\mathbb{Z}_{3}$. For instance 1 is a root; we also see that $1^2 + 47 = 48 = 3*2^4$. I am wondering if this is true in general, My guess is to look at a root of $x^2 – x + 12$ (call it $x_{o}$) and then find the corresponding root $2x_{o} – 1$ of $x^2 + 47$.
I also observed this for the prime 7; notice that 2 is a root of $x^2 – x + 12$ and 3 is a root of $x^2 + 47$. Notice that $3^2 + 47 = 56 = 7 * 2^3$. But still I am stuck. Any help?
Here is a link to my previous related question
Here is an update of my status:
(1) I was able to prove this when $(p) = PP^{*}$, where P and its conjugate are principal prime ideals.
(2) What do I do when the ideal P is not principal? The class number of $\mathbb{Z[\frac{1 + \sqrt{-47}}{2}]}$ is 5 so I cannot multiply P by a non principal ideal to make it principal, right? Or can I? If so, by what?
Best Answer
We can split $(2)$ as $\left(2,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1-\sqrt{-47}}{2}\right)$. Those are both nonprincipal because $2$ cannot be the norm of any principal element. If the class group has order $5$ like you claim, then either prime above $2$ is a generator because every non-identity element in a group of order $5$ is a generator. And in fact, $\left(2, \frac{1+\sqrt{-47}}{2}\right)^5=\left(\frac{9+\sqrt{-47}}{2}\right)$.
Now assume $p$ isn't prime in that ring. Say $x^2-x+12=(x-u)(x-v)\mod p$. Then the ideals $\left(p, \frac{1+\sqrt{-47}}{2}-u\right)$ and $\left(p, \frac{1+\sqrt{-47}}{2}-v\right)$ have norm $p$. If we look at them in the class group, since the prime above $2$ is a generator, we can multiply by it enough times so that we end up at the identity. This means that we have a principal ideal whose norm is some power of $2$ times $p$. But then we can complete the square and get $a^2+47b^2=2^kp$.
Your example with $3$ ends up being $(3)=\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(3,\frac{1-\sqrt{-47}}{2}\right)$. The prime $\left(3,\frac{1+\sqrt{-47}}{2}\right)$ is not principal, and neither is $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1+\sqrt{-47}}{2}\right)$, but $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1+\sqrt{-47}}{2}\right)^2$ is, as you showed: $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1+\sqrt{-47}}{2}\right)^2=\left(\frac{1+\sqrt{-47}}{2}\right)$. So this has norm $12$ so multiplying by $2$ gives $1+\sqrt{-47}$ which has norm $48$ as you need.
You can also find a number of norm $96$ by using the other prime above $2$. You get $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1-\sqrt{-47}}{2}\right)^3$ being principal, because it equals $\left(\frac{7+\sqrt{-47}}{2}\right)$. This has norm $24$ which means $7+\sqrt{-47}$ has norm $96$ as you wanted.
We could have also done the previous paragraph abstractly: if $\mathfrak{p}_1$ and $\mathfrak{p}_2$ are the two primes over $2$, and $\mathfrak{q}$ is the prime $\left(3, \frac{1+\sqrt{-47}}{2}\right)$ we talked about, then we know that in the group, $\mathfrak{q}+2\mathfrak{p}_1=0$ (where the group operation is addition instead of the natural multiplication) because we wrote $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1+\sqrt{-47}}{2}\right)^2$ as a principal ideal. But we also know that $\mathfrak{p}_1+\mathfrak{p}_2=0$ because the product of the primes $\mathfrak{p}_1$ and $\mathfrak{p}_2$ was exactly the principal ideal $(2)$. So then these two are negatives of each other and $\mathfrak{q}-2\mathfrak{p}_2=0$. But the group has order $5$ so we can add $5\mathfrak{p}_2=0$ to both sides to get $\mathfrak{q}+3\mathfrak{p}_2=0$. This is how we knew that we could get $\left(3,\frac{1+\sqrt{-47}}{2}\right)\left(2,\frac{1-\sqrt{-47}}{2}\right)^3$ to be principal.
Finally, to make sure you understand the ideas here, here's a question to leave you with: If $2^kp=m^2+47n^2$, for some pair of $k$ values $k_1$ and $k_2$, with $m$ and $n$ both odd, show that either $5|k_1-k_2$ or $5|k_1+k_2+1$. Also the converse is true: if $2^{k_1}p=m_1^2+47n_1^2$ for some odd $m$ and $n$, then for any $k_2\geq 2$ satisfying either of the divisibilities above, there are another pair $(m_2, n_2)$ that have $2^{k_2}p=m_2^2+47n_2^2$.
So for example, knowing what you wrote in your question, you should be able to find $m$ and $n$ with $2^6\cdot7=448=m^2+47n^2$ with $m$ and $n$ odd.