The following is a problem from my elementary number theory class, which I believe the solution has an error. I'd appreciate if someone could verify that my hunch is correct, or help me understand why I am incorrect.
We know that $\mathbb{Z}\left[\frac{1+\sqrt{-7}}{2}\right]$ is a Euclidean domain, with norm form given by $N\left(a+b \frac{1+\sqrt{-7}}{2}\right)=a^{2}+a b+2 b^{2}$. In lectures, we showed that the positive integers $n$ that can be expressed as $a^{2}+a b+2 b^{2}$ for $a, b$ integers, are those of the form $7^{a} p_{1}^{b_{1}} \ldots p_{r}^{b_{r}} q_{1}^{2 c_{1}} \ldots q_{s}^{2 c_{s}}$ where the $p_{i}$ are distinct primes congruent to $1,2,$ or $4 \bmod 7$ and the $q_{j}$ are distinct primes congruent to $3,5,$ or $6 \bmod 7$. ($a,b_i,c_j \in \mathbb{Z}_{\ge 0}$)
Now, question: Define an equivalence relation on pairs $(a, b)$, where the equivalence class of $(a, b)$ is the set $\{(a, b),(-a,-b),(a+b,-b),(-a-b, b)\}$. Describe the number of equivalence classes of pairs $(a, b)$ such that $a^{2}+a b+2 b^{2}=n$ in terms of the prime factorization of $n$.
Clearly, this is a counting argument. Here is the official solution.
The number of such pairs is the number of elements $z=x+y \frac{1+\sqrt{-7}}{2}$ such that $z \bar{z}=n$. The elements of the equivalence class of such a $z$ correspond to $\{z,-z, \bar{z},-\bar{z}\}$. If the prime factorization of $n$ is as in the statement of the question, then the prime factorization of such a $z$ must be of the form $$ \pm \sqrt{-7}^{a} \mathfrak{p}_{1}^{t_{1}} \overline{\mathfrak{p}}_{1}^{b_{1}-t_{1}} \ldots \mathfrak{p}_{r}^{t_{r}} \overline{\mathfrak{p}}_{r}^{b_{r}-t_{r}} q_{1}^{c_{1}} \ldots q_{s}^{c_{s}} $$ for some choice of $t_{i}$ between 0 and $b_{i}$, where $\mathfrak{p}_{i}$ is a prime in $\mathbb{Z}\left[\frac{1+\sqrt{-7}}{2}\right]$ with norm $p_{i}$. Replacing $z$ by $\bar{z}$ amounts to replacing each $t_{i}$ with $b_{i}-t_{i}$. Thus the number of possible choices for each $t_{i}$ is $b_{i}+1$, and the only choice that is fixed under complex conjugation is $t_{i}=\frac{b_{i}}{2}$ for all $i$ when all $b_{i}$ are even. Thus the total number of choices up to equivalence is $$ \frac{1}{2} \prod_{i}\left(b_{i}+1\right) $$ if some $b_{i}$ is odd, and $$ \frac{1}{2}\left[1+\prod_{i}\left(b_{i}+1\right)\right] $$ if all $b_{i}$ are even.
Here is my question. Doesn't $7$ also have a factorzation in $\mathbb{Z}\left[\frac{1+\sqrt{-7}}{2}\right]$? Namely, we have $N\left(1-2 \frac{1+\sqrt{-7}}{2}\right)=1^{2}-2+2 \times 2^{2} = 7$. So in our counting argument, shouldn't we include $7^a$ as well, so thus we have that the total number of choices up to equivalence is $$ \frac{1}{2} (a+1) \prod_{i}\left(b_{i}+1\right) $$ if some $a,b_{i}$ is odd, and $$ \frac{1}{2}\left[1+(a+1)\prod_{i}\left(b_{i}+1\right)\right] $$ if all $a,b_{i}$ are even.
Is this correct? Any answers helping me to understand this problem, whether or not they "fully" answer (whatever that means) the specific question, are very welcome! I will upvote them too.
[Comment promoted to answer at request of OP]
For $n=7^3$, I get only one pair (up to equivalence), which fits the formula in the offical solution, whereas your formula gives two.