For given positive integers $s$ and $t$, how many integer solutions are there to $x^2+7y^2=3^411^s23^t$?

Question

For given positive integers $s$ and $t$, how many solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ are there to $$x^2+7y^2=3^411^s23^t\,?$$

Working in $\mathbb{Z}[\sqrt{-7}]$, I know that I am trying to find how many $\alpha$ have norm equal to $3^411^s23^t$. I have so far found examples of $\alpha$ with norms of $11$ and $23$---these are $2 \pm \sqrt{-7}$ and $4 \pm \sqrt{-7}$, but this is as far as I have gotten.

Any help is appreciated, thanks!

Answer 1

Let $R:=\mathbb{Z}[\omega]$, where $\omega:=\dfrac{1+\sqrt{-7}}{2}$. Write $\bar{\omega}:=\dfrac{1-\sqrt{-7}}{2}$. Observe that $\omega$ and $\bar{\omega}$ are the roots of the quadratic polynomial $t^2-t+2$. For $x,y\in\mathbb{Z}$, the conjugate of $\alpha:=x+y\omega\in R$ is denoted by $\bar{\alpha}:=x+y\bar{\omega}$. Note that $R$ is a unique factorization domain. Define the norm $N:R\to\mathbb{Z}_{\geq 0}$ by $$N(x+y\omega ):=(x+y\omega)(x+y\bar{\omega})=x^2+xy+2y^2$$ for every $x,y\in\mathbb{Z}$. The group of units $R^\times$ of $R$ is $R^\times=\{-1,+1\}$.

First Observation. For each prime natural number $p$, there exist integers $u$ and $v$ such that $$p=u^2+uv+2v^2=N(u+v\omega)$$ if and only if $p\equiv 0,1,2,4\pmod{7}$. The prime $p=7$ ramifies in $R$ (that is, the only prime elements of $R$ that divides $7$ are $\pm (-1+2\omega)$). Primes $p\equiv 1,2,4\pmod{7}$ splits in $R$ into two coprime factors $\pi_p$ and $\bar{\pi}_p$, which are prime elements of $R$. Primes $p\equiv 3,5,6\pmod{7}$ remain inert in $R$ (that is, $p$ is also a prime element of $R$).

Second Observation. If $n\in\mathbb{Z}$ is an integer such that $n=N(\alpha)$ for some $\alpha\in R$, then there is a one-to-one correspondence between the representations of $n$ in the form $$n=u^2+uv+2v^2\text{ where }u,v\in\mathbb{Z}\text{ and }v\text{ is even}$$ and the representations of $n$ in the form $$n=x^2+7y^2\text{ where }x,y\in\mathbb{Z}\,.$$ This one-to-one correspondence is given by $$(u,v)\mapsto \left(u+\dfrac{v}{2},\dfrac{v}{2}\right)\,.$$ In particular, when $n$ is odd, $v$ is always even.

From the two observations above, the only thing we need to do is factorizing $m(s,t):=3^4\cdot 11^s\cdot 23^t$ in $R$: $$m(s,t)=3^4\cdot \pi_{11}^s \cdot \bar{\pi}_{11}^s\cdot \pi_{23}^t\cdot \bar{\pi}_{23}^t\,.$$ Each element $\alpha$ in $R$ of $m(s,t)$ such that $m(s,t)=N(\alpha)$ must take the form $$\alpha=\upsilon\cdot 3^2\cdot \pi_{11}^{a}\cdot \bar{\pi}_{11}^{s-a}\cdot \pi_{23}^b\cdot \bar{\pi}_{23}^{t-b}\,,$$ where $a$ and $b$ are integers such that $0\leq a\leq s$ and $0\leq b\leq t$, and $\upsilon\in R^\times$. Thus, there are precisely $$|R^\times|\cdot(s+1)\cdot(t+1)=2(s+1)(t+1)$$ possible values of $\alpha$. This means the number of representations of $m(s,t)$ in the form $u^2+uv+2v^2$ with $(u,v)\in\mathbb{Z}\times \mathbb{Z}$, which is the same as the number of representations of $m(s,t)$ in the form $x^2+7y^2$ where $x,y\in\mathbb{Z}$, is given by $2(s+1)(t+1)$.

In general, let $n\in\mathbb{Z}_{>0}$. Write $$n=7^h\,\prod_{i=1}^s\,p_i^{k_i}\, \prod_{j=1}^t\,q_j^{l_j}\,,$$ where

• $h,s,t,k_1,k_2,\ldots,k_s,l_1,l_2,\cdots,l_t$ are nonnegative integers,
• $p_1,p_2,\ldots,p_s$ are pairwise distinct prime natural numbers such that $p_i\equiv 1,2,4\pmod{7}$ for $i=1,2,\ldots,s$, and
• $q_1,q_2,\ldots,q_t$ are pairwise distinct prime natural numbers such that $q_j\equiv 3,5,7\pmod{7}$ for $j=1,2,\ldots,t$.

If $A(n)$ denotes the number of representations of $n$ in the form $u^2+uv+2v^2$ with $u,v\in\mathbb{Z}$, and $B(n)$ is the number of representations of $n$ in the form $x^2+7y^2$ with $x,y\in\mathbb{Z}$, then $$A(n)=B(n)=0$$ in the case where $l_j$ is odd for some $j=1,2,\ldots,t$. From now on, we assume that $l_j$ is even for every $j=1,2,\ldots,t$. In this case, $$A(n)=2\,\prod_{i=1}^s\,(k_i+1)\,.$$ When $n$ is odd, we get $$B(n)=A(n)=2\,\prod_{i=1}^s\,(k_i+1)\,.$$ When $n$ is even, we may assume without loss of generality that $p_1=2$, and so we have $$B(n)=2\,(k_1-1)\,\prod_{i=2}^s\,(k_i+1)\,.$$ In particular, if $2\mid n$ but $4\nmid n$, then $B(n)=0$.

Remark. Working in $\mathbb{Z}\big[\sqrt{-7}\big]$ is not a good choice. This is because $\mathbb{Z}\big[\sqrt{-7}\big]$ is not a unique factorization domain. Observe that $(1+\sqrt{-7})\cdot(1-\sqrt{-7})=2\cdot 2\cdot 2$ in $\mathbb{Z}\big[\sqrt{-7}\big]$, with $1\pm\sqrt{-7}$ and $2$ being irreducible elements of $\mathbb{Z}\big[\sqrt{-7}\big]$.