I think this is explained by the fact that $x^2-(n^2-3)y^2=-1$ with $3\not|n$ is soluble if and only if there are integers $p,q,r$ so that $n=\frac{3p^2+q^2-r^2\pm 2r^2}{2pr}$.
Proof for the forwards implication:
We have $x^2+2y^2\equiv 2 \pmod 3$ which implies $x\equiv 0$ and $y\not\equiv 0$ s...
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$$
f(s) = \sum_{n \in \Bbb{N}} \dfrac{\phi(n)}{n^s} = 0 \\
g(s) = \sum_{n \in \Bbb{N}} \dfrac{\psi(n)}{n^s} = 0 \implies
f(s) - g(s) = \sum_{n\in \Bbb{N}}\dfrac{\phi(n) - \psi(n)}{n^s} = 0
$$
Thus, for a fixed element $s \in \Bbb{C}$, the set of all coefficient functions $\psi : \Bbb{N}\to \Bbb{...
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$$
f(s) = \sum_{q \in \Bbb{Q}^{\times}} q^{s}
$$
Then the set of elements $G_s = \{q^{s} : q \in \Bbb{Q} \}$ is a subgroup of $\Bbb{C}^{\times}$. $|q^{s}| = \sqrt{ q^{s} q^{\overline{s}}}= \sqrt{q^{\overline{s} + s}} = q^{\text{Re}(s)}$, so that we can say that the series converges absolutely wh...
