We know that the set $B = \{(d \mid x) = \begin{cases} 1, d \mid x \\ 0, \text{ otherwise} \end{cases}, \ d \in \Bbb{N}\}$ forms a $\Bbb{Z}$-module Schauder basis for the module $M =\{ \Bbb{N} \to \Bbb{Z}\}$.
The proof that $B$ is a linearly independent set of functions over $\Bbb{Z}$ is to consider:
$$ f(x) = \sum_{d \in P} \alpha_{d} (d \mid x) = 0 $$
Where $P \subset \Bbb{N}$ is the set of all $k \in \Bbb{N} : \alpha_k \neq 0$. Now choose the minimum $\min{P} = d$ and you get that for all $e \gt d$ that $(e\mid d) = 0$ so that the sum $f(d)$ becomes $0 = \alpha_d (d \mid d) = \alpha_d\cdot 1 = \alpha_d$. But that contradicts the fact that $d \in P$ by definition of $P$.
That $B$ spans $M$ can be proven via Mobius Inversion. This is because if we let $f \in M$ be a general element, then putting:
$$ g(x) := \sum_{d \mid x} f(d) = \sum_{d \in \Bbb{N}} f(d) (d \mid x) $$
We get that taking the Mobius Inverse of $g$ that:
$$ f(x) = \sum_{d \mid x} \mu(d) g(\frac{x}{d} ) = \sum_{d \mid x}\mu(d)\sum_{c \mid d} f(d/c) = \sum_{d \in \Bbb{N}} (\mu(d)\sum_{c \mid d} f(d/c)(c\mid d))(d\mid x) $$
where for $c \nmid d$ we can put anything for the value of $f(d/c)$. So, we haven written any $f \in M$ as a (possibly infinite) linear combination of the $(d \mid \cdot)$'s or in other words they span $M$.
Therefore, $B$ is some type of Schauder $\Bbb{Z}$-module basis for the arithmetic functions.
What I'm wondering is if $B' = \{(d \mid h(x)) : d \in \Bbb{N}\}$ is also a basis for fixed function $h(x) = x^2 - 1$. For linear independence the same proof won't work because you'd need to choose $d \in P$ such that $x^2 - 1 = d$ which might not always be possible.
Question. Is the set $B'$ a $\Bbb{Z}$-linearly independent set?
Attempt.
Using the above formula for general $f(x) \in M$ we can describe $(d \mid h(x))$ as:
$$ (d\mid h(x)) = \sum_{e\in \Bbb{N}} \left(\mu(e) \sum_{c \mid e} (d \mid h(\frac{e}{c}))\right)(d \mid x) $$