Is it possible to simplify this further:
$$ n \equiv - \frac d 4\bigg(\frac 3 2\bigg)^{m-2} - 2 \pmod {3^m} \qquad \text{where } n, d \geq 0 \text{ and } m \geq 2 \text{ are integers} $$
or to glean some info on what $n$ should be to satisfy the equivalence?
I obtained the above as follows:
I want to know when
$$ \frac 1 {3^m} \big(2^m(n+2) - (2 - \Delta) 3^m \big) \qquad \text{where } \Delta = \frac d 9 \text{ for some integer } d \geq 0 $$
is an integer. We have
$$ \begin{align} 2^m (n+2) &\equiv (2 - \Delta) 3^m &\pmod {3^m}\\ 2^m (n+2) &\equiv (2 - \frac d 9) 3^m &\pmod {3^m}\\ 2^m (n+2) &\equiv 2 \cdot 3^m - \frac d {3^2} \cdot 3^m &\pmod {3^m}\\ 2^m (n+2) &\equiv - d \cdot 3^{m-2} &\pmod {3^m}\\ 4 \cdot 2^{m-2} (n+2) &\equiv - d \cdot 3^{m-2} &\pmod {3^m}\\ n+2 &\equiv - \frac d 4 \bigg(\frac 3 2\bigg)^{m-2} &\pmod {3^m}\\ n &\equiv - \frac d 4 \bigg(\frac 3 2\bigg)^{m-2} - 2 &\pmod {3^m} \end{align} $$
From $2^m (n+2) \equiv - d \cdot 3^{m-2} \pmod {3^m}$, I know that if $d \equiv 0 \pmod 9$, then $n \equiv -2 \pmod {3^m}$.
Also, by writing $d = 9k + (d \bmod 9)$, we have
$$ \begin{align} 2^m (n+2) &\equiv - (9k + (d \bmod 9)) \cdot 3^{m-2} &\pmod {3^m}\\ 2^m (n+2) &\equiv -9k \cdot 3^{m-2} + -(d \bmod 9) \cdot 3^{m-2} &\pmod {3^m}\\ 2^m (n+2) &\equiv -k \cdot 3^m + -(d \bmod 9) \cdot 3^{m-2} &\pmod {3^m}\\ &\boxed{ 2^m (n+2) \equiv -(d \bmod 9) \cdot 3^{m-2} \pmod {3^m} } \end{align} $$
Simplfy $n \equiv - \dfrac d 4\bigg(\dfrac 3 2\bigg)^{m-2} - 2 \pmod {3^m} \qquad \text{where } n, d \geq 0 \text{ and } m \geq 2 \text{ are integers}$.
Multiplying by $\dfrac94$ we get $$\dfrac94n\equiv -\dfrac d4 \left(\dfrac 32\right)^m-\dfrac 92\pmod{3^m}\Rightarrow 9n\equiv \left(\dfrac{-d}{2^m}\right)3^m-18\pmod{3^m}$$ Then $$\boxed{n\equiv-2\mod{3^m}}$$