<s>[INST] Are there some known results for solving this kind of modular arithmetic problem? Preliminaries

Consider a number in the form:
$$x = a^n b - c,$$
where $a, b, c$ and $n$ are natural numbers. For which $n$ we can say
  that $x$ is divisible by $d$ with $a$ and $d$ coprime?

WLOG, consider the following example with $a=2$, $c=1$ and $d = 3$.
\begin{array}{llll}
2^{n}b - 1 \equiv 0 &(\text{mod}~ 3) & (1)\\
2^{n}b - 1 + 3 \equiv 0 &(\text{mod}~ 3) & (\text{we add a multiple of $d=3$)}\\
2^{n}b  + 2 \equiv 0 &(\text{mod}~ 3) & (\text{we divide by a number (2) non divisible by $d=3$)}\\
2^{n-1}b  + 1 \equiv 0 &(\text{mod}~ 3) & (2) \\
2^{n-1}b  + 1 - 3\equiv 0 &(\text{mod}~ 3) & (\text{we subtract a multiple of $d=3$)}\\
2^{n-1}b  - 2\equiv 0 &(\text{mod}~ 3) & (\text{we divide by a number (2) non divisible by $d=3$)}\\
2^{n-2}b  - 1\equiv 0 &(\text{mod}~ 3) & (3).
\end{array}
Reiterating this schema until I get $2^{n-n} = 2^0 = 1$, I conclude that


*

*If $n$ is even (see equation (1) and (3)), then I need to prove


$$b  - 1\equiv 0 ~(\text{mod}~ 3). $$


*If $n$ is odd (see equation (2)), then I need to prove


$$b  + 1\equiv 0 ~(\text{mod}~ 3). $$
Now, depending on $b$, I can find the answer to the initial question.

My question
I am wondering if the presented procedure has a name, or I can arrive to the conclusion by simply applying some known results that I'm missing. 
 [/INST] Since $2+1=3$, it follows that $2+1\equiv0\pmod3$; i.e., $2\equiv-1\pmod3$.  
Therefore $2^n\equiv(-1)^n\pmod3$.  
$(-1)^n$ is $1$ when $n$ is even and $-1$ when $n$ is odd.
Therefore $2^n\equiv1\pmod3$ when $n$ is even and $2^n\equiv-1\pmod3$ when $n$ is odd.
So if $2^nb-1\equiv0\pmod3$ then $b-1\equiv0$ when $n$ is even and $-b-1\equiv0$ when $n$ is odd; 
i.e., $b\equiv1$ when $n$ is even and $b\equiv-1\equiv2\pmod3$ when $n$ is odd.
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