<s>[INST] Induction: How to prove that $ab^n+cn+d$ is divisible by $m$. 
If $a+d$, $(b-1)c$, $ab-a+c$ are divisible by $m$, prove that
  $ab^n+cn+d$ is also divisible by $m$.

I want to prove this by induction. For proving $ab^{k+1}+c(k+1)+d$ is divisible by $m$, i want to prove that $ab^k(b-1)+c$ is divisible by $m$ and then add it to $ab^{k}+ck+d$. Any idea how to prove $ab^k(b-1)+c$ is divisible by $m$? Or is there a better way to solve the problem?
Thanks in advance.
 [/INST] Without induction. Let's do some tagging 1st
$$m \mid a+d \tag{1}$$
$$m \mid (b-1)c \tag{2}$$
$$m \mid ab-a+c \tag{3}$$
then
$$m \mid ab^n+cn+d \iff m \mid a\left(b^n-1\right)+cn+\color{red}{a+d} \overset{(1)}{\iff}\\
m \mid a\left(b^n-1\right)+cn \iff \\
m \mid a(b-1)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \iff \\
m \mid (ab-a+c-c)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \iff \\
m \mid \color{red}{(ab-a+c)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)}-c\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \overset{(3)}{\iff}\\
m \mid c\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)-cn \iff \\
m \mid c\left(b^{n-1}-1\right)+c\left(b^{n-2}-1\right)+...+c\left(b^2-1\right)+\color{red}{c\left(b-1\right)} \overset{(2)}{\iff}\\ 
m \mid c\left(b^{n-1}-1\right)+c\left(b^{n-2}-1\right)+...+c\left(b^2-1\right)$$
which is true because for $\forall k\geq 2$ we have
$$c\left(b^k-1\right)=\color{red}{c\left(b-1\right)}\left(b^{k-1}+b^{k-2}+...+b^{2}+b+1\right)$$
and from $(2)$
$$m \mid c\left(b^k-1\right)$$
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