if $x_{n+1}=ax_{n}+b$ then How find this $a,b,c$? Question:


  Let there be a sequence $x_{n}$ such that $$x_{n+1}=ax_{n}+b,,x_{1}=c$$
  where $a,b$ and $c$ are positive integers.


Suppose, too, that for $n,m\in \mathbb{Z^{+}}$ we have that 
$$
\text{(*)   } ~~~n\mid m\implies x_{n}|x_{m}
$$


  show that:$$b=c$$


My try: if $a=1$, then we have
$$x_{n+1}-x_{n}=b,x_{1}=c\Longrightarrow x_{n}=c+(n-1)b=bn+c-b$$
So  if
$n|m$ then let $m=nk$, so
$$\dfrac{x_{m}}{x_{n}}=\dfrac{bkn+c-b}{bn+c-b}=k+\dfrac{c-b-kc+kb}{bn+c-b}\in N^{+}$$
So how do I find $a,b$ and $c$? 

Case $2$: if $a\neq1$, then we have
$$x_{n}+\dfrac{b}{a-1}=a(x_{n}+\dfrac{b}{a-1})$$
So
$$x_{n}=\left(c+\dfrac{b}{a-1}\right)a^{n-1}-\dfrac{b}{a-1}$$

By the way:

Sometime ago, I solved the following hard problem: (IMO shorsits)

If $a,b\in \mathbb{Z^{+}}$, and for any positive integer $n$ we have
$$\dfrac{a^n-1}{b^n-1}\in N^{+}$$

Then we also must have that there exists some positive integer $k$ so that $a=b^k$

Then I can't, thank you

This problem was created by a China IMO team student.
I'm going to restate the problem, and give a partial answer:

Find all positive-integer triples $a, b, c$ such that the arithmetic sequence
$$
x_n = a x_n + b; x_1 = c
$$
has the property that $n | m \implies x_n | x_m$.

Suppose that the sequence has this property. Then consider the case $n = 1$, which divides any other number $m$; we get that $x_1 | x_m$ for every $m$. In particular, the number $c = x_1$ must divide $x_2 = ac + b$; that means that 
$$
c | b.
$$
What about $x_3$? That's
$$
x_3 = a(ac + b) + b = a^2 c + ab + b
$$
which is also divisible by $c$ (since $ac$ and $b$ both are). A simple induction argument shows that $c | x_n$ for every $n$ once we know that $c | b$. 

The stuff above is a cleaner version of what I wrote earlier. But $c | b$, although necessary, is not a sufficient condition. Let's examine further. We have
$$
x_4 = a^3 c + a^2 b + ab + b = a^2(ac + b) + (a+1)b
$$
which must be divisible by $x_2 = ac + b$. Thus we must have $(ac + b) | (a+1) b$. Writeing $b = sc$, we get $(ac + sc) | (a+1) sc$, so $(a + s) | (a+1)s$, so $(a + s) | (as + s)$.

So just using the first few terms, we find that 

(1) $c | b$. Writing $b = cs$, we also have

(2) $(a+s) | (as + s)$

So those are a couple of constraints on $a, b, c$.
I'm going to restate the problem, and give a partial answer:

Find all positive-integer triples $a, b, c$ such that the arithmetic sequence
$$
x_n = a x_n + b; x_1 = c
$$
has the property that $n | m \implies x_n | x_m$.

Suppose that the sequence has this property. Then consider the case $n = 1$, which divides any other number $m$; we get that $x_1 | x_m$ for every $m$. In particular, the number $c = x_1$ must divide $x_2 = ac + b$; that means that 
$$
c | b.
$$
What about $x_3$? That's
$$
x_3 = a(ac + b) + b = a^2 c + ab + b
$$
which is also divisible by $c$ (since $ac$ and $b$ both are). A simple induction argument shows that $c | x_n$ for every $n$ once we know that $c | b$. 

The stuff above is a cleaner version of what I wrote earlier. But $c | b$, although necessary, is not a sufficient condition. Let's examine further. We have
$$
x_4 = a^3 c + a^2 b + ab + b = a^2(ac + b) + (a+1)b
$$
which must be divisible by $x_2 = ac + b$. Thus we must have $(ac + b) | (a+1) b$. Writeing $b = sc$, we get $(ac + sc) | (a+1) sc$, so $(a + s) | (a+1)s$, so $(a + s) | (as + s)$.

So just using the first few terms, we find that 

(1) $c | b$. Writing $b = cs$, we also have

(2) $(a+s) | (as + s)$

So those are a couple of constraints on $a, b, c$.