Find positive integers $(x,n)$ such that $x^{n} + 2^{n} + 1$ is a divisor of $x^{n+1} +2^{n+1} +1$ Find all positive integers $(x,n)$ such that $x^{n} + 2^{n} + 1$ is a divisor of $x^{n+1} +2^{n+1} +1$   

I encountered this question in one of my monthly assignments. Unfortunately, I don't know how to proceed about this question at all. Please help.
  Thanks in advance!

 A: Check that $n=1$ gives two solutions $x=4$ and $x=11$. From now on $n>1$.
For each single case $x=1$, $x=2$ check that there is no solution.
Now we will consider 
$$
x(x^n+2^n+1)-(x^{n+1}+2^{n+1}+1)=2^n(x-2)+x-1
$$
instead of $x^{n+1}+2^{n+1}+1$.
Check that $x=3$ gives no solution. From now on $x>3$ and $n>1$, hence
$$
x^{n-1}(x-2)\geq 2^n(x-2),
$$
$$
x^{n-1}\cdot 2\geq x-1,
$$
$$
2^n+1>0.
$$
Summing last three lines we get
$$
x^n+2^n+1>2^n(x-2)+x-1
$$
and the left hand side is not a divisor of the right hand side, there is no solution for $n>1$, $x>3$.
A: Famous Romania TST problem .
See the detailed solution here : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849335&sid=04dbb9363eed1be6ee4afb63bbebb787#p849335
A: Hints:  


*

*Consider $f(x,n)= \dfrac{x^{n+1} +2^{n+1} +1}{x^{n} + 2^{n} + 1}$ and the behaviour of $f(x,n)-x$

*Put bounds on $f(x,n) - x $ for $x \ge 2, n \ge 2$

*Put bounds on $f(x,1) - x $ i.e. for $n = 1$

*Find limit of $f(1,n) - 1 $ as $n\to \infty$ i.e. for $x=1$

*

*Consider cases where the absolute value of $\displaystyle f(1,n) - 1 - \lim_{n\to \infty} (f(1,n) - 1)$ is greater than or equal to $1$


