nicetry007 wrote: Let $ x = 1$. We have $ 2^{n}+2 \;|\; 2^{n+1}+2 \; \Rightarrow\; 2^{n-1}+1 \;|\; 2^{n}+1\; \Rightarrow\; 2^{n-1}+1 \;|\; 2^{n-1}+2^{n-1}+1$ $ \; \Rightarrow\; 2^{n-1}+1 \;|\; 2^{n-1}$. Not possible Let $ x = 2$. We have $ 2^{n+1}+1 \;|\; 2^{n+2}+1 \; \Rightarrow\; 2^{n+1}+1 \;|\; 2^{n+1}+2^{n+1}+1\; \Rightarrow\; 2^{n+1}+1 \;|\; 2^{n+1}$. This is not feasible. This implies $ x \;\geq \;3$. Let $ n = 1$. In this case, we have $ x+3 \;|\;x^{2}+5 \;\Leftrightarrow\; x+3 \;|\; x^{2}-9+14 \;\Leftrightarrow\; x+3 \;|\; 14 \;\Leftrightarrow\; x = 4$ or $ 11$. For $ n \geq 2$, $ \frac{ x^{n+1}+2^{n+1}+1}{x^{n}+2^{n}+1}\; \in\; Z\; \Leftrightarrow \; x-\frac{ (x-2)2^{n}+x-1}{x^{n}+2^{n}+1}\; \in\; Z\;\Rightarrow \; (x-2)2^{n}+x-1 \;\geq \;x^{n}+2^{n}+1$ $ \Leftrightarrow \; (x-2)(2^{n}+1) \;\geq \;x^{n}+2^{n}\;\Leftrightarrow \; (x-2)(1+\frac{1}{2^{n}}) \;\geq \;\left(\frac{x}{2}\right)^{n}+1\;$ Now, we have $ \left(\frac{x}{2}\right)^{2}+1\;\leq\;\left(\frac{x}{2}\right)^{n}+1\;\leq \;(x-2)(1+\frac{1}{2^{n}}) \;\leq \;(x-2)(1+\frac{1}{2^{2}}) \;=\; \frac{5}{4}(x-2)$ Rewriting the inequality, we get $ x^{2}-5x+14 \;\leq\; 0$. But the inequality has no real roots. Hence, there is no solution for $ n\; \geq \;2$ The only solutions are $ (x,n) = (4,1)$ and $ (11,1)$.

Posted here also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=76597

Putting $x=1,2,3$ we find that there is no solution. (Use inequalities) So assume $x\ge 4$. $x^n+2^n+1|x^{n+1}+2^{n+1}+1\Longleftrightarrow x^n+2^n+1|x(x^n+2^n+1)-(x^{n+1}+2^{n+1}+1)$ $\Longrightarrow x^n+2\leq (x-3)2^n+x$ By putting $x=4,5$ we find that this inequality is false if $n>1$. And by induction, \[(x+1)^n+2=x^n+n\cdot x^{n-1}+...+1+2> x^n+2^n+1+2>(x-3)2^n+2^n+x+1\]$=(x-2)2^n+x+1$ So we must have $n=1$ Now the first condition becomes $x+3|x^2+5\Longleftrightarrow x+3|(x^2+5)-(x^2-9)=14$ As $x\ge 4$, we have $x+3=7,14\Longleftrightarrow x=4,11$. It's easy to see both of them are correct. So all the solutions are $(x,n)=(4,1),(11,1)$

$ \frac{ x^{n+1}+2^{n+1}+1}{x^{n}+2^{n}+1}\; \in\; Z\; \Leftrightarrow \; x-\frac{ (x-2)2^{n}+x-1}{x^{n}+2^{n}+1}\; \in\; Z\;\Rightarrow \; (x-2)2^{n}+x-1 \;\geq \;x^{n}+2^{n}+1$ $ \Leftrightarrow \; (x-2)(2^{n}+1) \;\geq \;x^{n}+2^{n}\;\Leftrightarrow \; (x-2)(1+\frac{1}{2^{n}}) \;\geq \;\left(\frac{x}{2}\right)^{n}+1\;$ Now, we have $ \left(\frac{x}{2}\right)^{2}+1\;\leq\;\left(\frac{x}{2}\right)^{n}+1\;\leq \;(x-2)(1+\frac{1}{2^{n}}) \;\leq \;(x-2)(1+\frac{1}{2^{2}}) \;=\; \frac{5}{4}(x-2)$ Rewriting the inequality, we get $ x^{2}-5x+14 \;\leq\; 0$. But the inequality has no real roots. Hence, there is no solution for $ n\; \geq \;2$