Peter wrote: Let $\{u_{n}\}_{n \ge 0}$ be a sequence of positive integers defined by \[u_{0}= 1, \;u_{n+1}= au_{n}+b,\] where $a, b \in \mathbb{N}$. Prove that for any choice of $a$ and $b$, the sequence $\{u_{n}\}_{n \ge 0}$ contains infinitely many composite numbers. We can easily find that \[u_{n}=a^{n}+b\frac{a^{n}-1}{a-1}\] for all $n\in{N_{0}}$. Also it is obvious that in the case when $gcd(a;b)>1$ then all the members of the sequence would be composite numbers. Let's discusse the case when $gcd(a;b)=1$. \[u_{n}=a^{n}+b\frac{a^{n}-1}{a-1}=(a^{n}-a)+b\frac{a^{n}-a}{a-1}+(a+b)\] Let's denote by $p$ any prime divisor of $a+b$. We have that $(a;p)=1$=> \[p^{m}|a^{\varphi(p^{m})+1}-a\] Therefore there is a $m_{0}$ number ($m_{0}=\parallel a-1\parallel _{p}$) that for any $m>m_{0}$ \[p^{m}|u_{\varphi(p^{m})+1}\] and of course $p^{m}<u_{\varphi(p^{m})+1}$. So the infinite set of numbers $u_{\varphi(p^{m})+1}$ ,where $m>m_{0}$, includes only composed numbers.