Let $f(n)=\left [ an+b \right ], \, \forall n \in \mathbb{N}^*$ $$\Rightarrow f(n+1)=\left [ a(n+1)+b \right ]=\left[\begin{matrix}& \left [ an+b \right ]+ \left [ a\right ]=f(n)+\left [ a \right ] \\& \left [an+b \right ]+ \left [ a \right ]+1=f(n)+\left [ a \right ]+1\end{matrix}\right. \rightarrow 0 \leq f(n+1)-f(n) \leq \left [ a \right ]+1$$We know, $\underset{n \rightarrow + \infty}{\textrm{lim}}f(n)= + \infty$. So, for any $m \in \mathbb{N}^*$, $\exists n_0: \, f(n_0) \in \left\{m;m+1;...;m+\left [ a \right ]+1 \right\}.$ $\bullet$ Choose $p_1,p_2,...,p_{\left [ a \right ]+2}$ be $\left [ a \right ]+2$ dinstinct prime numbers, from Chinese Remainder Theorem, we know that exists infinitely $m_1$ such that: $$\left\{\begin{matrix}m_1 \equiv 0 \left ( \textrm{mod}\, p_1 \right ) & & \\ m_1 \equiv -1 \left ( \textrm{mod}\, p_2 \right ) & & \\... & & \\ & & \\ m_1 \equiv -\left [ a \right ]-1 \left ( \textrm{mod}\, p_{\left [ a \right ]+2} \right ) \end{matrix}\right.$$Then, $\exists \, n_1: \, f(n_1) \in \left\{m_1;m_1+1;...;m_1+\left [ a \right ]+1 \right\},$ which cant be a prime. Q.E.D