A general type of generator of the multiplicative group of a finite field

Question

Let $p>2$ be a prime number and $\alpha \in \overline{\mathbb{F}_p}$. It generates a finite field $\mathbb{F}_p(\alpha)$. Is there some $u \in \mathbb{F}_p$ such that $ \alpha + u$ is a generator of the cyclic group $\mathbb{F}_p(\alpha)^{\times}$? In other words, do we have $\mathrm{ord}(\alpha+u)=\deg(\alpha)-1$ for some $u \in \mathbb{F}_p$?

Answer 1

Not always. $$f(n)=|\{\alpha\in \Bbb{F}_{p^n},\not \in \bigcup_{d<n}\Bbb{F}_{p^d}\}|=\sum_{d | n} \mu(d) p^{n/d}$$ There are $\phi(p^n-1)$ generators of $\Bbb{F}_{p^n}^\times$. Thus if for all $\alpha\in \Bbb{F}_{p^n},\not \in \bigcup_{d<n}\Bbb{F}_{p^d}$ there is $u\in \Bbb{F}_p$ such that $\alpha+u$ is a generator of $\Bbb{F}_{p^n}^\times$ then $$ f(n) \le p\ \phi(p^n-1)$$ As $n\to \infty$, $f(n)= p^n+O(p^{n/2})\sim p^n-1$, thus it would contradict that $$\inf_n \frac{p \ \phi(p^n-1)}{f(n)}=\inf_np \ \phi(p^n-1)/(p^n-1)\le p\inf_{p\ \nmid\ m} \phi(m)/m=p\prod_{q\ne p} (1-1/q)=0$$