TITLE: generator of the cyclic group : $(\mathbb{F}_p^*, \times)$
QUESTION [8 upvotes]: I would like to prove that there exist an element $m$ in the cyclic group : 
$$(\mathbb{F}_p^*, \times)$$ with $p \geq 3$ prime such that $m$ and $4m$ are generator of this group. 
Yet I don't know at all how to proceed, any hint or ideas are welcome :)

REPLY [6 votes]: I think I exaggerated with Bunyakovsky's conjecture :-), the claim it's actually  an easy application of the chinesse reminder theorem.
In the notation of @DerekHolt comment, write $p-1=2^{n_0}p_{1}^ {n_1}\cdots p_{r}^{n_r}$ where $p_i$ is an odd prime for $i=1,2,\ldots,r$. Since each $\mathbb{Z}/p_i \mathbb{Z}$ has at least that $3$ elements exists $a_i$ such that $a_i \not\equiv0,-2j \,\,(\text{mod} \, p_i)$. Now take $k$ such that $k\equiv1 \,\,(\text{mod} \, 2)$ and $k\equiv a_i \,\,(\text{mod} \, p_i)$, then $k$ and $k+2j$ are coprime to $p-1$  (because $k\equiv a_i \not\equiv0,-2j \,\,(\text{mod} \, p_i)$ and both $k$ and $k+2j$ are odd ).