Does there exists for all prime $ p $ an integer $ n $ such that $ {2n \choose n} $ is a generator modulo $ p $?
Up to now, I could only check that small values of $ p $ indeed work.
Note that this is equivalent to asking whether all integers are congruent to some $ {2n \choose n} $ modulo $ p $ as:
- $ 2 \mid \displaystyle{4 \choose 2} $ and, for every $ p \ne 2 $, $ p \mid \displaystyle{p \choose \frac{p - 1}2} + {p \choose \frac{p + 1}2} = {p + 1 \choose \frac{p + 1}2} $
- Lucas Theorem ensures that binomial coefficients form a closed group under multiplication modulo any prime.
Maybe Wilson Theorem ($ (p - 1)! \equiv -1 \mod p $) might be of some help.
It is true. As stated above, middle binomial coefficients form a subgroup of $ (\mathbb Z/p\mathbb Z, *) $ because of Lucas Theorem.
For $ p = 2 $, take $ {0 \choose 0} = 1 $.
For odd $ p $, let's prove by induction that $ {2 \choose 1}, \dots, {2n \choose n} $ generate $ [\![1, 2n]\!] $ as long as $ 2n < p $:
$ \bullet $ Initialisation: For $ n = 1 $, we indeed have $ {2 \choose 1} = 2 $ and $$ {\overline{2\dots 2}^p \choose \overline{\underbrace{1\dots 1}_{p - 1 \text{ times}}}^p} \equiv {2 \choose 1}^{p - 1} \equiv 1 \mod p $$
$ \bullet $ Induction: Suppose $ {2 \choose 1}, \dots, {2n - 2 \choose n - 1} $ generate $ [\![1, 2n - 2]\!] $ for $ n \ge 2 $. We have $$ 2n - 1 = \frac n2\frac{2n \choose n}{2n - 2 \choose n - 1} \text{ and } 2n = 2\cdot n $$ Knowing that $ 2 $ and $ n $ are generated by $ {2 \choose 1}, \dots, {2n - 2 \choose n - 1} $ because $ 2, n \in [\![1, 2n - 2]\!] $, we find that $ 2n - 1 $ and $ 2n $ are generated by $ {2 \choose 1}, \dots, {2n \choose n} $, as wanted.