What's the intution behind this result:
Let $q$ be a positive integer greater than $1$, and $p \equiv 1 \mod q$ a prime. Let $G$ be a multiplicative subgroup $G$ of $\mathbb{F}_p$ with size exactly $q$. Call the prime $p$ nice if for any $a, b \in G$, we have $a \not \equiv b + 1 \mod p$.
It's not hard to prove that there's constant $c_q$ depending on $q$ such that all primes of the form $p \equiv 1 \mod q$ with $p > c_q$ are nice, but what's the intuition behind this result ?
I find this result totally counterintuive since as per my current crappy intuition I expect the $q$-th roots of unity in $\mathbb{F}_p$ to be distributed "randomly", and there's no guarantee that no two $q$-th roots are not placed distance $1$ from each other.
Let $\zeta_q$ denote a primitive (complex) $q$-th root of unity.
The idea here is that the $q$-th roots of unity are algebraic numbers (and, in fact, algebraic integers). More narrowly, they are elements of a number field we call $\mathbb{Q}(\zeta_q)$.
For many purposes, number fields have a lot of the same properties that the field of rational numbers does.
Each of the equations $(\zeta_q)^a = (\zeta_q)^b - 1$ is equivalent to the assertion
$$ (\zeta_q)^a - (\zeta_q)^b + 1 = 0 $$
The number on the left hand side is a nonzero algebraic integer, and thus only divisible by finitely many primes (more precisely, prime ideals), and it is precisely those primes for which we get an equation $(\zeta_q)^a \equiv (\zeta_q)^b - 1 \bmod \mathfrak{p}$.