You two post faster than I can prove theorems in algebra – which, as @DPRK has illustrated, is an ability I may have lost. (No joke, thanks for the feedback @DPRK, I’ve been wrong several times now).
There’s a lot going on here. In an attempt not to embarrass myself further, I shall now differentiate between theorems that have been proven true, and claims I make which I have not yet proven.
Definition: let a, k \in \mathbb{N}, a > 1, k>1, and let p be a prime such that \gcd(a, p)=1. We say that a is a k-th power residue modulo p if there exists b \in \mathbb{Z} such that b^k \equiv a (mod p). If a is not a k-th power residue modulo p, we say that a is a k-th power nonresidue modulo p.
It is a theorem that a is a primitive root modulo a prime p if and only if a is a q-th power nonresidue modulo p for every prime q dividing p-1. I claim that those lines that you see are in fact lines with slope 1/2, 1/3, 1/5, 1/7, … , 1/q for q prime (but I also claim that not every prime lies on such a line, more on this later). We know (that is, it is a theorem proven by an individual that is not myself) that the decimal representation of \frac{1}{p}, p prime has period \mbox{ord}_p(10), that \mbox{ord}_p(10) \mid \phi(p)=p-1, and that if \mbox{ord}_p(10)=p-1 then 10 is a primitive root modulo p by definition.
So, the red line at the top is the set of full reptend primes, which means that it is the set of all primes p such that 10 is a q-th power nonresidue for every prime q dividing p-1. I claim that the first orange line is the set of all primes p such that 10 is a quadratic residue modulo p, and 10 is a q-th power nonresidue for every prime q \neq 2 dividing p-1. I claim that the second orange line is the set of all primes p such that 10 is a cubic residue modulo p, and 10 is a q-th power nonresidue modulo p for every prime q \neq 3 dividing p-1. I claim that the third orange line (if one can make it out, things get hazy there at the bottom) is the set of all primes p such that 10 is a quintic residue modulo p and 10 is a quintic nonresidue modulo p for every prime q \neq 5 dividing p-1, and so on, for every prime q.
From here on out, let p and q always denote a prime, it makes things easier. It is a theorem that if a is a q-th power residue modulo p with q \mid p-1, then \mbox{ord}_p(a) \leq \frac{p-1}{q}. Furthermore, if 10 is a q-th power redisue modulo exactly one q \mid p-1, then \mbox{ord}_p(10)=\frac{p-1}{q}. (Personal claim now:) That is why those lines have slopes that are the reciprocals of primes. Remember, each of those points has coordinates (p, \mbox{ord}_p(10)). If we, say, consider the first orange line, these points have coordinates (p, \frac{p-1}{2}). These points lie on a line of slope 1/2, and so on for the remaining orange lines for all the primes q. The top red line consists of points (p, p-1), and so has slope 1.
However, I claim that not every prime lies on such a line. From the way we have defined those lines, if 10 is sumultaneously a q_1-th power residue modulo p and a q_2-th power residue modulo p, then p will not lie on any of those lines. They’ll be kind of an intermediate orange fuzz (no, not standard terminology).
That those lines are so clear and so solid is something of an artifact of the fact that your graph only shows tiny, tiny, tiny primes. (The lines are real, this is a real result, it’s just that things get more complicated.) As the upper bound for the x-axis approaches infinity, several things will start to happen. First, the primes become more sparse in \mathbb{N} as you consider larger primes. Your line will look like a dotted line, and eventually there will be points on a given line arbitrarily far apart.
Even beyond that fact, the line with slope 1/2 will look fuller than the line with slope 1/3 will look fuller than the line with slope 1/5, and so on. This is because “elementary” k-th power reciprocity theory is actually governed by rather deep algebraic phenomena: the splitting/ramification behavior of prime ideals sitting above p in algebraic number fields. That theory is a little bit too complicated to go into here, but in some very well defined sense (in the sense of a density statement given by the Chebotarev Density Theorem), for every q \mid p-1, \frac{1}{q} th of the p-1 incongruent residue classes modulo p are q-th power residues modulo p. So, in the sense described, the lines with slope \frac{1}{q} must be getting “sparser” as q increases.
Also, 2 \mid p-1 for every p, while only finitely many other primes divide p-1, and so, in some sense, 10 has a greater “chance” of being a quadratic residue modulo p than any other q th power residue modulo p for q \mid p-1. I use the word “chance” as an analytic number theorist does; these phenomena are deterministic, not probabilistic, that’s just how they roll. The number, on average, of distinct prime divisors of an integer n grows very very slowly as n does, so for small primes (however you want to define small), it can’t be a q-th power residue modulo p for very many q's.
And, finally, we come to perhaps the biggest issue if we seek a general theorem for identifying our “good numbers.” There is no getting around it. Or, perhaps one should say, if you get around it you will be a millionaire and the most famous number theorist on the planet: It is not currently known whether there exist infinitely many primes p such that 10 is a primitive root modulo p. Period. The infinitude has been proven under the assumption of a generalized Riemann Hypothesis (this is Artin’s Conjecture on Primitive Roots), which is believed by most to be true, but no one has proven it. So not only will that red line get sparser and sparser, for all we know, it may eventually stop. . .
