# Is there a version of Dirichlet's theorem for a system of power congruences?

I know that Dirichlet's theorem says that there are infinitely many primes $$p$$ such that $$p\equiv a$$ (mod $$n$$) when gcd$$(a,n)=1$$. I'm wondering more generally about a system of power congruences

$$x^{k_1}\equiv a_1$$ (mod $$n_1$$)

$$x^{k_2}\equiv a_2$$ (mod $$n_2$$)

:

$$x^{k_m}\equiv a_m$$ (mod $$n_m$$)

Given fixed positive integers $$k_i,a_i,n_i$$ as above, is there a condition, e.g., $$gcd(a_i,lcm(n_1,\ldots,n_m))=1$$, that guarantees the existence of infinitely many primes $$p$$ such that $$p^{k_i}\equiv a_i$$ (mod $$n_i$$) for all $$1\leq i\leq m$$?

Maybe this should be a separate post, but I'm also wondering about conditions that guarantee that the above system has any solutions $$x\in\{0,1,\ldots,L-1$$}, prime or not, where $$L=lcm(n_1,\ldots,n_m)$$.

• There are many variables here. Is $x$ the desired prime number ? Which of those variables are assumed to be given ? Commented Feb 2, 2023 at 9:00
• I'll edit the post. Commented Feb 2, 2023 at 9:02
• I don't think there's any simple condition on $n$ guaranteeing there's a solution to $x^3\equiv2\bmod n$. But if $n$ is odd, and if there is a solution, then Dirichlet guarantees there's a solution with $x$ being prime. Commented Feb 2, 2023 at 10:19
• Why did you delete your previous similar question? Your congruences constraints are equivalent to $x \bmod L \in S$ for some finite subset $S\subset 0\ldots L-1$ that you have to compute. Then the density of primes satisfying those congruences is $\frac{\# A}{\phi(L)}$ where $A = \{ s\in S, \gcd(s,L)=1\}$ and if $A=\emptyset$ then only finitely primes satisfy it (as they'll divide $L$). Commented Feb 2, 2023 at 11:00
• I deleted that one because it was a slight variant of - yet another - question that I'd posted previously. I think I understand everything now. Thank you. Commented Feb 2, 2023 at 12:18

(1) Assume $$(a_i,n_i)=1$$.
Solve each $$x^{k_i}\equiv a_i\pmod {n_i}$$ individually, see this.
Let $$x\equiv b_i\pmod {n_i}$$ be one of the solutions with $$(b_i,n_i)=1$$.
(2) Assume $$(n_i,n_j)=1$$, for all $$i\neq j$$.
Then it becomes \left\{\begin{align} x&\equiv b_1\pmod {n_1}\\ &\vdots\\ x&\equiv b_s\pmod {n_s} \end{align}\right. Then use the Chinese Remainder Theorem, the solution is unique with $$x\equiv c\pmod{ \prod_{i=1}^s n_i}.$$
(3) To have infinitely many prime solutions, we only need $$(c,\prod_{i=1}^s n_i)=1.$$ We see that $$(c,\prod_{i=1}^s n_i)=1$$ if and only if $$(c,n_i)=1\iff (b_i,n_i)=1,\quad i=1,\ldots,s,$$ which follows from (1).
(4)In conclution, with $$(a_i,n_i)=1$$ and $$(n_i,n_j)=1$$ $$(1\leq i\neq j\leq s)$$, if the system of congruences has a integer solution, then it has infinitely many prime solutions.