Consider an integer $ k $ that verify the following property: for all prime $ p $, $ k $ is a cubic residue module $ p $. This is, for all prime $ p $ the following statement is true: $$ k = x^3 \ \ (p) \mbox{ for some } x \in \mathbb{Z}_p $$
My question is: necesarily must be $ k = x^3 $ for some $ x \in \mathbb{Z} $?
Summarizing: to be a cubic residue for all prime $ p $ implies to be the cube of some integer?
At first, I think on Chinese Remainder Thorem: for each $ m \in \mathbb{Z} $ we can fix integers $ x_1^3, x_2^3, ... x_m^3 $ and primes $ p_{j_1}, p_{j_2}, ... , p_{j_m} $ and found an integer $ x $ such that $ x = x_i^3 \ \ (p_i) $ for $ i = 1, 2, ... , m $. The problem is basically that but wiht a infinite number of equations.
Even we can think on this problem in the following equivalent way: if $ k $ is an integer and there exists sucessions of integers $ (a_n)_{n \in \mathbb{N}} $, $ (b_n)_{n \in \mathbb{N}} $ such that $ p_n a_n + b_n^3 = k $ for all $ n \in \mathbb{N} $, is true that $ k = x^3 $ for some integer $ x $? . (Here $ p_n $ denotes the $ n $-th prime number and we suposse that the sucessions are not null).
To find a counterexample seems hard...
Yes, this is a special case of the Grunwald-Wang Theorem.