Find all $ a, b, c \in \mathbb{Z} $, $ c \ge 0 $ such that $ a^n + 2^n | b^n + c $ for all positive integers $ n $ where $ 2ab $ is non-square.
1. **Claim:** \( c = 0 \) or \( c = 1 \).

   **Proof of Claim:** 
   From the given condition \( a^n + 2^n \mid b^n + c \) for all positive integers \( n \), we can derive:
   \[
   a^n + 2^n \mid b^{3n} + c^3
   \]
   and
   \[
   a^n + 2^n \mid a^{3n} + 2^{3n} \implies a^n + 2^n \mid b^{3n} + c
   \]
   Combining these, we get:
   \[
   a^n + 2^n \mid c^3 - c \quad \forall n \in \mathbb{N}
   \]
   Since \( a^n + 2^n \) becomes arbitrarily large for large \( n \), it follows that:
   \[
   c^3 - c = 0 \implies c(c^2 - 1) = 0 \implies c(c-1)(c+1) = 0
   \]
   Therefore, \( c = 0 \) or \( c = 1 \).

2. **Case 1:** \( c = 0 \).

   Let \( a \neq 2 \). Then:
   \[
   a^n + 2^n \mid b^n \quad \forall n \in \mathbb{N}
   \]
   By Zsigmondy's Theorem, for \( n > 3 \), \( a^n + 2^n \) must have a primitive prime factor. However, the support of \( b^n \) is the same as the support of \( b \), which is a finite set. Therefore, for large enough \( n \), there exists a prime \( p \) such that \( p \mid a^n + 2^n \) but \( p \nmid b \), leading to a contradiction.

3. **Case 2:** \( c = 1 \).

   Suppose \( a \) is even. Then:
   \[
   2^n \mid a^n + 2^n \mid b^n + 1 \quad \forall n \in \mathbb{N}
   \]
   Clearly, \( b \) must be odd. Taking \( n = 2 \):
   \[
   4 \mid b^2 + 1 \implies b^2 \equiv -1 \pmod{4}
   \]
   This is a contradiction since \( b^2 \equiv 1 \pmod{4} \) for any odd \( b \). Therefore, \( a \) must be odd.

   Let \( \mathcal{P} \) be the set of primes in the canonical factorization of \( a \) with odd exponents. Suppose that for every prime \( p \) such that \( \left( \frac{2a}{p} \right) = -1 \), we also have \( \left( \frac{b}{p} \right) = -1 \). Since \( 2ab \) is not a perfect square, take the sets of primes \( \mathcal{P} = \{ p_1, p_2, \ldots, p_r \} \) and \( \mathcal{Q} = \{ q_1, q_2, \ldots, q_s \} \) such that \( v_{p_i}(2a) \) and \( v_{q_i}(b) \) are all odd. Consider \( \mathcal{A} = \mathcal{P} \setminus \mathcal{Q} \) and \( \mathcal{B} = \mathcal{Q} \setminus \mathcal{P} \). Clearly, both \( \mathcal{A} \) and \( \mathcal{B} \) cannot be empty.

   Now, using the Law of Quadratic Reciprocity, the Chinese Remainder Theorem, and Dirichlet's Theorem of Arithmetic Progressions, we derive a contradiction.

   Therefore, there exists some prime \( p \) such that \( \left( \frac{2}{p} \right) = -\left( \frac{a}{p} \right) \) and \( \left( \frac{b}{p} \right) = 1 \). Taking \( n = \frac{p-1}{2} \), by Euler's Criterion:
   \[
   2^n + a^n \equiv 0 \pmod{p}
   \]
   However:
   \[
   b^n + 1 \equiv 2 \not\equiv 0 \pmod{p}
   \]
   This leads to a contradiction.

The final answer is \( \boxed{ c = 0 } \) or \( c = 1 \).