Determine all integers $n \ge 3$ whose decimal expansion has less than $20$ digits, such that every quadratic non-residue modulo $n$ is a primitive root modulo $n$.

[i]An integer $a$ is a quadratic non-residue modulo $n$, if there is no integer $b$ such that $a - b^2$ is divisible by $n$. An integer $a$ is a primitive root modulo $n$, if for every integer $b$ relatively prime to n there is a positive integer $k$ such that $a^k - b$ is divisible by $n$.[/i]
To determine all integers \( n \ge 3 \) whose decimal expansion has less than 20 digits, such that every quadratic non-residue modulo \( n \) is a primitive root modulo \( n \), we need to analyze the conditions under which \( n \) has a primitive root and the properties of quadratic non-residues and primitive roots.

1. **Primitive Roots and Quadratic Non-Residues:**
   - An integer \( n \) has a primitive root if and only if \( n \) is of the form \( 2, 4, p^k, \) or \( 2p^k \) where \( p \) is an odd prime and \( k \) is a positive integer.
   - A quadratic non-residue modulo \( n \) is an integer \( a \) such that there is no integer \( b \) with \( a \equiv b^2 \pmod{n} \).
   - A primitive root modulo \( n \) is an integer \( g \) such that for every integer \( b \) relatively prime to \( n \), there exists a positive integer \( k \) such that \( g^k \equiv b \pmod{n} \).

2. **Case Analysis:**
   - **Case \( n = 2 \):** 
     - \( n = 2 \) has a primitive root (1), but since \( n \ge 3 \), this case is not applicable.
   - **Case \( n = 4 \):**
     - \( n = 4 \) does not have a primitive root, so this case is not applicable.
   - **Case \( n = p^k \) for \( k \ge 2 \):**
     - If \( k \ge 2 \), then \( p \) is a quadratic non-residue modulo \( p^k \) but not a primitive root modulo \( p^k \). Therefore, \( n = p^k \) with \( k \ge 2 \) does not satisfy the condition.
   - **Case \( n = p \) (where \( p \) is an odd prime):**
     - We need to count the number of quadratic non-residues and primitive roots modulo \( p \).
     - The number of quadratic non-residues modulo \( p \) is \( \frac{p-1}{2} \).
     - The number of primitive roots modulo \( p \) is \( \phi(p-1) \), where \( \phi \) is the Euler's totient function.
     - For every quadratic non-residue to be a primitive root, we need \( \frac{p-1}{2} \leq \phi(p-1) \).

3. **Inequality Analysis:**
   - The inequality \( \frac{p-1}{2} \leq \phi(p-1) \) holds if and only if \( p-1 \) has only small prime factors, specifically 2.
   - This implies \( p-1 \) must be a power of 2, i.e., \( p-1 = 2^m \) for some integer \( m \).
   - Therefore, \( p = 2^m + 1 \), which are Fermat primes.

4. **Fermat Primes:**
   - Fermat primes are of the form \( 2^{2^k} + 1 \).
   - Known Fermat primes are 3, 5, 17, 257, and 65537.
   - All these primes have less than 20 digits.

5. **Conclusion:**
   - The integers \( n \ge 3 \) whose decimal expansion has less than 20 digits, such that every quadratic non-residue modulo \( n \) is a primitive root modulo \( n \), are the Fermat primes.

The final answer is \( \boxed{ n = 3, 5, 17, 257, 65537 } \).