Find the natural numbers $ n\ge 2 $ which have the property that the ring of integers modulo $ n $ has exactly an element that is not a sum of two squares.
1. **Initial Setup and Definitions:**
   We need to find natural numbers \( n \ge 2 \) such that the ring of integers modulo \( n \), denoted \( \mathbb{Z}/n\mathbb{Z} \), has exactly one element that is not a sum of two squares.

2. **Multiplicative Closure of Sums of Two Squares:**
   The set of elements which, modulo \( n \), are sums of two squares is multiplicatively closed. This follows from the identity:
   \[
   (a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2.
   \]

3. **Units and Non-Sums of Two Squares:**
   If \( a \) is an element in \( \mathbb{Z}/n\mathbb{Z} \) that is not a sum of two squares and \( u \) is a unit in \( \mathbb{Z}/n\mathbb{Z} \) that is a sum of two squares, then \( u^{-1}a \) is also not a sum of two squares. This follows from the multiplicative closure property.

4. **Exclusion of Small Values:**
   It is clear that \( n = 2 \) and \( n = 3 \) do not satisfy the condition. For \( n = 2 \), the elements are \( 0 \) and \( 1 \), both of which are sums of two squares. For \( n = 3 \), the elements are \( 0, 1, \) and \( 2 \), where \( 2 \equiv -1 \pmod{3} \) and \( -1 \) is a sum of two squares.

5. **Prime Power Form:**
   **Claim 1:** \( n = p^k \) for some prime \( p \) and some positive integer \( k \).

   **Proof:**
   Let \( n = \prod_{i=1}^m p_i^{k_i} \) be the prime factorization of \( n \). By the Chinese Remainder Theorem:
   \[
   \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/p_1^{k_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_m^{k_m}\mathbb{Z}.
   \]
   Let \( (a_1, \cdots, a_m) \) be the unique element of \( \mathbb{Z}/n\mathbb{Z} \) that is not a sum of two squares. There exists \( 1 \le i \le m \) such that \( a_i \) is not a sum of two squares modulo \( p_i^{k_i} \). Without loss of generality, assume \( i = 1 \). If \( m > 1 \), then we have at least two elements that are not sums of two squares: \( (a_1, 0, \cdots, 0) \) and \( (a_1, 1, \cdots, 1) \), which is a contradiction. Hence, \( n = p^k \).

6. **Power of 2:**
   **Claim 2:** \( n = 2^k \) for some integer \( k \ge 2 \).

   **Proof:**
   By Claim 1, \( n = p^k \) for some prime \( p \) and some positive integer \( k \). Suppose \( p \) is odd. Let \( a \) be the unique element of \( \mathbb{Z}/n\mathbb{Z} \) that is not a sum of two squares. The element \( 2 \) modulo \( n \) is a unit in \( \mathbb{Z}/n\mathbb{Z} \) and is a sum of two squares. By the earlier property, \( 2^{-1}a \) is not a sum of two squares, implying \( 2^{-1}a = a \). Thus, \( 2a = a \), leading to \( a = 0 \), which is a contradiction. Therefore, \( p = 2 \).

7. **Specific Value of \( k \):**
   **Claim 3:** \( n = 4 \).

   **Proof:**
   By Claim 2, \( n = 2^k \) for some integer \( k \ge 2 \). Suppose \( k > 2 \) and let \( a \in \mathbb{Z}/n\mathbb{Z} \) be the unique element that is not a sum of two squares. The element \( 5 \) modulo \( n \) is a unit in \( \mathbb{Z}/n\mathbb{Z} \) and is a sum of two squares. By the earlier property, \( 5^{-1}a \) is not a sum of two squares, implying \( 5^{-1}a = a \). Thus, \( 5a = a \), leading to \( 4a = 0 \). Therefore, modulo \( n \), \( a \in \{2^{k-2}, 2^{k-1}, 3 \times 2^{k-2}\} \). By the multiplicative closure property, both \( 2^{k-2} \) and \( 2^{k-1} \) are sums of two squares (since \( 2 \) is). Thus, \( a = 3 \times 2^{k-2} \). For \( k > 2 \), \( 3 \) is a sum of two squares, contradicting the uniqueness of \( a \). Hence, \( k = 2 \) and \( n = 4 \).

The final answer is \( \boxed{4} \).