Find the values of $a\in [0,\infty)$ for which there exist continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$, such that $f(f(x))=(x-a)^2,\ (\forall)x\in \mathbb{R}$.
1. **Injectivity and Monotonicity**:
   - Given the function \( f: \mathbb{R} \rightarrow \mathbb{R} \) such that \( f(f(x)) = (x-a)^2 \) for all \( x \in \mathbb{R} \).
   - On the intervals \( [a, +\infty) \) and \( (-\infty, a] \), the function \( f \) is injective. This, combined with continuity, implies that \( f \) is strictly monotone on each of these intervals.

2. **Behavior on Intervals**:
   - \( f \) is strictly decreasing on \( (-\infty, a] \) and strictly increasing on \( [a, +\infty) \). This implies that \( f(a) \) is the minimum value of \( f(x) \).

3. **Minimum Value**:
   - Since \( f(a) \) is the minimum value, we have \( f(f(a)) = 0 \ge f(a) \).

4. **Contradiction for \( f(a) < 0 \)**:
   - Assume \( f(a) < 0 \). Take \( u \) such that \( u > a \) and \( f(u) < 0 \). By the intermediate value theorem (Darboux's property), there exists a number \( v \) in \( (a, +\infty) \) such that \( u = f(v) \).
   - Substituting \( v \) into the initial condition, we get \( f(f(v)) = (v-a)^2 = f(u) \ge 0 \), which contradicts \( f(u) < 0 \).

5. **Conclusion for \( f(a) \)**:
   - Therefore, \( f(a) \ge 0 \). Since \( f(f(a)) = 0 \), we must have \( f(a) = 0 \).

6. **Behavior at Zero**:
   - Given \( f(a) = 0 \), we have \( f(f(a)) = f(0) = 0 \). From the initial condition, \( f(f(0)) = f(0) = 0 = a^2 \). Thus, \( a = 0 \) is the only solution.

7. **Verification**:
   - The function \( f(x) = |x|^{\sqrt{2}} \) is continuous and satisfies \( f(f(x)) = x^2 \), which confirms that \( a = 0 \) is a valid solution.

\(\blacksquare\)

The final answer is \( \boxed{ a = 0 } \).