Problem 1529

1. Initial Setup and Substitution:
Given the functional equation:
$$f(x+2y) = 2f(x)f(y)$$
for all $x, y \in \mathbb{R}$, we start by substituting $x = 0$ and $y = 0$:
$$f(0+2 \cdot 0) = 2f(0)f(0) \implies f(0) = 2f(0)^2$$
Let $f(0) = c$. Then:
$$c = 2c^2 \implies 2c^2 - c = 0 \implies c(2c - 1) = 0$$
Therefore, $c = 0$ or $c = \frac{1}{2}$.

2. **Case 1: $f(0) = 0$:**
If $f(0) = 0$, then substituting $x = 0$ in the original equation:
$$f(2y) = 2f(0)f(y) = 2 \cdot 0 \cdot f(y) = 0$$
Thus, $f(2y) = 0$ for all $y \in \mathbb{R}$. By induction, we can show that $f(x) = 0$ for all $x \in \mathbb{R}$. Hence, $f$ is the zero function, which is constant.

3. **Case 2: $f(0) = \frac{1}{2}$:**
If $f(0) = \frac{1}{2}$, then substituting $x = 0$ in the original equation:
$$f(2y) = 2f(0)f(y) = 2 \cdot \frac{1}{2} \cdot f(y) = f(y)$$
Thus, $f(2y) = f(y)$ for all $y \in \mathbb{R}$. By induction, we can show that $f(2^n y) = f(y)$ for all $n \in \mathbb{Z}$.

4. Continuity and Limit Argument:
Since $f$ is continuous, we can take the limit as $n \to \infty$:
$$f(y) = f\left(\frac{y}{2^n}\right) \quad \text{for all } n \in \mathbb{Z}$$
Taking the limit as $n \to \infty$:
$$f(y) = f\left(\lim_{n \to \infty} \frac{y}{2^n}\right) = f(0) = \frac{1}{2}$$
Therefore, $f(y) = \frac{1}{2}$ for all $y \in \mathbb{R}$. Hence, $f$ is the constant function $\frac{1}{2}$.

5. Conclusion:
In both cases, $f$ is shown to be constant. Therefore, $f$ is a constant function.

The final answer is $\boxed{ f(x) = 0 }$ or $f(x) = \frac{1}{2}$.