Determine the function satisfying `f^2(x+y)=f^2(x)+f^2(y)AAx ,y in Rdot`

Explanation

To determine the function satisfying the equation \( f^2(x+y) = f^2(x) + f^2(y) \) for all \( x, y \in \mathbb{R} \), we can follow these steps:

### Step 1: Analyze the given functional equation
The equation we have is:
\[ f^2(x+y) = f^2(x) + f^2(y) \]
This resembles the form of the Pythagorean theorem, suggesting that \( f^2 \) could represent some squared function.

### Step 2: Substitute specific values
Let's substitute \( x = 0 \) and \( y = 0 \):
\[ f^2(0 + 0) = f^2(0) + f^2(0) \]
This simplifies to:
\[ f^2(0) = 2f^2(0) \]
From this, we can rearrange to find:
\[ f^2(0) - 2f^2(0) = 0 \implies -f^2(0) = 0 \implies f^2(0) = 0 \]
Thus, we conclude:
\[ f(0) = 0 \]

### Step 3: Substitute \( y = -x \)
Next, we substitute \( y = -x \):
\[ f^2(x + (-x)) = f^2(x) + f^2(-x) \]
This simplifies to:
\[ f^2(0) = f^2(x) + f^2(-x) \]
Since we found that \( f^2(0) = 0 \), we have:
\[ 0 = f^2(x) + f^2(-x) \]
Since both \( f^2(x) \) and \( f^2(-x) \) are non-negative, the only way for their sum to be zero is if both terms are zero:
\[ f^2(x) = 0 \quad \text{and} \quad f^2(-x) = 0 \]
This implies:
\[ f(x) = 0 \quad \text{for all } x \in \mathbb{R} \]

### Step 4: Conclusion
Thus, the only function that satisfies the given functional equation is:
\[ f(x) = 0 \quad \text{for all } x \in \mathbb{R} \]