Given: `f: R to R` such that `f(x+f(x)) =4f(x)` and `f(0)=1`. Find `f(1), f(5), f(21)`.

Answer

To solve the problem, we start with the functional equation given:

1. **Given**: \( f(x + f(x)) = 4f(x) \)
2. **Also given**: \( f(0) = 1 \)

We need to find \( f(1) \), \( f(5) \), and \( f(21) \).

### Step 1: Find \( f(1) \)

Let's substitute \( x = 0 \) into the functional equation:

\[ f(0 + f(0)) = 4f(0) \]

Since \( f(0) = 1 \), we have:

\[ f(0 + 1) = 4 \cdot 1 \]

This simplifies to:

\[ f(1) = 4 \]

### Step 2: Find \( f(5) \)

Now, we substitute \( x = 1 \) into the functional equation:

\[ f(1 + f(1)) = 4f(1) \]

We already found \( f(1) = 4 \), so we substitute that in:

\[ f(1 + 4) = 4 \cdot 4 \]

This simplifies to:

\[ f(5) = 16 \]

### Step 3: Find \( f(21) \)

Next, we substitute \( x = 5 \) into the functional equation:

\[ f(5 + f(5)) = 4f(5) \]

We found \( f(5) = 16 \), so we substitute that in:

\[ f(5 + 16) = 4 \cdot 16 \]

This simplifies to:

\[ f(21) = 64 \]

### Summary of Results

Thus, we have:

- \( f(1) = 4 \)
- \( f(5) = 16 \)
- \( f(21) = 64 \)