Alice sends a secret message to Bob using her RSA public key $n=400000001$. Eve wants to listen in on their conversation. But to do this, she needs Alice's private key, which is the factorization of $n$. Eve knows that $n=pq$, a product of two prime factors. Find $p$ and $q$.
1. We are given that \( n = 400000001 \) and need to find its prime factors \( p \) and \( q \) such that \( n = pq \).

2. Notice that \( n \) can be rewritten as \( 4 \cdot 10^8 + 1 \). This form suggests that we can use the Sophie-Germain identity, which states:
   \[
   a^4 + 4b^4 = (a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab)
   \]

3. We rewrite \( n \) as \( 1^4 + 4 \cdot 100^4 \). Here, we set \( a = 1 \) and \( b = 100 \).

4. Applying the Sophie-Germain identity:
   \[
   1^4 + 4 \cdot 100^4 = (1^2 + 2 \cdot 100^2 - 2 \cdot 1 \cdot 100)(1^2 + 2 \cdot 100^2 + 2 \cdot 1 \cdot 100)
   \]

5. Simplify the expressions inside the parentheses:
   \[
   1^2 + 2 \cdot 100^2 - 2 \cdot 1 \cdot 100 = 1 + 20000 - 200 = 19801
   \]
   \[
   1^2 + 2 \cdot 100^2 + 2 \cdot 1 \cdot 100 = 1 + 20000 + 200 = 20201
   \]

6. Therefore, the factors of \( n \) are \( 19801 \) and \( 20201 \).

The final answer is \( \boxed{19801 \text{ and } 20201} \).