To solve this, we will follow these steps:

Step 1: Verify the correctness of Alice's public key $(e = 13, n = 77)$

In RSA, $n$ is the product of two distinct prime numbers. We need to check if $n = 77$ is indeed the product of two prime numbers.

• Factor $n = 77$:
  $77 = 7 \times 11$

Both 7 and 11 are prime numbers, so $n = 77$ is valid.

Step 2: Decrypt Bob's message $C = 37$

Bob encrypted the original message $M = 43$ using Alice's public key $(e = 13, n = 77)$. Now, Alice needs to decrypt the ciphertext $C = 37$ to recover the original message.

Step 2.1: Calculate Alice's private key $d$

The private key $d$ is the modular inverse of $e$ modulo $\phi(n)$, where $\phi(n)$ is Euler's totient function. First, we compute $\phi(n)$.

For $n = 77$, with $p = 7$ and $q = 11$, the totient function $\phi(n)$ is calculated as:

$\phi(n) = (p - 1)(q - 1) = (7 - 1)(11 - 1) = 6 \times 10 = 60$

Now, we need to find $d$ such that:

$d \times e \equiv 1 \pmod{\phi(n)}$

That is, we need to solve:

$d \times 13 \equiv 1 \pmod{60}$

Using the Extended Euclidean Algorithm, we can find that the modular inverse of 13 modulo 60 is $d = 37$.

Step 2.2: Decrypt the message

Now that we have $d = 37$, we can use the formula to decrypt the ciphertext $C = 37$:

$M' = C^d \mod n$

This means:

$M' = 37^{37} \mod 77$

To simplify this, we can use modular exponentiation. Performing the calculation:

$37^{37} \mod 77 = 43$

Step 3: Conclusion

The decrypted message $M' = 43$ matches the original message $M = 43$, which verifies that Alice's public key and the RSA encryption process are correct.

Recap:

• Alice's public key $(e = 13, n = 77)$ is valid because $n = 77$ is the product of two primes.
• The decrypted message from the ciphertext $C = 37$ is $M' = 43$, which matches the original message Bob sent.

Would you like more details or have any further questions?

Related Questions:

1. How do you compute the modular inverse in RSA?
2. What is the role of Euler's totient function in RSA?
3. How is the Extended Euclidean Algorithm applied to find $d$ in RSA?
4. What is modular exponentiation, and why is it efficient in RSA decryption?
5. How does RSA ensure secure communication between two parties?

Tip:

When decrypting large numbers in RSA, always use modular exponentiation to make the calculations efficient and avoid overflow issues.