In RSA public key cryptosystem suppose 𝑛=𝑝∗𝑞 where 𝑝 and 𝑞 are primes.\(…

Correct statements and explanation:

• Encryption and decryption formulas: Ciphertext is computed as C ≡ M^e (mod n), and decryption is M ≡ C^d (mod n). These operations work in the ring of integers modulo n, so exponents are applied and results reduced modulo n.

• Relation between e and d: e and d must satisfy ed ≡ 1 (mod φ(n)). This makes d the modular inverse of e modulo φ(n).

• Why these two conditions give correct decryption: If ed ≡ 1 (mod φ(n)), then ed = 1 + kφ(n) for some integer k. For messages M with gcd(M,n)=1, Euler's theorem gives M^{φ(n)} ≡ 1 (mod n), so M^{ed} = M^{1+kφ(n)} ≡ M (mod n), and decryption recovers M.

• Why the other statements are incorrect:

  • ed ≡ 1 (mod n) is wrong: requiring e and d to be inverses modulo n is far too strong and not the condition used in RSA; it generally will not hold and is unnecessary for decryption to work.

  • Using modulo φ(n) in the encryption/decryption formulas is wrong: encryption and decryption produce values modulo n (the modulus of the public key), so exponentiation should be reduced modulo n, not modulo φ(n).

• Practical note: Euler's theorem guarantees correctness for messages coprime to n. To ensure decryption for all messages 0 < M < n, implementations often rely on the stronger condition ed ≡ 1 (mod lcm(p−1,q−1)) or use Chinese Remainder Theorem-based proofs, but the standard RSA statement taught uses ed ≡ 1 (mod φ(n)).