This is a 3-person encryption scheme based on RSA. One manager(can be trusted in this case) generates two large primes $p$ and $q$, calculates both $n$ and $\phi(n)$. The manager also chooses $k_1$, $k_2$ and $k_3$ such that $\gcd(k_i, n) = 1$ and $k_1k_2k_3 \equiv 1 \pmod{\phi(n)}$. Keys are securely distributed to three others as follows:
A: $\langle n, k_1, k_2 \rangle$
B: $\langle n, k_2, k_3 \rangle$
C: $\langle n, k_3, k_1 \rangle$
Question: A has a message $M_1$ for B. Give the encryption function for A as well as the decryption function for B, so that the message won’t be seen by anyone else.
My answer: Given that $k_1$, $k_2$, and $k_3$ are such that $\gcd(k_i, n) = 1$ and that $k_1k_2k_3 \equiv 1 \pmod{\phi(n)}$. Since , A had a message for B. Encryption formula for A: $C_1 = M_1^{k_1} \pmod{n}$
Decryption formula for B: $M_1 = C_1^{k_2k_3} \pmod{n}$
Because $k_1k_2k_3 \equiv 1 \pmod{\phi(n)}$, $M_1^{k_1k_2k_3} \equiv M_1 \pmod{n}$. Thus, $C_1^{k_2k_3} \pmod{n} = M_1^{k_1k_2k_3} \pmod{n}$ $= M_1$.
It looks OK. Can you also argue why in this message from A to B, C cannot eavesdrop? and nor can anyone not A,B or C?