Okay so I wanted to know if my logic is correct for the approach of this practice problem that I'm doing in preparation for my final and see if I truly understand how the 3 pass protocol works so:

Alice wants to send a message to Bob using the 3-pass protocol. She decides to use the
prime p = 17, and picks her key, a = 11. Bob picks his key, b = 13.
(a) What are Alice and Bob’s decryption keys?

So the decryption function for Alice is $D_A \equiv C^{a^-1}(\mod p)$ and for Bob $D_B \equiv C^{b^-1}(\mod p)$ which I assume we use these functions to generate the keys correct? If we do how do we find C?


If $x \equiv y \space(mod \space\phi(p))$ then $n^x \equiv n^y \space(mod\space p)$ (1)

With encryption key $e$, decryption $d$, and message $m$, you have $E(m) = m^e \space(mod \space p)$.

For decryption you do $D(E(m)) \equiv E(m)^d \equiv (m^e)^d \equiv m^{ed} \space (mod \space p)$.

You want $m^{ed} \equiv m^1 \space (mod \space p)$ so from (1) you can determine you want to choose $d$ such that $de \equiv 1 \space (mod \space p-1)$.

(Note: $p$ is prime so $\phi(p) = p - 1$.)

Since $ed \equiv 1 \space (mod \space p-1)$, $e$ and $d$ are multiplicative inverses of each other mod $p - 1$. So if you can calculate the inverse, you can calculate one of the keys from the other.

I got 3 for Alice's decryption key and 5 for Bob's.

$11 * 3 \equiv 33 \equiv 1 \space(mod \space 17 - 1)$

$13 * 5 \equiv 65 \equiv 1 \space(mod \space 17 - 1)$

  • $\begingroup$ Thank you that makes complete sense! so you find the inverse of 11 and 13 which finds the numbers that make the condition true? $\endgroup$ – AceNinja1101 May 19 '18 at 21:18

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