# RSA Protocol behind Yaksha Security System

So I'm reading over the Yaksha Security System and see it is based on the RSA cryptosystem and a centralized server, easy enough. What I'm slightly confused on is the math behind the related keys.

It states

nAlice - Alice(some user) modulus
dAliceA - First private key only known to Alice
dAliceY - Second Private key only known to Yaksha server
eAlice - Alice's public key

The keys are related by the following:

dAliceA*dAliceY mod phi(mod nAlice) = dAlice

It states that the server sends Alice:

Calice = (Ksession) ^ dAliceY * eAlice mod nAlice

And Alice can determine the session key as:

(Calice) ^ dAliceA mod nAlice = Ksession

My question is, Can Alice obtain the session key due to the multiplicative properties of the modulus function and the basis of which RSA is built on? Or am I missing something here?

Edit: Also, is dAlice just the combined private key of the server and Alice to verify that the keys are correct?

Can Alice obtain the session key due to the multiplicative properties of the modulus function and the basis of which RSA is built on?

Essentially, yes. One way of looking why RSA works (that is, why the encryption and decryption are inverses of each other) is because of two mathematical identities:

$$(M^a \bmod N)^b \bmod N = M^{a \cdot b} \bmod N$$

$$\text{If}\ a \cdot b \equiv 1 \pmod{\phi(N)}\ \text{then}\ M^{a\cdot b} \equiv M \pmod{N}$$

(Note: in the second one, you can replace $\pmod{\phi(N)}$ with $lcm(p-1,q-1)$, which gives you a stronger identity; however what I have written is good enough for now).

For normal RSA, this is sufficient; however it is easy to extend these identities to three exponents:

$$((M^a \bmod N)^b \bmod N)^c \bmod N = M^{a \cdot b \cdot c} \bmod N$$

$$\text{If}\ a \cdot b \cdot c \equiv 1 \pmod{\phi(N)}\ \text{then}\ M^{a\cdot b \cdot c} \equiv M \pmod{N}$$

This is what Yaksha is relying on; the have $a$ = the public exponent (eAlice), $b$ = the private key known to the server (dAliceY), and $c$ = Alice's private key (dAliceA).

We can see that the relationship between eAlice, dAliceY and dAliceA holds, based on the fact that:

$dAliceA \cdot dAliceY \equiv dAlice \pmod{ \phi(mod nAlice)}$

If we multiply both sides by eAlice (and rely on the fact that $dAlice \cdot eAlice \equiv 1 \pmod{ \phi(modnAlice)}$, we get:

$$dAliceA \cdot dAliceY \cdot eAlice \equiv 1 \pmod{ \phi(mod nAlice)}$$

which is precisely the relationship we needed to assume.

• Thank you, This is what I was thinking but wanted some clarification. Dec 14 '14 at 19:12