MHB How Can You Multiply Ciphertexts in ElGamal Without Decrypting?

[mathmari]

Hey!

At an ElGamal encryprion system at the group $\mathbb{Z}_{786}^{\star}$ with base $g=\overline{2}$ ( it is a generator ) the public key of Alice is $y=\overline{5}$. We see to encryptions $(r_1, c_1)=(318, 191)$ of the unknown message $m_1$ and $(r_2, c_2)=(79, 118)$ of the unkown message $m_2$. Show how you can calculate the encryption of the message $m_1 \cdot m_2 \pmod p$(without calculations $m_1$ and $m_2$).

Could you give me some hints what we could do ?? (Wondering)

 

[Sudharaka]

Hey!

At an ElGamal encryprion system at the group $\mathbb{Z}_{786}^{\star}$ with base $g=\overline{2}$ ( it is a generator ) the public key of Alice is $y=\overline{5}$. We see to encryptions $(r_1, c_1)=(318, 191)$ of the unknown message $m_1$ and $(r_2, c_2)=(79, 118)$ of the unkown message $m_2$. Show how you can calculate the encryption of the message $m_1 \cdot m_2 \pmod p$(without calculations $m_1$ and $m_2$).

Could you give me some hints what we could do ?? (Wondering)

Hi mathmari, :)

Elgamal is a Partially Homomorphic encryption scheme which means that there's a correspondence between the operations over the cipertext and the operations over the plaintext. The correspondence is multiplication. So when you multiply two ciphertexts the underlying plaintexts will be multiplied.

Hope this helps. :)