ElGamal Decryption variant

I am trying to do some ElGamal encryption but having a different encryption formula. For that I doing the following steps:

The key generator:

1. Choosing value $$p = 107$$ and $$a = 2$$
2. Random number $$d = 67$$, and $$b = a^d \bmod p$$ where $$b = 2^{67} \bmod 107 = 94$$
3. $$k_{priv} = 67$$ and $$k_{pub} = (p,a,b) = (107,2,94)$$

Encryption

1. Random value $$v = 45$$ and $$C_1 = a^v \bmod p = 2^{45} \bmod 107 = 28$$
2. We have the message $$m = 66$$; $$C_2 = m \cdot b^v \cdot a^v \bmod 107 = 66 \cdot 94^{45} \cdot 2^{45} \bmod 107 = 38$$
3. Finally, $$C = (C_1, C_2)$$

My problem comes when I try to decrypt the message, maybe I am totally wrong. But I am doing:

$$C_1 = a^v$$
$$C_2 = m \cdot a^v \cdot (a^d)^v$$ $$C_2 = m \cdot C_1 \cdot (a^d)^v$$

I trying to do that is where I am a little bit lost. If someone can help me with a clue to decrypt the message I would be nice

• What is the source of this question? Commented Oct 28, 2020 at 18:10
• You forgot to use the $d$, it is your secret, right? Commented Oct 28, 2020 at 19:35
• my question is how i can decrypt the message. I not sure if my approach is the best Commented Oct 29, 2020 at 8:32
• I'm assuming this is homework question there we only provide hints. Since you show some effort I'll direct you. Use the fact that $b=a^d$ since the message sent to you and $d$ is your secret. Commented Oct 29, 2020 at 9:57
• Thanks, this helped me. I didn't solved yet. I don't know how to solve the $(a^d)^v$ Commented Oct 30, 2020 at 19:34

Finally, the decryption is: $$m = C_2 \cdot C_1^{-1} \cdot C_1^{-d}$$