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We know that if the decisional Diffie-Hellmann problem is hard then the Elgamal encryption is CPA-secure. One show that if the chosen group is $\mathbb{Z}_p^*$ then the decisional Diffie-Hellman problem is not hard. For instance, one realizes that $g^{ab}$ is a square with probability $3/4$ with an argument based on the parity of $a,b$. Therefore, it makes self to ask the following question:

Is there a CPA-attack on Elgamal when used with $\mathbb{Z}_p^*$?

I need some help on the way to go here. Should I try to build a distinguisher using the above fact about the Diffie-Hellman problem? Or does is just come from the algebraic structure of the group?

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HINT

You already have a hint given in the problem statement:

$g^{ab}$ is a square with probability $3/4$ with an argument based on the parity of $a,b$.

You should start by making sure you understand why is this the case. Then, once you do that, think about what it means to break DDH: it means distinguishing things that look like $g^c$ for a random $c$ from things of the form $g^{ab}$ for random $a,b$. Given the hint above, can you find a feature that differs between these two?

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  • $\begingroup$ Thanks for the extended explanation of the hint. Let me see if i can exploit it $\endgroup$ – Rodrigo Jan 28 at 17:57
  • $\begingroup$ Do you think asking Alice and Bob to encrypt either the generator or 1 would suffice? Because then in the second component i get an even exponent or an odd exponent and i can test whether they are quadratic residues efficiently $\endgroup$ – Rodrigo Jan 28 at 18:05
  • $\begingroup$ Yes! That would be a good way of doing it. To make sure I got you: you ask them to encrypt either $m_0 = g^0 = 1$ or $m_1 = g^1 = g$, then the encryption you get is $g^{ab}\cdot g^i = g^{ab+i}$ for either $i=0$ or $i=1$. Then you check whether this ciphertext is a quadratic residue... and what do you do next? $\endgroup$ – Daniel Jan 28 at 18:22
  • $\begingroup$ If it is quadratic residue i ouput 0, otherwise 1 $\endgroup$ – Rodrigo Jan 28 at 18:29
  • $\begingroup$ Right. Just make sure to write down precisely what is the exact success probability, which won't be negligible. Good work! $\endgroup$ – Daniel Jan 28 at 19:04

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