# Why do we need to hash both the message and the $h$ value in ElGamal signature?

The professor left us a question on ElGamal signatures:

Given the hash function $H$ and message $M$, choose a random $r$ and compute $h=g^r$ and $H(M||h)$. Show that, if $H(M)$ is used instead of $H(M||h)$, the signature can be existentially forged.

I am struggling with the problem, and I think that maybe we can get one known message signed twice with different $(M, c_1, h_1)$ and $(M, c_2, h_2)$, and then use that to sign some other messages $M'$, but I have no idea how to proceed. Can anyone give me some hints?

-
I tried to clean up your question a bit, but there were some places where I wasn't really sure what you meant. Could you please check that I didn't introduce any mistakes, and maybe try to clarify your question further. – Ilmari Karonen Feb 27 '13 at 12:22
It would useful to clarify how the question relates to the ElGamal signature scheme – fgrieu Feb 27 '13 at 12:40

Daniel Bleichenbacher has described such kind of attacks in his article Generating ElGamal signatures without knowing the secret key. (PDF)

He noticed that if verifier would accept signatures where $r$ is larger than $p$ then any signature $(r,s)$ on $H(M)$ could be used to generate a signature $(r2, s2)$ on arbitrary hash value $H(M2)$.

For that attacker should calculate $u=H(M2)H(M)^{-1} \bmod (p-1)$. This implies $g^{H(M2)}=g^{uH(M)}=y^{ru}r^{su}$. $s2$ could be setting as $s2=su\bmod (p-1)$ and $r2$ could be computing by using Chinese Remainder Theorem for two equations $r2=ru\bmod (p-1)$ and $r2=r\bmod p$.

-

It's not a complete answer because an adversary needs control on the random choice of a signing algorithm.

1. First let me define ElGamal signature to not get lost in notation. $x \in N$ is the secret key. $p$ is a prime, it defines $Z_p^*$. $g$ is a generator of $Z_p^*$. $y=g^x$ and the public key is $(p, g, g^x)$. Then $k$ is picked at random from $0<k<p-1$, $r = g^k \bmod p$, $s = (H(m) - xr)k^{-1} \bmod (p-1)$ and the signature is $(r,s)$.

2. Suppose we know two signatures of the message $m$ for different $k$-s: $\sigma_1 = (g^{k_1} \bmod p, (H(m) - xr)k_1^{-1} \bmod (p-1)) = (r_1, s_1)$ and $\sigma_2 = (g^{k_2} \bmod p, (H(m) - xr)k_2^{-1} \bmod (p-1)) = (r_2, s_2)$.

Let $\sigma_3=(r_1r_2 \bmod p, s_1 + s_2 \bmod (p-1))$.

1. Verification of $\sigma_3$ leads us to an equation $(k_1+k_2)(k_1^{-1}+k_2^{-1})\equiv 1 \bmod (p-1)$. If an attacker can choose $k_1, k_2$, then he can probably find proper $p$ and $k_1, k_2$. Otherwise - still a problem.
-
How is this an answer to the question? – poncho Jun 14 '15 at 1:15