Do $v_1=\alpha\cdot r_1$ and $v_2=\alpha\cdot r_2$ leak information about $\alpha$ Please consider we have finite field $\mathbb{F}_p$ for large prime number $p$. We have a fixed field element $\alpha$. By $r_i\leftarrow \mathbb{F}_p$ we mean we pick $r_i$ uniformly random from the field. We assume the values $r_i$ are distinct so $r_i\neq r_j$



Question: Given $n$ values $v_1=\alpha \cdot r_1 \bmod p,..., v_n=\alpha \cdot r_n \bmod p$ for a large $n$ can the adversary learn the value $\alpha$?



Intuitively one may say the adversary can compute $gcd(v_1,...v_n)$ to extract value $\alpha$.

Am I right that the adversary cannot learn anything; otherwise, ElGamal encryption would not be secure?
Question: Given $n$ values $v_1=\alpha \cdot r_1 \bmod p,..., v_n=\alpha \cdot r_n \bmod p$ for a large $n$ can the adversary learn the value $\alpha$?


Answer: assuming that the $r_i$ values are random (that is, equidistributed and uncorrelated), then the attacker gets absolutely no information about $\alpha$ (other than whether or not it's 0).

We can see this because, for any sets of values $v_i$, there are a unique set of $r_i$ values that is consistent with $\alpha$ being a specific nonzero value, namely, $r_i = \alpha^{-1} v_i$.  So, if someone cannot distinguish the $r_i$ values from random, they cannot determine which nonzero $\alpha$ value is any more likely than any other.

You state that the $r_i$ values are distinct, and so technically that is a correlation.  However, we can show that distinct $v_i$ values translate to distinct $r_i$ values (for any fixed nonzero $\alpha$), and hence any $\alpha$ value will preserve uniqueness.  So, the attacker cannot use this correlation to gain any information.