Let $\mathbb{Z}_p$ be a prime field, $a, b$ be two random elements independently drawn from $\mathbb{Z}_p^*$, and $$ c = a* b \mod p,$$ is $c$ uniformly random? and why?
How is this question related to the discrete-log assumption? Because if $a =b$, then my understanding is that the discrete-log assumption asserts $$ c =a^n \mod p$$ is indeed uniformly random.
Let $\mathbb{Z}_p$ be a prime field, $a, b$ be two random elements independently drawn from $\mathbb{Z}_p^*$, and $$ c = a* b \mod p,$$ is $c$ uniformly random?
No; if either $a = 0$ or $b = 0$, then $c = 0$. Out of the $p^2$ possible pairs $(a, b)$, that gives $2p-1$ pairs that results in 0. As the probability of each possible pair is equal, that is, $p^{-2}$, that gives a probability of a 0 output to be $2/p - 1/p^2$, which is larger than the probability $1/p$ it would have if $c$ was uniformly distributed.
Now, if we modify the distribution so that $a, b$ were uniformly and independently distributed over nonzero values, then it turns out that $c$ would be uniformly distributed over nonzero values.
Because if $a =b$, then my understanding is that the discrete-log assumption asserts $$ c =a^n \mod p$$ is indeed uniformly random.
Well, I don't know what $a=b$ has to do with it, and that assertion would appear to be unrelated to the discrete-log problem. However, if $n$ (which you have not specified) is relatively prime to $p-1$, then yes, $c$ would be uniformly distributed (from the simple observation that $a^n \bmod p$ is a permutation in that case), and if $n$ is not relatively prime to $p-1$, then it is not.
No. For example, let $a$ be a uniformly sampled element and let $b=a$ in which case $c=a^2\mod p$ is a quadratic residue and only half of the elements of $\mathbb Z_p^\times$ are quadratic residues.
You may wish to rephrase the question using the word "independent".
This question is not related to whether the discrete logarithm is hard to compute.
