Implications of proving NP=RP on complexity theory
Prelude: the below is just one consequence of $\mathsf{RP}=\mathsf{NP}$ and probably not the most important, e.g. compared to collapse of the polynomial hierarchy. There was a great and more comprehensive answer than this, but its author removed it for some reason. Hopefully the question can continue to get more answers.
$\mathsf{P}/\mathsf{poly}$ is the set of decision problems solvable by polynomial-size circuits. We know $\mathsf{RP} \subseteq \mathsf{BPP}$ and, by Adleman's theorem, $\mathsf{BPP} \subseteq \mathsf{P}/\mathsf{poly}$. So among the only mildly shocking implications of $\mathsf{RP}=\mathsf{NP}$ would be $\mathsf{NP} \subseteq \mathsf{P}/\mathsf{poly}$.
Another way to put it is that instead of each &quot;yes&quot; instance of an $\mathsf{NP}$ problem having its own witness, there would exist for each $n$ a single witness string that can be used to verify, in polynomial time, membership of any instance of size $n$.