Encryption in the original Rabin scheme took a message $x$ and computed $x(x + b) \bmod n$, where $0 \le b \lt n$ and $n$ is the product of two secret primes $p$ and $q$. The private key is defined as $(p,q)$ and the public key as $(b,n)$. Modern Rabin encryption however is defined as $x^2 \bmod n$, which is equivalent to the original scheme with $b = 0$. Why is $b$ now ignored?
This question is based on another question which I found interesting but which was mysteriously self-deleted.