I believe another way I could phrase the question is:
Given $f(m,n)\equiv(m-1)(n-1)$ where $m$ and $n$ are two primes:
Is $f$ one-to-one for pairings of $(m,n)$?
Are there only two distinct primes $(p,q)$ which make $f=A$?
I conjecture the answer is yes, there is only one pairing which exists for each output, but I can still somewhat envision there being some way to get non-distinct mappings from $A\to(m,n)$