Given two distinct primes $p,q$ such that $(p-1)(q-1)=A$, are $p$ and $q$ uniquely determined by $A$? 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 $(m,n)$?



Or:

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 $A\to(m,n)$
No, for example for $A=72$, the equation $(p-1)(q-1)=A$ has $4$ prime solutions:
$$(p,q)\in\{ (2, 73),(3, 37), (5, 19), (7,13)\}.$$
For $A=1080$ there are $6$ prime solutions:
$$(p,q)\in\{ (3, 541), (5, 271), (7,181), (11,109), (19,61), (31,37)\}.$$
It's quite easy to hunt for counter examples. For example

Consider the factors of $36$.
\begin{array}{rr|rr}
p-1 &amp; q-1 &amp; p &amp; q \\
\hline
1   &amp;  36 &amp; 2 &amp; 37 \\
2   &amp;  18 &amp; 3 &amp; 19 \\
3   &amp;  12 \\
4   &amp;   9 \\
6   &amp;   6 &amp; 7 &amp; 7 \\
\hline
\end{array}

So $f(2,37)=f(3,19)=36$.

Consider the factors of $60$.
\begin{array}{rr|rr}
p-1 &amp; q-1 &amp; p &amp; q \\
\hline
1   &amp;  60 &amp; 2 &amp; 61 \\
2   &amp;  30 &amp; 3 &amp; 31 \\
3   &amp;  20 \\
4   &amp;  12 &amp; 5 &amp; 13\\
6   &amp;  10 &amp; 7 &amp; 11 \\
\hline
\end{array}

So $f(2,61)=f(3,31)=f(5,13)=f(7,11)=60$.